TITLE: For what values of $n > 1$ and a is $x^n - a$ irreducible in $\mathbb Q[x]$?
QUESTION [2 upvotes]: $$x^n-a$$
So $n$ is any integer greater than 1, and $a$ is any integer.  $a$ being any integer is where I am running into trouble.  I have already shown and worked out a proof for this being irreducible when a is prime.  Now I am just working through examples, trying to figure out a pattern.
The exercise is asking for application of Eisenstein's Criterion only, so I am assuming that the value of n should have no effect on the irreducibility.

REPLY [4 votes]: Theorem $\ $ Suppose $\,F\,$ is a field and  $\:a\in F\:$   and $\:0 < n\in\mathbb Z.\ $ Then
$\ \ \ x^n - a\ $ is irreducible over $F \iff a \not\in F^{\large p}\,$ for all primes $\,p\mid n,\,$ and $\ a\not\in -4F^4$ when $\: 4\mid n $
For a proof see e.g.  Karpilovsky, Topics in Field Theory, Theorem 8.1.6, excerpted below, or see Lang's Algebra (Galois Theory).


TITLE: How to prove $(1+i)$ is a prime in $\mathbb{Q}(i)$?
QUESTION [1 upvotes]: I know the definition of prime in $\mathbb{Q}(\sqrt{m})$ that is an algebraic integer is a prime if it is divisible by its associates and units of the field. Units of $\mathbb{Q}(i)$ are $+1,-1,i,-i$
 How to solve this example?please help me.

REPLY [4 votes]: It's a bit hard to tell what you're actually asking, here. $\Bbb Q(i)$ is a field, so every non-zero element of $\Bbb Q(i)$ is a unit (and talking about primes doesn't really make sense). After all, let's suppose $a+bi\in\Bbb Q(i)$ with $a,b$ not both zero. Then the following are equivalent: $$(a+bi)(c+di)=1$$ $$ac-bd+(ad+bc)i=1$$ $$ac-bd=1\text{ and }ad+bc=0\tag{1}$$ Since $a,b$ are not both zero, then we can solve the left equation in $(1)$ for one of $c,d$ and the right equation for the other one.
I suspect, instead, that you're interested in $\Bbb Z[i]:=\{a+bi:a,b\in\Bbb Z\}.$ In that case, the units are $1,-1,i,-i,$ as you say, and $1+i$ is, indeed, prime.
To prove it, let's suppose that $$1+i=(a+bi)(c+di)$$ for some integers $a,b,c,d.$ Then $$|1+i|=|a+bi||c+di|$$ $$|1+i|^2=|a+bi|^2|c+di|^2$$ $$2=\left(a^2+b^2\right)\left(c^2+d^2\right),$$ but since $a^2+b^2$ and $c^2+d^2$ are both integers--in fact, positive integers--and $2$ is a prime number, then one of them must be equal to $1.$
Without loss of generality, say $a^2+b^2=1.$ Can you take it from there?


TITLE: Why is the sample interferogram given by the following integral of the intensity pattern?
QUESTION [0 upvotes]: I am trying to understand equation (2) in the paper by J. E. Greivenkamp with the title Generalized data reduction for heterodyne interferometry from year  1984. If you dont have acces to the article, don't worry; the equation appears in the post below.
The general expression for the intensity pattern of a sinusoidal fringe is given by
$$I(x,y,\delta) = I'(x,y) + I''(x,y) \cos(\phi(x,y) - \delta)$$
$\qquad$ where $I'$ is the average background Intensity, $I''$ is the modulation coefficient, $\phi$ is the unknown phase and $\delta$ is the phase shift.
The data required for heterodyne interferometry (or equivalently for digital fringe profilometry) are a series of interferograms (or fringe images) with different phase shifts. To be more specific we need atleast 3 phase-shifts since there are 3 unknowns  $I', I''$ and $\phi$  in the above equation.
In Fig.1 below is a camera which captures images of the fringes displayed on the LCD monitor. For  total number of 4 phase-shifts as shown in fig.2,  one obtains 4 images $I_1,\dots, I_4$, where the phase-shifts are $\delta_1 = 0, \delta_2 = \dfrac{\pi}{2}, \delta_3 = \pi, \delta_4 = \dfrac{3\pi}{2}$ respectively.

Now comes the part which I dont fully grasp. The Author mentions:
"In practice, the phase shift is often allowed to vary linearly over a range of $\Delta$ during the measurement, so that a sample interferogram is the integral of the intensity pattern from $\delta_i - \frac{\Delta}{2}$ to $\delta_i + \frac{\Delta}{2}:$
$$ I_i (x,y) = \int_{\delta_i - \frac{\Delta}{2}}^{\delta_i + \frac{\Delta}{2}} I(x,y,\delta) \, \mathrm{d}\delta\,.$$
This is the equation (2) in the paper I mentioned in the begining. I don't fully understand

what exactly is $\Delta$- is it $2\pi$?
Why is  the lower and upper bound of the integral chosen this way or in other words what is the intuition behind the upper and lower bounds?

Evaluating the integral and exploiting trigonometric identities gives
$$I_i(x,y) = I'(x,y) + I''(x,y) \, \mathrm{sinc}\bigg(\dfrac{\Delta}{2}\bigg)\cos(\phi(x,y) - \delta_i).$$
I am familiar with the $\mathrm{sinc}$ function but I can't grasp why its appearance makes sense in this  context. Maybe I am missing a key point here. Any comments will be kindly appreciated.

REPLY [1 votes]: Without having read the paper, it looks like what they mean is that when you take a measurement, you are not getting $I(x,y,\delta)$ with infinite precision in $\delta$, rather you choose some fixed phase parameter $\Delta$ (either fixed by you or your equipment) and sample across the range $\delta-\Delta/2$ to $\delta+\Delta/2$. In this sense, $\Delta$ is a free parameter.
The signal you get out is then given by the integral in your Eq. (2), rather than just $I(x,y,\delta)$.
The intuition behind this is that sampling over the range given by $\Delta$ can have an effect on the signal you get out, so it is better to have a theoretical expression which takes it into account. The appearance of the sinc function is just a consequence of evaluating the integral.
A better way of understanding $\Delta$ could also be to reverse the experiment: If you assume that $\Delta$ is a parameter representing the accuracy of your equipment, you can measure the signal and fit it to your Eq. (2) to find $\Delta$.


text:
\begin{document}

\title[Absolute ruin in the Ornstein-Uhlenbeck type risk model]{Absolute ruin in the Ornstein-Uhlenbeck type risk model}
\author{R.L. Loeffen}
\address{Weierstrass Institute for Applied Analysis and Stochastics,Mohrenstrasse 39
10117 Berlin, Germany} \email{RonnieLambertus.Loeffen@wias-berlin.de}

\author{P. Patie}
   \address{D\'epartement de Math\'ematiques, Universit\'e Libre de Bruxelles\\
Boulevard du Triomphe,  B-1050, Bruxelles, Belgique.}
\email{ppatie@ac.ulb.be}

\begin{abstract}
We start by showing that the finite-time absolute ruin probability in the classical risk model with constant interest force can be expressed in terms of the transition probability of a positive Ornstein-Uhlenbeck type process, say $\hat{X}$. Our methodology  applies to the case when the dynamics of the aggregate claims process is  a subordinator. From this expression,  we easily deduce necessary and sufficient  conditions for the infinite-time absolute ruin to occur. We proceed by showing that,  under some technical conditions, the transition density of  $\hat{X}$ admits a spectral type representation involving merely the limiting distribution of the process. As a by product, we obtain a series expansions for the finite-time absolute ruin probability. On the way, we also derive, for the aforementioned risk process, the Laplace transform of the first-exit time from an interval from above.  Finally, we illustrate our results by detailing some examples.

\bigskip

\noindent{\it Key words:} Risk theory, absolute ruin, Ornstein-Uhlenbeck type processes, first-passage time, spectral representation.

\bigskip

\noindent{\it 2000 Mathematics Subject Classification:} 60G18, 60G51, 60B52
\end{abstract}

\thanks{Research partially funded by  the AXA research fund for the project Modern Actuarial Risk Theory. Both authors would like to thank H. Albrecher for fruitful discussions. The second author is also grateful to F. Avram for valuable comments and the warm hospitality during his stay at the University of Pau.}

\date{}

\maketitle

\bibliographystyle{plain}

\section{Introduction}
Let $Z=(Z_t, t\geq0)$ be a driftless subordinator defined on a filtered probability space $(\Omega,\mathcal{F},(\mathcal F_t)_{t\geq0},\mathbb{P})$, that is $Z$ is a $\R^+-$valued process with stationary and independent increments.  It is well known that the law of $Z$ is characterized by its Laplace exponent which admits the following L\'evy-Khintchine representation
\begin{equation} \label{eq:le_s}
\phi(\beta)=-\log\mathbb{E}(\mathrm{e}^{-\beta Z_1})=\int_0^\infty (1-\mathrm{e}^{-\beta y})\nu( \mathrm{d}y), \quad \beta\geq0,
\end{equation}
where the L\'evy measure satisfies the integrability condition  $\int_0^\infty (1 \wedge y) \nu( \mathrm{d}y)<\infty$ and we refer to the monograph of Kyprianou \cite{Kyprianou-06} for background on subordinators. By $\widehat{\mathbb{P}}$ we denote the law of the dual of $Z$, i.e. the law of $-Z$ under $\mathbb{P}$.
For $x,r,c\in\mathbb{R}$,  we denote   by $\mathbb{P}_x^{(r,c)}$ the law of the process $X$  defined, for any $t\geq0$, by
\begin{equation} \label{eq:def-X}
X_t= \mathrm{e}^{rt} \left( x+\int_0^t\mathrm{e}^{-rs}\mathrm{d}(cs-Z_s) \right),
\end{equation}
where $Z\sim \mathbb{P}$. Similarly,  $\widehat{\mathbb{P}}_x^{(r,c)}$ stands for the law of the process $X$ as defined in \eqref{eq:def-X} with $Z\sim \widehat{\mathbb{P}}$. We simply write $\mathbb{P}_x^{(r)}$ (resp. $\widehat{\mathbb{P}}_x^{(r)}$) for $\mathbb{P}_x^{(r,0)}$ (resp. $\widehat{\mathbb{P}}_x^{(r,0)}$).  Note that equivalently, $X$ is the unique strong solution to the stochastic differential equation
\begin{equation}
\label{SDE}
\mathrm{d}X_t=(rX_t+c)\mathrm{d}t -\mathrm{d}Z_s,\quad X_0=x.
\end{equation}

The process $X$ under $\mathbb{P}_x^{(0,c)}$ with $c>0$ has been used in the literature to model the reserves of an insurance company, the parameter $c$ standing for the premium rate, the jumps of $Z$ standing for the claims and $x$ standing for the initial value of the reserves. In particular, when $\nu(0,\infty)<\infty$, $X$ under $\mathbb{P}_x^{(0,c)}$  is the classical risk process where the claims arrive according to a Poisson process with intensity parameter $\nu(0,\infty)$ and the claim distribution is given by $\nu(\mathrm{d}x)/\nu(0,\infty)$.
When $r>0$, the process $X$ under $\mathbb{P}_x^{(r,c)}$ has been suggested as a  risk process where  the cost of lending/borrowing money are  taken into account.  In this model the insurer earns (credit) interest when the surplus is positive and when the surplus becomes negative, the insurer can cover the deficit by a loan for which he has to pay a (debit) interest. Although in practice the debit interest is much higher than the credit interest, we restrict ourselves to the case where both rates are equal since this choice is particularly tractable and allows us to use techniques which can no longer be used in the general case. Processes of the form \eqref{eq:le_s} with $Z$ a general L\'evy process are known in the literature as processes of Ornstein-Uhlenbeck type (for short OU-type) and therefore in this paper we call the  process $X$ under $\mathbb{P}_x^{(r,c)}$ with $r>0$, the OU-type risk process.

 From \eqref{SDE}, it is clear that when the OU-type risk process reaches the interval $(-\infty,-c/r]$, the premium rate $c$ can no longer compensate the interest payments and so the surplus will decrease to minus infinity. Following Gerber  \cite{Gerber-71}, we say that in this case absolute ruin occurs.
This model and in particular the event of absolute ruin, has been the focus of many research in insurance mathematics, its first appearance can be traced back to Segerdahl \cite{Segerdahl-42}. For more recent investigations and substantial refinements, we mention Gerber \cite{Gerber-71},  Dassios and Embrechts \cite{Embrechts-Dassios-89}, Embrechts and Schmidli \cite{Embrechts-Schmidli-94}, Schmidli \cite{schmidli1994}, Sundt and Teugels \cite{Sundt-Teugels-95}, Albrecher et al.~\cite{Albrecher-01}, Gerber and Yang \cite{Gerber-Yang-07} and Cai \cite{Cai-07}. We refer to the survey papers of Paulsen \cite{paulsen_survey1} and \cite{Paulsen-08} for an overview of ruin models with interest. For general background in ruin theory, we refer to Gerber \cite{ge79} and Albrecher and  Asmussen \cite{Albrecher-Asmussen-10}.

We also point out that the process $X$ under $\widehat{\mathbb{P}}_x^{(-r)}$ with $r>0$, is well known in the literature and has appeared in various settings. For instance, in mathematical finance, this process has been used by Barndorff-Nielsen and Shephard for the modeling of  the stochastic volatility of stock prices, see \cite{Barndorff-Shephard-01}. It   also  belongs to the class of one factor affine term structure models, see e.g.~Filipovic~\cite{filipovic-01}.
Moreover, when the L\'evy measure is finite, it is a specific example of a Poisson shot noise process, see e.g. Perry et al. \cite{Perry-01} and  Iksanov and Jurek \cite{iksanovjurek}.

 The remainder of the paper is organized as follows. The next section is devoted to the statement of our main results which are then proved in Section \ref{section:proofs}. Examples illustrating our approach are presented in Section \ref{section:examples}.

\section{Main results}
We start by providing a representation of the law of the absolute ruin time. To this end, we introduce some notation. First, let, for any $a\leq x$,
\begin{equation*}
\tau_{a}=\inf\{s>0:\: X_s<a\}
\end{equation*}
be the first-passage time below the level $a$ for $X$.
  Henceforth, we shall assume that $r>0$.
We are interested in the  distribution  of $\tau_{-c/r}$ under $\mathbb{P}_x^{(r,c)}$.  We call this random variable the absolute ruin time. As mentioned in the introduction, the reason for the adjective `absolute' is that once the OU-type risk process  goes below the level $-c/r$, it will never go back above this critical level, i.e.
\begin{equation}
\label{def_absoluteruin}
\mathbb{P}^{(r,c)}_x \left(X_t\geq -c/r, t>\tau_{-c/r}\right)=0.
\end{equation}
We first note that from \eqref{eq:def-X} we immediately see that the process $X$ under $\mathbb{P}_x^{(r,c)}$ has the same law as the process $(X_t-c/r,t\geq0)$ under $\mathbb{P}_{x+c/r}^{(r)}$. In particular
\begin{equation*}
\mathbb{P}_x^{(r,c)} \left(\tau_{-c/r}\in\mathrm{d}t\right) = \mathbb{P}_{x+c/r}^{(r)} \left(\tau_0\in \mathrm{d}t\right), \quad t\geq0.
\end{equation*}
Based on this observation, we state all the results in the paper for the $c=0$ case only; the analogue for $c\neq0$ is then obvious.
The first theorem gives the link between the distribution of $\tau_0$ under $\mathbb{P}^{(r)}_x$ and the distribution of $X$ under $\widehat{\mathbb{P}}^{(-r)}_0$, which leads to an explicit expression for the Laplace transform in space of the finite-time absolute ruin probability. This is in contrast with the the finite-time ruin probability in the $r=0$ case, where only an explicit expression for the double Laplace transform in space and time exists (cf. Theorem 8.1(ii) of Kyprianou \cite{Kyprianou-06}).
\begin{thm}  \label{thm:1}
For any $x>0$ and $t\geq0$, we have
\begin{equation*}
 \mathbb{P}^{(r)}_x\left(\tau_{0} \leq t\right) =  \widehat{\mathbb{P}}^{(-r)}_0\left(X_t> x \right).
\end{equation*}
Consequently, for any $\beta>0$, we have
\begin{equation*}
\mathbb{P}^{(r)}_{{\bf{e}}_{\beta}}(\tau_{0}> t )   =  \exp \left(  -\int_0^t\phi(\beta \mathrm{e}^{-rs}) \mathrm{d}s \right)
 \end{equation*}
where ${\bf{e}}_{\beta}$ stands for the exponential distribution of parameter $\beta>0$ (where we used the notation ~$\mathbb{P}^{(r)}_{{\bf{e}}_{\beta}} (A)=\int_{\R}\mathbb{P}^{(r)}_x (A){\bf{e}}_{\beta}(dx)$).
\end{thm}

Theorem  \ref{thm:1} in combination with Theorem 17.5 of Sato \cite{Sato-99} leads to the following result about the infinite time absolute ruin probability. We refer to Sato \cite{Sato-99} for background on self-decomposable random variables.
\begin{cor} \label{cor:1} \
\begin{itemize}
 \item [(i)] If  $\int_1^\infty \log(y)\nu(\mathrm{d}y)<\infty$, then, under $\widehat{\mathbb{P}}_0^{(-r)}$, $X_t$ converges in distribution, as $t\rightarrow \infty$, to a positive self-decomposable random variable $X_{\infty}$ and
\begin{equation*}
\mathbb{P}^{(r)}_x\left(\tau_{0} <\infty\right) =  \widehat{\mathbb{P}}^{(-r)}_0 \left(X_{\infty}\geq x \right).
\end{equation*}
Moreover, for any $\beta>0$, we have

\[\mathbb{P}^{(r)}_{{\bf{e}}_{\beta}}(\tau_{0}= \infty ) = \exp \left( -\int_0^{\beta}\frac{\phi(u)}{ru}\mathrm{d}u \right).\]

\item[(ii)]
If  $\int_1^\infty\log(y)\nu( \mathrm{d}y) =\infty$, then, for any $x>0$,
\begin{equation*}
 \mathbb{P}^{(r)}_x\left(\tau_{0} < \infty\right) =  1.
\end{equation*}
\end{itemize}
\end{cor}

It is interesting to note that for the risk process without interest ($r=0$), ruin is certain when the safety loading is negative that is whenever $c\leq \int_1^\infty y\nu(\mathrm{d}y)$, whereas for $r>0$ the premium rate does not have any influence on whether ruin is certain or not.

\bigskip

Before stating  the next two theorems we need to introduce a little further notation. Let  $\xi=\mathbb{I}_{\{\int_1^\infty \log(y)\nu(\mathrm{d}y)=\infty\}}$ and define the measure $W$ on $[0,\infty)$ via its Laplace transform as follows
\begin{equation}\label{eq:lt_w}
\int_0^\infty \mathrm{e}^{-\beta x} W(\mathrm{d}x) = \mathrm{e}^{- \int_{\xi}^\beta \frac{\phi(u)}{ru} \mathrm{d}u}, \quad \beta\geq \xi.
\end{equation}
Noting, by an integration by parts, that
\begin{equation*}
\int_{\xi}^\beta \frac{\phi(u)}{ru} \mathrm{d}u = \int_0^\infty (1-\mathrm{e}^{(\xi-\beta) x}) \frac{\mathrm{e}^{-\xi x}\nu(x,\infty)}{rx} \mathrm{d}x,
\end{equation*}
 it follows readily that
\[W(\mathrm{d}x)=\mathrm{e}^{\xi x} \mu(\mathrm{d}x),\] where $\mu$ is the law of a positive, self-decomposable random variable with infinite L\'evy measure $\frac{\mathrm{e}^{-\xi x}\nu(x,\infty)}{rx} \mathrm{d}x$ (cf. \cite[Corollary 15.11]{Sato-99}). It follows that the function $W(x):=W[0,x]$ is well-defined, increasing and by \cite[Theorem 27.4]{Sato-99} that $W(x)$ is continuous. Note that if $\int_1^\infty \log(y)\nu(\mathrm{d}y)<\infty$, then $W$ is simply the distribution function of the random variable $X_\infty$ from Corollary \ref{cor:1}. We extend $W$ to the whole real line by setting $W(x)=0$ on $(-\infty,0)$.
Further, for all $n<\nu(0,\infty)/r$, the $n$-th derivative of $W$, denoted by $W^{(n)}$, exists and is continuous on $(-\infty,\infty)$ (cf. \cite[Theorem 28.4]{Sato-99}).

The next theorem concerns a discrete spectral type representation of the transition distribution  of the process $X$ under $\widehat{\mathbb{P}}_0^{(-r)}$. We stress  that the spectral theory for self-adjoint operators  in an Hilbert space structure is well established. In particular, McKean~\cite{McKean-56} discusses in details the nature of the spectrum of the semigroup of linear diffusions. However, this general theory does not apply here since we are dealing with non self-adjoint operators.  In this context, there are little examples in the literature where such a spectral decomposition has been given. One notable exception is the paper of  Ogura \cite{Ogura-70}. Therein, the author  provides conditions under which the semigroup of continuous state branching processes with immigration (for short CBI) admits a discrete or continuous spectral representation. Furthermore, in a very elegant fashion, he manages to characterize through Laplace transform, the eigenmeasure and eigenfunctions associated to the  semigroup. To be more precise, in \cite[Theorem 3.1]{Ogura-70}, Ogura shows, under some conditions, that for some $t_0\geq0$ the semigroup  $P_t(x,\mathrm{d}y)$  of a CBI process satisfies the discrete spectral representation
\begin{equation}
\label{spectral_representation}
P_t(x,\mathrm{d}y)= \sum_{n=0}^\infty \eta_n(x) \zeta_n(\mathrm{d}y)\mathrm{e}^{-\lambda_n t}, \quad t\geq t_0, \: x,y\geq0,
\end{equation}
where $\eta_n$, resp.~$\zeta_n$, are eigenfunctions, resp.~eigenmeasures, of $P_t$ corresponding to the eigenvalue $\mathrm{e}^{-\lambda_n t}$ with $\lambda_n\geq0$.
 Although the process $X$ under $\widehat{\mathbb{P}}_x^{(-r)}$ does belong to the class of CBI processes, they were excluded in \cite{Ogura-70}. Moreover, the  methodology of \cite{Ogura-70} does not extend to our case; in particular the suggestion made in footnote 3) on p.309 of \cite{Ogura-70} does not lead to the right direction. In Theorem \ref{thm:sr} below we are able to give, under some technical conditions on the L\'evy measure, a spectral representation similar to  \eqref{spectral_representation} for the semigroup of this process when $x=0$.
We refer to Chapter 2.7 of Bingham et al.~\cite{Bingham-Goldie-Teugels-89} for the definition of  a quasi-monotone function and remark  that the theorem remains valid if  quasi-monotone is replaced by ultimately monotone. Recall that a function $f:(0,\infty)\rightarrow (0,\infty)$ is ultimately monotone at infinity if there exists $A>0$ such that $f$ is monotone on $(A,\infty)$ and that $f$ is slowly varying at infinity if $\lim_{x\rightarrow\infty} \frac{f(tx)}{f(x)}=1$ for all $t>0$.

 \begin{thm} \label{thm:sr}
 Assume that the L\'evy measure $\nu$ satisfies
 \begin{enumerate}[i)]
 \item $\int_1^\infty \mathrm{e}^{-\beta x}\nu(\mathrm{d}x)<\infty$ for all $\beta\in\mathbb R$, \item $\nu(x,\infty)= \ell(1/x)x^{-\alpha}$ with $0<\alpha<1$ and $\ell$ a quasi-monotone slowly varying function at infinity.
     \end{enumerate}
 Then for any $t>t_\alpha=-\frac1{r\alpha}\log\cos\left(\frac{\pi\alpha}2\right)$  and $x\in\mathbb{R}$, the transition distribution  of the process  $X$ under $\widehat{\mathbb{P}}^{(-r)}_x$  is given by
\begin{equation*}
\widehat{\mathbb P}^{(-r)}_x\left(X_t\in\mathrm{d}y\right)=\sum_{n=0}^\infty \mu_n \mathrm{e}^{-rnt} W^{(n+1)}(y-x\mathrm{e}^{-rt})\mathrm{d}y, \quad y\in\mathbb R,
\end{equation*}
 where
\begin{equation*}
\mu_n=\widehat{\mathbb{E}}^{(-r)}[X_{\infty}^n]/n!= \frac1{n!} \left( \frac{\mathrm{d}^n}{\mathrm{d}v^n} \exp \left( -\int_0^{v}\frac{\phi(u)}{ru}\mathrm{d}u \right) \right)_{v=0}.
\end{equation*}
Consequently, for any $x>0$,
 \begin{equation} \label{eq:main}
 \mathbb{P}^{(r)}_{x}(\tau_{0}> t )  = \sum_{n=0}^\infty \mu_nW^{(n)}(x)\mathrm{e}^{-rnt},
 \quad  t>t_{\alpha}.
 \end{equation}
Moreover, for any $n$, $W^{(n+1)}(y)\mathrm{d}y$ is an eigenmeasure of $\widehat{\mathbb P}^{(-r)}_x(X_t\in\mathrm{d}y)$ corresponding to the eigenvalue $\mathrm{e}^{-rn t}$ in the sense that
 \begin{equation*}
 \int_{-\infty}^\infty \widehat{\mathbb P}^{(-r)}_x(X_t\in\mathrm{d}y) W^{(n+1)}(x)\mathrm{d}x = \mathrm{e}^{-rn t} W^{(n+1)}(y)\mathrm{d}y, \quad y\in\mathbb R.
 \end{equation*}
 \end{thm}

Our final theorem concerns the two-sided exit problem for the OU type process. More specifically, we compute the Laplace transform of the stopping time
\begin{equation*}
\tau_a^+=\inf\{t>0: X_t>a\},
\end{equation*}
on the event that $X$ exits the interval $[0,a]$ at $a$.
When $X$ is a spectrally negative L\'evy process, this quantity is given by $f_q(x)/f_q(a)$ with $q\geq0$ being the parameter of the Laplace transform and where $f_q$ (resp.~$f_0$) is the so-called $q$-scale  function (resp.~scale function) of the spectrally negative L\'evy process, cf. \cite[Theorem 8.1]{Kyprianou-06}. In Theorem \ref{thm:scale}, we derive a similar expression for the case  where $X$ is the OU-type risk process and show that the `corresponding $q$-scale function' is given in terms of the fractional integral of  the function $W$, which we denote by $W_q$. Hence if one has an explicit expression for $W$, one automatically gets $W_q$ for $q>0$ in closed-form; this is in contrast to the case of spectrally negative L\'evy processes, see the discussion on p.1674 of Kyprianou and Rivero \cite{Kyprianou-Rivero-08}.

\begin{thm}\label{thm:scale}
For $q\geq0$, define the function $W_q:(-\infty,\infty)\rightarrow[0,\infty)$ by $W_q(x)=0$ on $(-\infty,0)$ and on $[0,\infty)$ by the Riemann-Liouville fractional integral of order $q$ of $W,$ i.e.
\begin{equation}\label{eq:dfi}
 W_{q}(x)= \frac{1}{\Gamma(q)}\int_0^x(x-y)^{q-1}W(y)\mathrm{d}y, \quad x\geq0.
\end{equation}
Then, for any $q\geq 0$, $x\leq a$ and $a>0$,
we have
\begin{equation*}
\mathbb{E}^{(r)}_x \left[  \mathrm{e}^{-q\tau_a^+} \mathbf{1}_{\{\tau_a^+ <\tau_0\}} \right] = \mathbb{E}^{(r)}_x \left[  \mathrm{e}^{-q\tau_a^+} \mathbf{1}_{\{\tau_a^+ <\infty\}} \right] = \frac{ W_{q/r}(x)}{ W_{q/r}(a)}.
\end{equation*}

\end{thm}
We point out that Theorem \ref{thm:scale} simultaneously gives the first-passage time above $a$ of the OU-type risk process. For a process $X$ defined by \eqref{eq:def-X} with $Z$ a general L\'evy process with no negative jumps  and with $r<0$, Hadjiev \cite{Hadjiev-84} (under an extra assumption) and Novikov \cite{Novikov-03} provided an explicit expression for the Laplace transform of the first-passage time of $X$ above a fixed level. Although not considered in these papers, it can be checked that their methodology and expression extend to the case where $r>0$ and $Z$ a L\'evy process with no downward jumps and paths of unbounded variation. However, when $r>0$ and $Z$ is a subordinator (that is, the OU-type risk process), the situation is completely different due to the possibility of absolute ruin.

We remark here that we cannot expect to find a more explicit expression like the one given in \cite{Hadjiev-84} and \cite{Novikov-03} for the $r<0$ case, since this would lead to an explicit expression for the distribution function of any positive self-decomposable random variable.

\section{Examples}\label{section:examples}
Before studying some examples let us recall the definition of a transformation recently introduced by  Kyprianou and Patie \cite{Kyprianou-Patie-08}, which will be helpful in our context. For any Laplace exponent $\phi$ of a subordinator, that is of the form \eqref{eq:le_s}, we write for any $\gamma\geq 0$
\[\mathcal{T}_{\gamma}\phi(\beta)=\frac{\beta}{\beta+\gamma}\phi(\beta+\gamma).\]
Then, in \cite{Kyprianou-Patie-08}, it is proved that $\mathcal{T}_{\gamma}\phi$ is  the Laplace exponent of the driftless subordinator with L\'evy measure $\nu_\gamma(\mathrm{d}x)=\mathrm{e}^{-\gamma x}(\nu(\mathrm{d}x)+\gamma\nu(x,\infty)\mathrm{d}x)$. In particular, $\nu_\gamma(x,\infty)=\mathrm{e}^{-\gamma x}\nu(x,\infty)$. Obviously, $\mathcal{T}_0\phi=\phi$. We assume throughout this section that $\int_1^\infty \log(y)\nu(\mathrm{d}y)<\infty$ and therefore the quantity $\varphi_r(\beta)$ defined by
\begin{equation*}
\varphi_{r}(\beta) = \frac{1}{r}\int_0^{\beta}\frac{\phi(s)}{s}\mathrm{d}s,
\end{equation*}
is well-defined for $\beta\geq0$. It is easily seen, after a change of variable, that
\begin{equation*}
\varphi^{(\gamma)}_{r}(\beta) := \frac{1}{r}\int_0^{\beta}\frac{\mathcal{T}_{\gamma}\phi(s)}{s}\mathrm{d}s \\
= \varphi_{r}(\beta+\gamma)-\varphi_{r}(\gamma).
\end{equation*}
In other words, the action of the mapping $\mathcal{T}_\gamma$ on the backward Laplace exponent is equivalent to the action of the Esscher transform on the Laplace exponent of the limiting distribution.
In particular, if $W(x;\gamma)$  stands for the limiting distribution function associated to the backward Laplace exponent $\mathcal{T}_{\gamma}\phi$, then we have the following simple relationship
\begin{equation} \label{eq:Tgamma}
W'(x;\gamma) = \mathrm{e}^{-\varphi_{r}(\gamma)-\gamma x} W'(x;0), \quad x>0.
\end{equation}
(Note that by \cite[Theorem 28.4]{Sato-99}, $W$ is always differentiable on $(0,\infty)$.)

Next we detail some examples where the function $W$ and/or the distribution of the absolute ruin time can be given in closed form. For more examples of cases where $W$ is explicit, we refer to Iksanov and Jurek \cite{iksanovjurek}. We end this section by giving an example that illustrates Theorem \ref{thm:sr}.

\subsection{The compound Poisson case with exponential jumps}
Let us start with the case where for $x>0$,
\begin{equation*}
\nu(x,\infty)=\eta\mathrm{e}^{-\delta x}, \quad \delta,\eta>0.
\end{equation*}
Hence the subordinator $Z$ is a compound Poisson process with exponentially distributed jumps.

We have
\begin{equation*}
\varphi_r(\beta)= \frac{\eta}r\log \left( 1+\beta/\delta  \right)
\end{equation*}
and so the Laplace transform of $X$ under $\widehat{\mathbb{P}}^{(-r)}_x$ is easily computed via \eqref{laplacetransform}. By inverting this Laplace transform, Perry et al. \cite{Perry-01} shows that the transition function of $X$ under $\widehat{\mathbb{P}}^{(-r)}_0$ admits the following form
\begin{equation*}
\widehat{\mathbb{P}}^{(-r)}_0 (X_t \in \mathrm{d}y) = \mathrm{e}^{-\eta t} \delta_0(\mathrm{d}y) + \frac{\eta \delta}{r}\mathrm{e}^{-\eta t}(\mathrm{e}^{rt}-1) \mathrm{e}^{-\delta x} {}_1F_1\left(1-\frac{\eta}{r};2;\delta(1-\mathrm{e}^{rt})y\right)\mathrm{d}y,\quad y>0,
\end{equation*}
where $\delta_0$ stands for the dirac point mass  at $0$ and  ${}_1F_1$ is the confluent hypergeometric function, see e.g.~Lebedev \cite[Section 9.9]{Lebedev-72}. Hence, from Theorem \ref{thm:1}, we deduce that
\begin{equation*}
{\mathbb{P}}^{(r)}_x (\tau_{0}>t) = \mathrm{e}^{-\eta t} \left(1 + \frac{\eta \delta}{r}(\mathrm{e}^{rt}-1)\int_0^{x}\mathrm{e}^{-\delta y}{}_1F_1\left(1-\frac{\eta}{r};2;\delta(1-\mathrm{e}^{rt})y \right) \mathrm{d}y\right).

\end{equation*}
From Corollary \ref{cor:1}, it follows that ${\mathbb{P}}^{(r)}_x (\tau_{0}=\infty)$ (or equivalently $W(x)$) equals the gamma distribution with shape parameter $\eta/r$ and scale parameter $1/\delta$; this fact was first established by Gerber \cite{Gerber-71}.

\subsection{The compound Poisson process with Linnik distribution }
 We consider an example found in Iksanov and Jurek \cite{iksanovjurek}. Assume that the L\'evy measure $\nu$ of $Z$ is given by
\begin{equation*}
\nu(x,\infty)=\eta E_\alpha(-\delta x^\alpha ), \quad x>0,

\end{equation*}
where $ \eta,\delta>0, 0<\alpha\leq1$ and
\[E_{\alpha}(x)=\sum_{k=0}^\infty \frac{x^k}{\Gamma(1+\alpha k)}\] is the Mittag-Leffler function and $\Gamma$ is the gamma function.
Hence the subordinator $Z$ is a compound Poisson process with arrival rate $\eta$ and with jumps distributed according to a  positive Linnik distribution. We have $\phi(\beta)=\frac{\eta \beta^\alpha}{\delta+\beta^\alpha}$ and thus
\[\varphi_r(\beta)=\frac{\eta}{r\alpha}\log(1+\beta^\alpha/\delta).\]
We point out that in the case $\alpha=1$, the Linnik distribution boils down to the exponential distribution and hence this example can be seen as a generalization of the previous. Next, we deduce, from the identity \eqref{laplacetransform} below, that
\begin{equation*}
\widehat{\E}^{(-r)}_0\left[ \mathrm{e}^{-\beta X_t}\right] =  \left(\frac{\delta^{-1}\beta^{\alpha} \mathrm{e}^{-\alpha rt}+1}{\delta^{-1}\beta^{\alpha}+1}\right)^{\frac{\eta}{\alpha r}}.
\end{equation*}
Denote for $\kappa>0$ by  ${}_{\kappa}W(x)$ the increasing function on $[0,\infty)$ characterized through its Laplace transform which is given by
\begin{equation*}
\int_0^\infty \mathrm{e}^{-\beta x}{}_{\kappa}W(\mathrm{d}x)=\left({\delta^{-1}\beta^{\alpha}+1}\right)^{-\kappa}, \quad \beta\geq0.
\end{equation*}
Note that  ${}_{\frac{\eta}{\alpha r}}W(x)=\mathbb{P}_x^{(r)}(\tau_0=\infty)$.
By means of the binomial
formula, we get, for any $\beta^{\alpha}>1$,
\begin{eqnarray*}
(1+\delta^{-1}\beta^{\alpha})^{-\kappa} &=& \sum_{n=0}^{\infty} (-1)^{n}
\frac{\Gamma(\kappa+n)}{n!\Gamma(\kappa)}
(\delta^{-1}\beta^{\alpha})^{-(n+\kappa)},
\end{eqnarray*}
and hence we obtain by a term-by-term inversion
\begin{eqnarray*}
{}_{\kappa}W'(x)  &=& \delta^{\kappa} x^{\alpha\kappa-1}\sum_{n=0}^\infty (-1)^n\frac{\Gamma(\kappa+n)}{\Gamma(\kappa)\Gamma(\alpha(n+\kappa))n!}(\delta x^{\alpha})^{n}\\
&=& \delta^{\kappa} x^{\alpha\kappa-1} \fx{1}{1} \left( \left.\begin{array}{c}
                  (1,\kappa) \\
                  (\alpha,\alpha\kappa
                  )
                \end{array} \right|
 \: - \delta x^{\alpha}\right)
 \end{eqnarray*}
where $\fx{1}{1}$ is a Wright hypergeometric function, see e.g.~Braaksma \cite[Chap.~12]{Braaksma-64}. Note that for $\alpha=1$,
\begin{eqnarray*}
{}_{\kappa}W'(x)  &=& \frac{\delta^{\kappa}}{\Gamma(\kappa)} x^{\kappa-1}\mathrm{e}^{-\delta x},
 \end{eqnarray*}
recovering the exponential jumps case studied above. Next, observing that for $\beta$ big enough by the binomial theorem,
\begin{equation*}
\begin{split}
\left(\frac{\delta^{-1}\beta^{\alpha} \mathrm{e}^{-\alpha rt}+1}{\delta^{-1}\beta^{\alpha}+1}\right)^{\frac{\eta}{\alpha r}}
= & \left( 1-\frac{1-\mathrm{e}^{-\alpha rt}}{\delta^{-1}\mathrm{e}^{-\alpha rt}\beta^{\alpha}+1} \right)^{-\frac{\eta}{\alpha r}} \\
= & \mathrm{e}^{-\eta t} + \sum_{n=1}^\infty \frac{\Gamma\left(n+\frac{\eta}{\alpha r}\right)}{\Gamma\left(\frac{\eta}{\alpha r}\right)n!} \frac{(1-\mathrm{e}^{-\alpha rt})^n} {(1+\delta^{-1}\mathrm{e}^{-\alpha rt}\beta^{\alpha})^n} ,
\end{split}
\end{equation*}
we get, by means of Laplace transform inversion and Theorem \ref{thm:1},
\begin{equation}\label{eq:ser}
\mathbb{P}^{(r)}_x (\tau_{0}>t) = \mathrm{e}^{-\eta t} \left( 1 +
 \mathrm{e}^{rt}\sum_{n=1}^\infty \frac{\Gamma\left(n+\frac{\eta}{\alpha r}\right)}{\Gamma\left(\frac{\eta}{\alpha r}\right)n!}  (1-\mathrm{e}^{-\alpha rt})^n {}_{n}W(\mathrm{e}^{rt}x) \right).
\end{equation}
Note that, for any $\kappa>0$, the power series
\[\sum_{n=0}^\infty \frac{\Gamma(\kappa+n)}{n!}x^n\]
is analytic in the disc of radius $1$. Since for any $t>0$, $|1-\mathrm{e}^{-rt}|<1$ and, clearly  ${}_nW(y)\leq 1$ for any $y>0$ and $n=1,2\ldots$, we deduce that the series on the right-hand side of  \eqref{eq:ser} is uniformly convergent on $\R^+\times\R^+$.
Note also that the identity \eqref{eq:ser} simplifies when $\frac{\eta}{\alpha r}$ is an integer. In particular, when ${\eta}={\alpha r}$, the finite-time ruin probability is simply a weighted sum of $1$ and the infinite-time ruin probability, i.e.
\begin{equation*}
\mathbb{P}^{(r)}_x (\tau_{0}>t) = \mathrm{e}^{-\eta t} + (1-\mathrm{e}^{-\eta t}){}_{1}W(x).
\end{equation*}
Moreover, when $\alpha=1$, we have
\begin{equation*}
\begin{split}
 \mathrm{e}^{rt}\sum_{n=1}^\infty \frac{\Gamma\left(n+\frac{\eta}{  r}\right)}{\Gamma\left(\frac{\eta}{  r}\right)n!}  (1-\mathrm{e}^{-rt})^n {}_{n}W(\mathrm{e}^{rt}x) = &
\int_0^x \sum_{n=1}^\infty \frac{\Gamma\left(n+\frac{\eta}{  r}\right)}{\Gamma\left(\frac{\eta}{  r}\right)n!} (\mathrm{e}^{rt}-1)^n  \frac{\delta^{n}}{\Gamma(n)} y^{n-1}\mathrm{e}^{-\delta\mathrm{e}^{rt} y}\mathrm{d}y \\
= & \delta(\mathrm{e}^{rt}-1)\int_0^x \mathrm{e}^{-\delta \mathrm{e}^{rt} y} {}_1F_1 \left( 1+\frac{\eta}{r};2;(\mathrm{e}^{rt}-1)\delta y \right)  \mathrm{d}y  \\
= & \delta(\mathrm{e}^{rt}-1)\int_0^x \mathrm{e}^{-\delta y} {}_1F_1 \left( 1-\frac{\eta}{r};2;(1-\mathrm{e}^{rt})\delta y \right) \mathrm{d}y
\end{split}
\end{equation*}
where the last equality follows from the identity $\mathrm{e}^{-y}{}_1F_1\left(a;b;y\right)={}_1F_1\left(b-a;b;-y\right)$, see,  after correcting the obvious typo, formula 9.11.2 in \cite{Lebedev-72}. This is  consistent with the previous example. We finally note that when $0<\alpha<1$, we can use \eqref{eq:Tgamma} to get a new example where the infinite-time ruin probability is explicit.

\subsection{The stable subordinator}
We assume that $Z$ under $\mathbb{P}$ is a stable subordinator of index $0<\alpha<1$.  Its
Laplace exponent takes the form
$\phi(\beta)=\frac{\beta^{\alpha}}{\cos(\pi\alpha/2)}$, $\beta \geq0$.
It is easy to verify that in this case
$\varphi_r(\beta) =\frac{ \beta^{\alpha}}{\cos(\pi\alpha/2)\alpha r}$, $\beta \geq0$.
One can check by taking Laplace transforms and using the scaling property of $Z$ that,
\begin{equation*}
\widehat{\mathbb{P}}^{(-r)}(X_t\in\mathrm{d}y) = {\mathbb{P}} \left( v^{1/\alpha}(t)Z_{1} \in\mathrm{d}y \right) \quad \text{where $v(t) =\frac{1-\mathrm{e}^{-\alpha rt}}{\alpha r}$.}
\end{equation*}
Hence, by writing $P_{\alpha}$ for the distribution of a positive stable random variable of index $\alpha$, we deduce  in combination with Theorem \ref{thm:1} that
\begin{equation*}
{\mathbb{P}}^{(r)}_x (\tau_{0}>t) = P_{\alpha}\left( xv^{-1/\alpha}(t) \right)
\end{equation*}
and by taking the limit as $t\rightarrow \infty$, we get from Corollary \ref{cor:1},
\begin{equation*}
{\mathbb{P}}^{(r)}_x (\tau_{0}=\infty) = P_{\alpha}\left((\alpha r)^{1/\alpha}x \right).
\end{equation*}
We mention that a series representation of stable densities can be found in Sato \cite{Sato-00}. Additional interesting representations has also been derived by Schneider \cite{Schneider-86}. In particular,  in the case $\alpha=1/2$, we recall that the stable distribution admits the well known simple expression
\begin{equation}
P_{1/2}(x)=\int_0^{x}\frac{1}{\sqrt{2\pi}} y^{-3/2}\mathrm{e}^{-1/(2y)}\mathrm{d}y.
\end{equation}
We consider now the image of $\phi$ by the mapping $\mathcal{T}_{\gamma}$ which gives
\begin{eqnarray*}
\mathcal{T}_{\gamma} \phi(\beta) = \frac{(\beta+\gamma)^{\alpha}-\gamma(\beta+\gamma)^{\alpha-1}}{\cos(\pi\alpha/2) }.
\end{eqnarray*}

This Laplace exponent is a specific instance
of the family of characteristic exponents of truncated L\'evy processes constructed by Boyarchenko and Levendorskii \cite{Boyarchenko-L-00}, also called tempered stable processes.
From the identity \eqref{eq:Tgamma}, we deduce that
\[W(x;\gamma)=\mathrm{e}^{-\gamma^{\alpha}/(\cos(\pi\alpha/2) \alpha r)}(\alpha r)^{1/\alpha}\int_0^x \mathrm{e}^{-\gamma y} P'_{\alpha}\left( (\alpha r)^{1/\alpha} y\right) \mathrm{d}y.\]

\subsection{Example from Theorem \ref{thm:sr}}

We now give an example where we use the representation \eqref{eq:main} for the finite-time absolute survival probability in Theorem \ref{thm:sr}. Assume $\nu(x,\infty)=\ell(1/x)x^{-\alpha}$ with $0<\alpha<1$ and
\begin{equation*}
\ell(x)= C \mathbb I_{\{x\geq A\}}, \quad \text{where $C,A>0$.}
\end{equation*}
Then $\nu$ is a L\'evy measure that satisfies all the conditions of Theorem \ref{thm:sr}. Note that this L\'evy measure corresponds to a subordinator with stable-like jumps near the origin and no jumps larger than $A$. This model with claims bounded in size, might for instance be used in the case where the insurance company  has excess-of-loss reinsurance. A straightforward computation involving integration by parts, shows that
\begin{equation*}

\varphi_r(\beta)= \frac{C}{r\alpha}  \left( \beta^\alpha \Gamma(1-\alpha;0,\beta A) - A^{-\alpha} \left(1-\mathrm{e}^{-\beta A}\right) \right),
\end{equation*}
where  $\Gamma(a;z_0,z_1) = \int_{z_0}^{z_1}\mathrm{e}^{-t}t^{a-1}\mathrm{d}t$ is the incomplete gamma function. We want to check how well the spectral representation \eqref{eq:main} performs.
Since we do not know an explicit expression for $W$, we proceed by numerical inversion. Let $N\in\mathbb Z_{+}$. For $n=0,1,2\ldots,N-1$ we compute $W^{(n+1)}$ by inverting the right-hand side of \eqref{nderiv} below using the (inverse) fast Fourier transform. We then numerically integrate $W'$ to get $W$ and consequently we get an approximation for the right-hand side of \eqref{eq:main} where the upper boundary $\infty$ is replaced by $N$.
We remark here that as $n$ increases, it gets more difficult to obtain accurate approximations for $W^{(n+1)}$ since these functions oscillate more and more as $n$ increases. In particular, for $n\geq0$, $W^{(n+1)}$ has $n$ distinct zeros, cf. \cite[Theorem 5.1]{satoyamazato}.
In order to get an approximation for the left-hand side of \eqref{eq:main}, note that from \eqref{laplacetransform}, it follows that the Fourier transform of the transition density of $X_t$ under $\widehat{\mathbb P}_0^{(-r)}$ is given by
\begin{equation}
\label{direct_fourier}
\widehat{\mathbb{E}}_0^{(-r)} \left[ \mathrm{e}^{-\mathrm{i}u X_t} \right]
 = \exp \left(  \varphi_r(\mathrm{i}u\mathrm{e}^{-rt}) -\varphi_r(\mathrm{i}u)  \right), \quad u\in\mathbb R,\:t\geq0,
\end{equation}
and by numerical Fourier inversion and integration, one gets (cf. Theorem \ref{thm:1}) an approximation of the finite-time absolute survival probability at a fixed time $t$.
Hence we  obtain numerical approximations of the truncation error
corresponding to \eqref{eq:main} which we define by
\begin{equation*}
e_{N,t}=\max_{i=0,1,\ldots,M} \left|\mathbb{P}^{(r)}_{hi}(\tau_{0}> t )  - \sum_{n=0}^N \mu_nW^{(n)}(hi)\mathrm{e}^{-rnt}\right|.
\end{equation*}
For our example we take $r=0.2$, $\alpha=0.5$, $C=A=1$, $h=0.2$ and $M=125$. All the calculations are done in Matlab.
In Table \ref{table}, numerical approximations of the truncation error are displayed for different values of $N$ and $t$. One sees that convergence takes place for $t=7,10,15$ as $N$ grows and that for a fixed $N$, the error becomes smaller as $t$ grows. We point out that the numbers in the table, besides the truncation error, also consists of integration error due to the Fourier inversion. In particular, the integration error will dominate the truncation error for high values of $N$ (and low values of $t$) due to the highly oscillating nature of $W^{(n)}$ for large values of $n$; this might explain the large value in the table for $N=16$ and $t=5$.
The spectral representation performs badly for $t=3$, but note that $3<t_\alpha\approx 3.46$ and hence this case is not covered by Theorem \ref{thm:sr}. The convergence of $\sum_{n=0}^N \mu_n W^{(n)}(x)\mathrm{e}^{-rnt}$ to $\mathbb{P}^{(r)}_{x}(\tau_{0}> t )$ for $t=7$ is graphically displayed in Figure 1.

We remark that the benefit of computing finite-time absolute survival/ruin probabilities via \eqref{eq:main} is that one can
quickly calculate these probabilities for a whole range of time points $t$, whereas if ones uses the method via \eqref{direct_fourier}, one has  to perform a separate Fourier inversion for each  $t$. Moreover, by using the spectral representation, one can take advantage of \eqref{eq:Tgamma} to quickly obtain the distribution of the absolute ruin time in the case where the tail of the L\'evy measure equals $\mathrm{e}^{-\gamma x}\nu(x,\infty)$ with $\gamma>0$.

\begin{table}[ht]
\centering
\begin{small}
\begin{tabular}{|c|c c c c c|}
\hline
\backslashbox{$N$}{$t$}   &          3 &          5 &          7 &         10 &         15 \\
\hline \hline
         0 &      0.905 &      0.718 &      0.526 &      0.312 &      0.130 \\

         1 &      0.768 &      0.461 &      0.244 &      0.091 &      0.027 \\

         2 &      1.283 &      0.453 &      0.139 &      0.025 &      0.020 \\

         3 &      1.587 &      0.385 &      0.090 &      0.024 &      0.021 \\

         4 &      2.424 &      0.426 &      0.088 &      0.025 &      0.021 \\

         6 &      4.080 &      0.349 &      0.039 &      0.022 &      0.021 \\

         9 &      9.320 &      0.237 &      0.033 &      0.022 &      0.021 \\

        12 &     22.508 &      0.167 &      0.031 &      0.022 &      0.021 \\

        16 &    582.088 &      0.887 &      0.030 &      0.022 &      0.021 \\ \hline

\end{tabular}
\vspace{0.4cm}

\caption{Numerical approximations of $e_{N,t}$ for different values of $N$ and $t$.}
\label{table}
\end{small}
\end{table}

\begin{figure}
\label{figure}
\centering
\includegraphics[scale=.6]{tis7plot5lines_edit2_psfile}
\begin{small}
\caption{Graph of $x\mapsto\sum_{n=0}^N \mu_n W^{(n)}(x)\mathrm{e}^{-rnt}$ for $t=7$ and  $N=0,1,3,6,\infty$.}
\end{small}
\end{figure}

\section{Proofs}\label{section:proofs}

\begin{proof}[\textbf{Proof of Theorem \ref{thm:1}}]
First, by means of a change of variables and the duality lemma for L\'evy processes, see e.g.~\cite[p.~45]{Bertoin-96}, we obtain
\begin{eqnarray}
\label{little_duality}

\mathbb{P}\left(\int_0^t  \mathrm{e}^{-rs} \mathrm{d}Z_s\in  \mathrm{d}y \right) &=&  \mathbb{P}\left(-\int_0^t  \mathrm{e}^{-r(t-u)} \mathrm{d}Z_{t-u}\in  \mathrm{d}y \right)  \nonumber \\
&=& \mathbb{P}\left(-\int_0^t  \mathrm{e}^{-r(t-u)} \mathrm{d}(Z_{t-u}-Z_t) \in  \mathrm{d}y\right)  \nonumber  \\
&=& \mathbb{P}\left(\int_0^t  \mathrm{e}^{-r(t-u)} \mathrm{d}Z_u \in  \mathrm{d}y \right).

\end{eqnarray}
Let $x>0$. Now using that the ruin time is absolute, \eqref{eq:def-X} and \eqref{little_duality},
\begin{eqnarray*}

 \mathbb{P}_x^{(r)}(\tau_0\leq t) &= & \mathbb{P}_x^{(r)}\left(X_t<0\right)\\
&=&  \mathbb{P}\left(-\int_0^t  \mathrm{e}^{-rs} \mathrm{d}Z_s <-x\right) \\
&= &   \mathbb{P}\left(-\int_0^t  \mathrm{e}^{-r(t-u)} \mathrm{d}Z_{u} <-x\right)\\
&=&   \widehat{\mathbb{P}}^{(-r)}_0\left(X_t>x\right),

\end{eqnarray*}
which gives the first assertion. The second follows  directly from the well-known expression for the Laplace transform of $X_t$ under $\widehat{\mathbb P}^{(-r)}_x$ which reads,
\begin{equation}
\label{laplacetransform}
\widehat{\mathbb{E}}_x^{(-r)} \left[ \mathrm{e}^{-\beta X_t} \right]
 = \exp \left( -\beta x \mathrm{e}^{-rt} +\int_0^t \phi(\lambda\mathrm{e}^{-rs})\mathrm{d}s  \right), \quad \beta,t\geq0, x\in\mathbb R,
\end{equation}
see e.g.~Sato \cite{Sato-99}.

\end{proof}

Denote by $P^{(r)}_t$ (resp.~$\widehat{P}^{(-r)}_t$) the semigroup of the process $X$ under $\mathbb{P}^{(r)}$  (resp.~$\widehat{\mathbb{P}}^{(-r)}$), i.e.~for any  positive measurable function $f$ on $\R$,
\begin{equation*}
P^{(r)}_t f(x)= \mathbb{E}_x^{(r)}\left[f(X_t)\right],\quad x\in \R, t\geq0.
\end{equation*}
For the proof of Theorem \ref{thm:scale} we need the following lemma; note that part (ii) is a (weak) duality result between the two semigroups.
\begin{lemma}\label{lemma_interval} \
\begin{itemize}
 \item[(i)] For any $t\geq0$ and $x\in\R$,
\begin{equation}
\label{asmussen}
(X_t,\mathbb{P}_x^r)\stackrel{(d)}{=}( \mathrm{e}^{rt}x+X_t,\mathbb{P}^r_0),
\end{equation}
where $\stackrel{(d)}{=}$ stands for the identity in distribution.
\item[(ii)] For any two positive measurable functions $f$ and $g$,
\begin{equation} \label{big_duality}
\langle P^{(r)}_tf,g\rangle\: = \: \mathrm{e}^{-rt}\langle f,\widehat{P}^{(-r)}_t g\rangle,
\end{equation}
where  $\langle f,g\rangle =\int_{\mathbb{R}} f(x)g(x) \mathrm{d}x$.

\item[(iii)] For all $a\geq0$, $\mathbb{P}_x^{(r)}\left(X_t\in[-a,a] \textrm{ for all } t\geq0\right)=0$.
\end{itemize}

\end{lemma}
\begin{proof}
Item (i)  follows directly from \eqref{eq:def-X}. Next, using successively Tonelli, \eqref{asmussen}, a change of variables, \eqref{little_duality} and \eqref{asmussen} again, we get
\begin{equation*}
\begin{split}
\langle P^{(r)}_tf,g\rangle  = & \int_{\mathbb{R}}\mathbb{E}^{(r)}_x[f(X_t)]g(x) \mathrm{d}x
= \mathbb{E}^{(r)}_0 \left[ \int_{\mathbb{R}} f(X_t+ \mathrm{e}^{rt}x)g(x) \mathrm{d}x \right] \\
= & \mathbb{E}^{(r)}_0 \left[ \int_{\mathbb{R}}  \mathrm{e}^{-rt} f(z)g((z-X_t) \mathrm{e}^{-rt}) \mathrm{d}z \right] \\

= &   \mathrm{e}^{-rt} \int_{\mathbb{R}}  f(z)\mathbb{E} \left[ g\left(z \mathrm{e}^{-rt} - \int_0^t  \mathrm{e}^{-rs} \mathrm{d}Z_s  \right) \right]  \mathrm{d}z  \\

= &  \mathrm{e}^{-rt} \int_{\mathbb{R}}  f(z)\mathbb{E} \left[ g\left( \mathrm{e}^{-rt} \left(z + \int_0^t  \mathrm{e}^{rs} \mathrm{d}(-Z_s)\right)  \right) \right]  \mathrm{d}z \\

= &  \mathrm{e}^{-rt} \int_{\mathbb{R}}  f(z)\widehat{\mathbb{E}}^{(-r) }_z \left[ g(X_t) \right]  \mathrm{d}z \\
=  & \mathrm{e}^{-rt}\langle f,\widehat{P}^{(-r)}_t g\rangle,
\end{split}
\end{equation*}
yielding (ii). Finally, choosing $f(x)=\mathbf{1}_{\{x\in[-a,a]\}}$ and $g(x)=\mathbf{1}_{\{x\geq0\}}$ with $a\geq0$ in \eqref{big_duality} leads to
$\int_0^\infty \mathbb{P}_x^{(r)}(X_t\in[-a,a])\mathrm{d}x=2a\mathrm{e}^{-rt}$
and now by applying standard argument from measure theory (cf. Exercise 3.1.12 of \cite{stroock}) and Fatou's lemma, we get
\begin{eqnarray*}

\int_0^\infty \mathbb{P}_x^{(r)}(\liminf_{t\rightarrow\infty} \{ X_t\in[-a,a]\} )\mathrm{d}x &\leq& \int_0^\infty \liminf_{t\rightarrow\infty} \mathbb{P}_x^{(r)}(X_t\in[-a,a])\mathrm{d}x \\
&\leq& \liminf_{t\rightarrow\infty} \int_0^\infty \mathbb{P}_x^{(r)}(X_t\in[-a,a])\mathrm{d}x =0.
\end{eqnarray*}
Hence  $\mathbb{P}_x^{(r)}(X_t\in[-a,a]$ for all $t\geq0)=0$ and the proof is completed.
\end{proof}

\begin{proof}[\textbf{Proof of Theorem \ref{thm:scale}}]
The first equality follows directly from Lemma \ref{lemma_interval}(iii) and \eqref{def_absoluteruin}. In order to prove the second equality,
we first show that for all $q\geq0$,
\begin{equation}
\label{invariant}
\mathrm{e}^{-qt} \mathbb{E}_x^{(r)} \left[ W_{q/r} (X_t) \right] =W_{q/r} (x).
\end{equation}

Let $\beta>0$, $q\geq0$ and recall that $\xi=\mathbb{I}_{\{\int_1^\infty \log(y)\nu(\mathrm{d}y)=\infty\}}$. From  the identity \eqref{eq:lt_w} and observing the convolution in \eqref{eq:dfi}, we get
\begin{equation}
\label{eq:lt_wq}
\int_{-\infty}^\infty \mathrm{e}^{-\beta x}W_q(x)\mathrm{d}x  = \beta^{-q-1}\mathrm{e}^{-\int_{\xi}^\beta \frac{\phi(u)}{ru} \mathrm{d}u}.
\end{equation}
On the other hand, using successively \eqref{big_duality}, \eqref{laplacetransform} and \eqref{eq:lt_wq}, we obtain, denoting $e_{\beta}(x)=\mathrm{e}^{-\beta x}$, $x\in\mathbb R$, that the Laplace transform of the left-hand side of \eqref{invariant} is given by
\begin{eqnarray*}
\mathrm{e}^{-qt} \langle e_{\beta},{P}^{(r)}_t W_{q/r}\rangle  &=& \mathrm{e}^{-(q+r)t} \langle \widehat{P}^{(-r)}_te_{\beta},W_{q/r}\rangle  \\
&=& \mathrm{e}^{-(q+r)t} \mathrm{e}^{-\int_{\xi}^\beta \frac{\phi(u)}{ru} \mathrm{d}u + \int_{\xi}^{\beta\mathrm{e}^{-rt}} \frac{\phi(u)}{ru}\mathrm{d}u } \langle e_{\beta \mathrm{e}^{-rt} },W_{q/r}\rangle  \\
&=& \beta^{-q/r-1}  \mathrm{e}^{-\int_{\xi}^\beta \frac{\phi(u)}{ru} \mathrm{d}u}.
\end{eqnarray*}
Hence by uniqueness of the Laplace transform and the continuity of $W_q$, \eqref{invariant} follows.

Now with the aid of \eqref{invariant} we can use the Dynkin formula \cite[Theorem 12.4]{Dynkin-65}, to derive
\begin{equation}
\mathbb{E}^{(r)}_x\left[ \mathrm{e}^{-q(t\wedge \tau_a^+)}W_{q}(X_{t\wedge \tau_a^+}) \right]= W_{q/r} (x).
\end{equation}
Since the mapping $W_{q}$ is increasing and continuous, the proof is completed by an argument of dominated convergence and  the property that under $\mathbb{P}_x^{(r)}$, $X_{\tau_a^+}=a$ on $\{\tau_a^+<\infty\}$ which follows by absence of positive jumps.
\end{proof}

\begin{proof}[\textbf{Proof of Theorem \ref{thm:sr}}]
Denote $\varphi_r(v)=\int_0^{v}\frac{\phi(u)}{ru}\mathrm{d}u$, $v\in\mathbb C$. By assumption (i), $\varphi_r$ and consequently $\exp(\varphi_r(\cdot))$ are entire functions (see e.g.~\cite[Section 25]{Sato-99})  and hence we can write
\begin{equation}
\label{entire}
\mathrm{e}^{\varphi_r(v)} = \sum_{n=0}^\infty \mu_n v^n, \quad v\in\mathbb{C}.
\end{equation}
By assumption (ii), $\nu(0,\infty)=\infty$ and thus from  Sato and Yamazato \cite[Equation (2.20)]{satoyamazato}, we get
\begin{equation}
\label{nderiv}
W^{(n+1)}(y)=\frac{1}{2\pi}\int_{-\infty}^\infty (\mathrm{i}u)^n \mathrm{e}^{-\varphi_r(\mathrm{i}u)+\mathrm iuy}\mathrm{d}u, \quad n=0,1,2,\ldots.
\end{equation}
From the above identity, we deduce that
\begin{equation}
\label{fourierinversion}
\sum_{n=0}^\infty \mu_n W^{(n+1)}(y)v^n = \frac{1}{2\pi}\sum_{n=0}^\infty \int_{-\infty}^\infty \mu_n (\mathrm{i}u v)^n \mathrm{e}^{-\varphi_r(\mathrm{i}u)+\mathrm iuy}\mathrm{d}u
\end{equation}
and now we want to use Fubini to  switch the sum and the integral.
We have for $v>0$ by \eqref{entire},
\begin{equation}
\label{fubini}
\begin{split}
 \int_{-\infty}^\infty  \sum_{n=0}^\infty  \left|   \mu_n (\mathrm{i}u v)^n \mathrm{e}^{-\varphi_r(\mathrm{i}u)+\mathrm iuy}  \right| \mathrm{d}u
\leq & \sqrt{2}\int_{-\infty}^\infty \sum_{n=0}^\infty  \mu_n(|u| v)^n \mathrm{e}^{\Re(-\varphi_r(\mathrm{i}u))} \mathrm{d}u \\
= & \sqrt{2}\int_{-\infty}^\infty     \mathrm{e}^{\varphi_r(|u|v)} \mathrm{e}^{-\Re(\varphi_r(\mathrm{i}u))} \mathrm{d}u.
\end{split}
\end{equation}

Using Fubini, we have
\begin{eqnarray*}
r\Re(\varphi_r(\mathrm{i}u)) &=& \int_0^\infty (1-\cos(ux))\frac{\nu(x,\infty)}x\mathrm{d}x \\ &=& \int_0^\infty \int_0^u x\sin(xs)\mathrm{d}s \frac{\nu(x,\infty)}x\mathrm{d}x \\
&=&  \int_0^u \int_0^\infty \sin(xs) {\nu(x,\infty)} \mathrm{d}x \mathrm{d}s.
\end{eqnarray*}
Because of condition (ii), it follows by \cite[Equation (4.3.8)]{Bingham-Goldie-Teugels-89} that
\begin{equation*}
\int_0^\infty \sin(xs) {\nu(x,\infty)} \mathrm{d}x \sim s^{\alpha-1} \ell(s)\Gamma(1-\alpha)\cos\left(\frac{\pi\alpha}2\right), \quad \text{as $s\rightarrow\infty$}
\end{equation*}
and hence by Karamata's Theorem \cite[Theorem 1.5.11]{Bingham-Goldie-Teugels-89},
\begin{equation*}

r\Re(\varphi_r(\mathrm{i}u)) \sim u^{\alpha} \ell(u)\frac{\Gamma(1-\alpha)}{\alpha}\cos\left(\frac{\pi\alpha}2\right), \quad \text{as $u\rightarrow\infty$}.
\end{equation*}
Here $f(x)\sim g(x)$ as $x\rightarrow\infty$ stands for $\lim_{x\rightarrow\infty} \frac{f(x)}{g(x)}=1$. Similarly for $u>0$ by Fubini,
\begin{eqnarray*}
r\varphi_r(u)&=&\int_0^\infty (1-\mathrm{e}^{-ux})\frac{\nu(x,\infty)}x\mathrm{d}x \\&=& \int_0^\infty \int_0^u x\mathrm{e}^{-sx}\mathrm{d}s \frac{\nu(x,\infty)}x \mathrm{d}x \\
&= & \int_0^u \int_0^\infty  \mathrm{e}^{-sx} {\nu(x,\infty)} \mathrm{d}x \mathrm{d}s
\end{eqnarray*}
and thus  by Karamata's theorem and Karamata's Tauberian theorem \cite[Theorem 1.7.1]{Bingham-Goldie-Teugels-89},
\begin{equation*}
r\varphi_r(u) \sim \ell(u)\frac{\Gamma(1-\alpha)}{\alpha}u^{\alpha}, \quad \text{as $u\rightarrow\infty$.}
\end{equation*}
In particular $r\varphi_r(u)$ is regularly varying at infinity with parameter $\alpha$ and thus
\begin{equation*}
\lim_{u\rightarrow\infty} \frac{r\varphi_r(vu)}{\ell(u)\frac{\Gamma(1-\alpha)}{\alpha}u^{\alpha}}=
\lim_{u\rightarrow\infty} \frac{r\varphi_r(vu)}{r\varphi_r(u)}\frac{r\varphi_r(u)}{\ell(u)\frac{\Gamma(1-\alpha)}{\alpha}u^{\alpha}}=v^\alpha.
\end{equation*}
It follows that $r[\Re(\varphi_r(\mathrm{i}u))-\varphi_r(vu)]\sim [\cos\left(\frac{\pi\alpha}2\right)-v^\alpha]\ell(u)\frac{\Gamma(1-\alpha)}{\alpha}u^{\alpha}$ as $u\rightarrow\infty$ if $v^\alpha\neq \cos\left(\frac{\pi\alpha}2\right)$.
Assume now that $\cos\left(\frac{\pi\alpha}2\right)>v^\alpha$ and let $\delta=\frac{\alpha}2$. Then $\Re(\varphi_r(\mathrm{i}u)) -\varphi_r(vu)$ is regularly varying at infinity with parameter $\alpha$ and it follows by e.g. \cite[Theorem 1.5.6(iii)]{Bingham-Goldie-Teugels-89} that there exists some $U\geq1$ and $A>0$ such that for all $u\geq U$, $\Re(\varphi_r(\mathrm{i}u))-\varphi_r(vu)\geq Au^{\alpha-\delta}$. Since $\Re(\varphi_r(\mathrm{i}u))$ and $\varphi_r(vu)$ are even and continuous functions, we can now conclude that the right-hand side of \eqref{fubini} is finite.

This allows us to use Fubini in \eqref{fourierinversion} for $v^\alpha<\cos\left(\frac{\pi\alpha}2\right)$ to get with the aid of \eqref{entire},
\begin{equation*}
\begin{split}
\sum_{n=0}^\infty \mu_n W^{(n+1)}(y)v^n = & \frac{1}{2\pi} \int_{-\infty}^\infty \sum_{n=0}^\infty \mu_n (\mathrm{i}u v)^n \mathrm{e}^{-\varphi_r(\mathrm{i}u)+\mathrm iuy}\mathrm{d}u \\
= & \frac{1}{2\pi} \int_{-\infty}^\infty   \mathrm{e}^{\varphi_r(\mathrm{i}vu)-\varphi_r(\mathrm{i}u)+\mathrm iuy}\mathrm{d}u  .
\end{split}
\end{equation*}
The right-hand side of the previous identity is the inverse Fourier transform of the Fourier transform of $\widehat P^{(-r)}_t(0,\mathrm{d}y)$ with $v=\mathrm{e}^{-rt}$. From earlier considerations, we see that this Fourier  transform is integrable when $t>0$. This implies in particular that for $t>0$, $\widehat P^{(-r)}_t(0,\mathrm{d}y)$ has a continuous density $\widehat p^{(-r)}_t(0,y)$, see e.g. the first line of Section 28 in Sato \cite{Sato-99}. Since $\widehat p^{(-r)}_t(0,y)$ is trivially integrable as well, we can now use the Fourier inversion theorem (see e.g.~\cite[Theorem 8.26]{folland}) to conclude
\begin{equation*}
\widehat p^{(-r)}_t(0,y)=\sum_{n=0}^\infty \mu_n W^{(n+1)}(y)\mathrm{e}^{-rnt}, \quad t>-\frac1{r\alpha}\log\cos\left(\frac{\pi\alpha}2\right).
\end{equation*}
The first identity follows then from the identity \eqref{asmussen}.

To prove the second identity, noting that  the right-hand side of \eqref{fubini} does not depend on $y$, we see that for any $f:\mathbb{R}\rightarrow\mathbb{R}$ measurable and integrable (i.e. $\int_{\mathbb{R}}|f(y)|\mathrm{d}y<\infty$), we can use Fubini to deduce
\begin{equation*}
\int_{\mathbb{R}} f(y)\widehat p^{(-r)}_t(0,y)\mathrm{d}y = \sum_{n=0}^\infty \mu_n\mathrm{e}^{-rnt} \int_{\mathbb{R}} f(y) W^{(n+1)}(y)\mathrm{d}y, \quad t>-\frac1{r\alpha}\log\cos\left(\frac{\pi\alpha}2\right).
\end{equation*}
Choosing $f(y)=\mathbf{1}_{\{y\leq x\}}$ and applying Theorem \ref{thm:1} shows \eqref{eq:main}. The last part follows by taking Laplace transforms on both sides, hereby making use of Fubini, \eqref{laplacetransform} and \eqref{eq:lt_w}.  Note that by p.190/191 of \cite{Sato-99},  $W^{(n+1)}(x)$ goes to zero as $x$ goes to $\pm\infty$ and hence $W^{(n+1)}$ is bounded on $(-\infty,\infty)$ which allows one to use Fubini.

\end{proof}

\bibliographystyle{plain}

config: arxiv


TITLE: The map $A \to A'$ is a linear isometry from $B(X,Y) \to B(Y',X')$.
QUESTION [2 upvotes]: Let $X$ and $Y$ be normed linear spaces. Show that the map $A \to A'$ is a linear isometry from $B(X,Y) \to B(Y',X')$.
$A'$ is the dual of $A$.
Need some hints for the problem.

REPLY [1 votes]: Recall that the norm on the dual of a normed vector space $X$ is given (for $L\in X^\star$) by:
$$ \Vert L \Vert_{X^\star} = \sup_{\Vert x \Vert_X \leq 1} \vert L(x) \vert.$$
Let
$$H: L(X,Y) \rightarrow L(Y^\star, X^\star), A \mapsto \left(A^\star: B \mapsto B\circ A\right).$$
One easily checks that this is a linear map. Using the definition of the operator norm and the norm on the dual of $X$, we get for $A\in L(X,Y)$
$$ \Vert H(A) \Vert_{L(Y^\star, X^\star)}
= \sup_{\Vert B \Vert_{Y^\star}\leq 1, \ \Vert x \Vert_X \leq 1} \vert (B \circ A)(x) \vert \leq \Vert A \Vert_{L(X,Y)} .$$
Now use the Hahn-Banach Theorem zu construct a suitable sequence $(B_n)_{n\geq 1} \subseteq Y^\star$ to prove
$$ \Vert H (A) \Vert_{L(Y^\star, X^\star)} \geq \Vert A \Vert_{L(X,Y)}$$
and conclude that $H$ is an isometry.


TITLE: Let X and Y be two random variables such that the vector (X,Y) is uniformly distributed over the region R = {(x,y)∈ℝ^2 : 0<x<y<1} . Find P(X+Y < 1)
QUESTION [0 upvotes]: Im having a hard time beginning because I don't know how to find the distribution function, and thus don't understand how to find the density function. I drew the graph and found the triangle. What do I do from there on?

REPLY [1 votes]: Let X and Y be two random variables such that the vector (X,Y) is uniformly distributed over the region $R = \{(x,y)\in\Bbb R^2 : 0<x<y<1\}$ . Find P(X+Y < 1)

I'm having a hard time beginning because I don't know how to find the distribution function, and thus don't understand how to find the density function.

If you wish to use it, you should find the probability density function first.
You don't strictly need it, but anyway...

I drew the graph and found the triangle.

Indeed, the region $R$ is a triangle of area $1/2$.  The random points are uniformly distributed over this.
The probability density function is therefore : $f_{X,Y}(x,y) = 2\cdot\mathbf 1_{0<x<y<1}$ $$f_{X,Y}(x,y) = \begin{cases}2&:& 0<x<y<1\\0&:&\textsf{otherwise}\end{cases}$$

What do I do from there on?

Draw the line $x+y=1$ on your graph.  Notice how it divides the triangle.  The ratio of areas will give the probability.
$$\mathsf P(X+Y<1) =\dfrac{\lvert R\cap\{(x,y)\in\Bbb R^2:x+y<1\}\rvert}{\lvert R \rvert}$$
Or you might do it the long way.  If you really want to:
$$\begin{align}\mathsf P(X+Y\lt 1)&=\int_0^1\int_0^{\min(y,1-y)}2~\mathrm d x~\mathrm d y\\[1ex]&=\int_0^{1/2}\int_0^y 2\mathsf d x\mathsf d y+\int_{1/2}^1\int_0^{1-y} 2\mathsf d x\mathsf d y\\[1ex]&~~\vdots\end{align}$$


TITLE: Rank-1 convexity is equivalent to the Legendre-Hadamard condition for a twice continuously differentiable function
QUESTION [0 upvotes]: Problem : Let $f:\mathbb{R}^{2\times2}\to\mathbb{R}$ be a twice continuously differentiable function and rank-$1$ convex function. Prove that $$\sum\limits_{i,j,k,l=1}^{2}\frac{\partial^2f(\xi)}{\partial\xi_{ij}\partial\xi_{kl}}\lambda_{i}\lambda_{k}\mu_{j}\mu_{l}\ge0$$ for every $\lambda,\mu\in\mathbb{R^2}$, $\xi=(\xi_{ij})_{1\leq i,j\leq2}\in\mathbb{R}^{2\times2}$. That is, the rank-$1$ convexity is equivalent to the Legendre-Hadamard condition.
I know that we have to set $\phi(t)=f(\xi+t\lambda\otimes\mu)$, where $t\in\mathbb{R}$, $\lambda=(\lambda_1,\lambda_2)$, $\mu=(\mu_1,\mu_2)$ and $\lambda\otimes\mu=\begin{pmatrix} \lambda_1\mu_1 & \lambda_1\mu_2\\
\lambda_2\mu_1 & \lambda_2\mu_2 \end{pmatrix}$ and use the convexity of $\phi$ to deduce the required inequality, that is $\phi^{\prime\prime}(t)\ge0$ but I am struggling to bring it in the required form. Also, how can I show that the converse is true?

REPLY [1 votes]: By the chain rule, we have:
$$\phi'(t)=\sum_{k,l=1}^2\frac{\partial f}{\partial\xi_{kl}}(\xi+t\lambda\otimes\mu)\lambda_k\mu_l,$$
$$\phi''(t)=\sum_{i,j,k,l=1}^2 \frac{\partial^2 f}{\partial\xi_{ij}\partial\xi_{kl}}(\xi+t\lambda\otimes\mu)\lambda_i\lambda_k\mu_j\mu_l.$$
Setting $t:=0$, we obtain the Legendre-Hadamard condition, since $\phi''(0)\geq 0$ by the rank-1 convexity of $f$.
Conversely, if the L.-H. condition is satisfied at $\xi\in\mathbb{R}^{2\times 2}$, then $\phi''(0)\geq 0$ (with any choice of $\lambda$ and $\mu$), hence $f$ is rank-1 convex at $\xi$. This can be done for any $\xi$, so $f$ is rank-1 convex.


TITLE: Algebra word problem : magician
QUESTION [0 upvotes]: A magician is playing mind tricks with the audience. He asks one member of the crowd to think of two numbers, write them on pieces of paper and throw them into his hat. Meanwhile, the magician also thinks of a number and drops it into the hat. He then waves his wand and a shiny paper comes out of the hat. The numbers the member of the audience picked are both less than the number the magician dropped into the hat. The first number is 40 percent less than the magician's number, while the second number is 50 percent less than the same. The shiny paper has the percentage comparison of the two numbers from the audience member. What is the percentage on the shiny paper?
I took the numbers as  $x$, $0.6x$ and $0.5x$, but by that logic we get the answer as 120.
Answer given is 20

REPLY [1 votes]: You seem to have done everything correctly other than the fact that you need to find the percentage difference of both numbers given by the audience and not there percentage ratio. So rather than doing, $\frac{0.6}{0.5}*100\%=120\%$ do $\frac{0.6-0.5}{0.5}*100\%=20\%$.
Hope this helps!!!


TITLE: Double-pinched washer surface
QUESTION [2 upvotes]: Take a flat, flexible ring, such as a rubber washer.  It has an inner radius for its inner circle, and an outer radius for its outer circle.  Fold it so that the inner circle of the washer joins up with itself in a half-circle, and the outer circle remains loose.  Then pinch the inner circle shut so that it becomes a line segment, still leaving the outer circle loose.

Is there a name for the resulting saddle-like surface?
A version of this surface can be plotted in polar coordinates as $f(r, \theta) = sin(2  \theta)$.
A 3d plot on sage cell (click and drag to move the figure):  here

REPLY [2 votes]: This is Plucker's conoid (with $n = 2$).

See also https://mathcurve.com/surfaces/plucker/plucker.shtml


TITLE: Calculating no. of solutions within some bounds
QUESTION [0 upvotes]: Find the total number of non-negative integral ordered triplets for-
$$x_1+x_2+x_3=11$$
under the bounds,

$x_1\in (2,6)$ and
$x_2 \in (3,7)$.

I was generally able to solve such problems involving bounds by introducing a new variable but that was only in case of a single boundary condition.
How do I proceed when there are two?

REPLY [1 votes]: I have figured out an approach using Multinomial Theorem-
We can find the coefficient of x^11 in (x^3+x^4+x^5)(x^4+x^5++x^6)(x^1+x^2+x^3+x^4).
This comes out to be 8.


TITLE: Quadratic reciprocity via generalized Fibonacci numbers?
QUESTION [60 upvotes]: This is a pet idea of mine which I thought I'd share.  Fix a prime $q$ congruent to $1 \bmod 4$ and define a sequence $F_n$ by $F_0 = 0, F_1 = 1$, and
$\displaystyle F_{n+2} = F_{n+1} + \frac{q-1}{4} F_n.$
Then $F_n = \frac{\alpha^n - \beta^n}{\alpha - \beta}$ where $\alpha, \beta$ are the two roots of $f(x) = x^2 - x - \frac{q-1}{4}$.  When $q = 5$ we recover the ordinary Fibonacci numbers.  The discriminant of $f(x)$ is $q$, so it splits $\bmod p$ if and only if $q$ is a quadratic residue $\bmod p$.
If $\left( \frac{q}{p} \right) = -1$, then the Frobenius morphism $x \mapsto x^p$ swaps $\alpha$ and $\beta$ (working over $\mathbb{F}_p$), hence $F_p \equiv -1 \bmod p$.  And if $\left( \frac{q}{p} \right) = 1$, then the Frobenius morphism fixes $\alpha$ and $\beta$, hence $F_p \equiv 1 \bmod p$.  In other words,
$\displaystyle F_p \equiv \left( \frac{q}{p} \right) \bmod p.$
Quadratic reciprocity in this case is equivalent to the statement that
$\displaystyle F_p \equiv \left( \frac{p}{q} \right) \bmod p.$
Question:  Does anyone have any ideas about how to prove this directly, thereby proving quadratic reciprocity in the case that $q \equiv 1 \bmod 4$?
My pet approach is to think of $F_p$ as counting the number of ways to tile a row of length $p-1$ by tiles of size $1$ and $2$, where there is one type of tile of size $1$ and $\frac{q-1}{4}$ types of tiles of size $2$.  The problem is that I don't see, say, an obvious action of the cyclic group $\mathbb{Z}/p\mathbb{Z}$ on this set.  Any ideas?

REPLY [25 votes]: The following paper seems to answer your question: P. T. Young, "Quadratic reciprocity via Lucas sequences", Fibonacci Quart. 33 (1995), no. 1, 78–81.
Here's its MathSciNet Review by A. Grytczuk:

Let $\{\gamma_n\}^\infty_{n=0}$ be a given Lucas sequence defined by $\gamma_0=0$, $\gamma_1=1$, $\gamma_{n+1}=\lambda \gamma_n+\mu \gamma_{n-1}$, $n\geq 1$, $\lambda, \mu\in{\bf Z}$, and let $q$ be an odd prime such that $D=(\frac{-1}q)q=\lambda^2+4\mu$. Then the author proves that there is a unique formal power series $\Phi$ with integer coefficients and constant term zero such that (1) $\sum^\infty_{n=1}\gamma_n\Phi^n(t)/n=\sum^\infty_{n=1}(\frac nq)t^n/n$ holds, where $(\frac nq)$ is the Legendre symbol.
     From this result follows the Gauss law of quadratic reciprocity in the following form: (2) $(\frac pq)=(\frac Dp)$, where $p$, $q$ are distinct odd primes and $D=(\frac{-1}q) q=\lambda^2+4\mu$.

Here's the direct link to the paper.


TITLE: Evaluate $\sum_{n=1}^\infty (-1)^{n+1}({1 - \cos(2n-1) \over 2n-1})$ using Fourier Series
QUESTION [2 upvotes]: Evaluate
$$\sum_{n=1}^\infty (-1)^{n+1} \left({1 - \cos(2n-1) \over 2n-1}\right)$$
Given
$f(x) = 1$ if $|x| < 1$ and $f(x) = 0$ if $1 \leqslant\ |x| \leqslant\ \pi$.
I calculated
$$Sf(x) = {1\over\pi} + {2\over\pi}\sum_{n=1}^\infty{{\sin(n)\over n} \cos(nx)}$$
And also calculated
$$\sum {{\sin(n) \over n} = {1\over 2}(\pi - 1)}$$
Which is asked to be used as an "inspiration" to solve the summation that I want.
But now I'm kinda lost on how to get to the sum in the title. Can someone at least point me in the right direction?
Edit:
From Yuriy S's suggestion I got
$${1\over\pi}(1 + 1\sum{\sin(1+x)+\sin(1-x)\over n})$$
But again, I don't know what $x$ to choose that will make me get to the desired sum.
Edit 2:
After A LOT of thinking about this series, I was finally able to answer it using the Fourier Series, simply by using the odd extension of the function, like so:
$f(x) = 1$ if $0 < x < 1$ and $f(x) = -1$ if $-1 < x < 0$.
This function will now give
$$Sf(x) = {2\over\pi}\sum_{n=1}^\infty{{1-\cos(n)\over n} \sin(nx)}$$
Now we have to force n to be an odd number, so we have
$$Sf(x) = {2\over\pi}\sum_{j=1}^\infty{{1-\cos(2j-1)\over 2j-1} \sin((2j-1)x)}$$
$$(n = 2j-1)$$
Choosing $x = {\pi\over 2}$ will give
$$Sf(x) = {2\over\pi}\sum_{j=1}^\infty{{1-\cos(2j-1)\over 2j-1} \ (-1)^{j+1}}$$
$Sf({\pi\over 2}) = 0$ will give

$$\sum_{j=1}^\infty{{1-\cos(2j-1)\over 2j-1} \ (-1)^{j+1}} = 0$$

Which is the same result as Felix Marin's solution to the problem.

REPLY [2 votes]: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

Besides the 'Fourier Technique', it's interesting to explore another ways !!!:

\begin{align}
&\bbox[10px,#ffd]{\ds{%
\sum_{n = 1}^{\infty}\pars{-1}^{n + 1}\,{1 - \cos\pars{2n - 1} \over 2n - 1}}} =
\Re\sum_{n = 1}^{\infty}\pars{-1}^{n + 1}\,
{1 - \expo{\pars{2n - 1}\ic} \over 2n - 1}
\\[5mm] = &\
\Re\bracks{\ic^{3}\sum_{n = 1}^{\infty}\ic^{2n - 1}\,
{1 - \expo{\pars{2n - 1}\ic} \over 2n - 1}} =
\Im\sum_{n = 1}^{\infty}\ic^{n}\,
{1 - \expo{n\ic} \over n}\,{1 - \pars{-}^{n} \over 2}
\\[5mm] = &\
\Im\sum_{n = 1}^{\infty}
{\ic^{n} - \pars{\ic\expo{\ic}}^{n} \over n}\,{1 - \pars{-1}^{n} \over 2}
=
\Im\sum_{n = 1}^{\infty}{\ic^{n}\over n} -
{1 \over 2}\Im\sum_{n = 1}^{\infty}{\pars{\ic\expo{\ic}}^{n} \over n} +
{1 \over 2}\Im\sum_{n = 1}^{\infty}{\pars{-\ic\expo{\ic}}^{n} \over n}
\\[5mm] = &\
-\,\Im\ln\pars{1 - \ic} + {1 \over 2}\,\Im\ln\pars{1 - \ic\expo{\ic}} -
{1 \over 2}\,\Im\ln\pars{1 + \ic\expo{\ic}}
\\[5mm] = &\
{\pi \over 4} + {1 \over 2}\,\Im\ln\pars{1 + \sin\pars{1} - \cos\pars{1}\ic} -
{1 \over 2}\,\Im\ln\pars{1 - \sin\pars{1} + \cos\pars{1}\ic}
\\[5mm] = &\
{\pi \over 4} + {1 \over 2}\,\arctan\pars{-\cos\pars{1} \over 1 + \sin\pars{1}} -
{1 \over 2}\,\arctan\pars{\cos\pars{1} \over 1 - \sin\pars{1}}
\\[5mm] = &\
{\pi \over 4} - {1 \over 2}\,\arctan\pars{\cos\pars{1} \over 1 + \sin\pars{1}} -
{1 \over 2}\,\mrm{arccot}\pars{1 - \sin\pars{1} \over \cos\pars{1}}
\\[5mm] = &\
{\pi \over 4}\
\underbrace{- {1 \over 2}\,\arctan\pars{\cos\pars{1} \over 1 + \sin\pars{1}} -
{1 \over 2}\,\mrm{arccot}\pars{\cos\pars{1} \over 1 + \sin\pars{1}}}
_{\ds{-\,{\pi \over 4}\quad\mrm{because}\quad
{\cos\pars{1} \over 1 + \sin\pars{1}} > 0}}\ =\ \bbx{\large 0}
\end{align}


TITLE: Why does the light from stars / satellites tremble?
QUESTION [7 upvotes]: Clear Sky.
1 ---> Light from stars visible.
2 ---> Light from satellites visible.
3 ---> Light from airplanes (passing by) visible.
Apart from the change in color, the light from 1 and 2 seems like its trembling but not 3.
Why is that so?

REPLY [13 votes]: The light from stars comes through the atmosphere and gets distorted like light passing through a pool. Stars are particularly prone to this effect because they are single points and the light only follows one path to your eye.
Planets don't twinkle as much because they are tiny circles and the light from either edge of the planet travels slightly different paths to your eye.
So for a planet to twinkle multiple paths would have to be blocked at once which is less likely.
Satellites are also less prone to twinkling because they are not points. Planes on the other hand will never twinkle because they are large objects whose light doesn't even need to pass through the entire atmosphere.


TITLE: Is this closed-form summation a special case of known Lerch zeta function formulas?
QUESTION [6 upvotes]: With some Poisson summation manipulations (credit: Michał Pacholski) I have convinced myself of a closed form expression for this conditionally convergent series:
$$\sum_{n=-\infty}^\infty \frac{e^{in\alpha}}{z+n}=\frac{2\pi  i}{e^{i\alpha z}-e^{i(\alpha-2\pi) z}},\;\;\alpha\in(0,2 \pi),\;\;z\in\mathbb{C}\backslash\mathbb{Z}.$$
Mathematica returns a Lerch zeta function (more precisely: a sum of two Hurwitz-Lerch transcendents), without further simplification. The above formula implies for the Lerch zeta function $L(a,b,s)=\sum_{n=0}^\infty \frac{e^{2\pi ina}}{(n+b)^s}$ the reflection formula
$$L(a,b,1)-L(-a,-b,1)=\frac{1}{b}+\frac{2\pi  i}{e^{2\pi iab}-e^{2\pi i(a-1) b}},\;\;a\in(0,1),\;\;b\in\mathbb{C}\backslash\mathbb{Z},$$
or more generally for $s\in\mathbb{N}$
$$L(a,b,s)+(-1)^{s}L(-a,-b,s)=\frac{1}{b^{s}}+\frac{(-1)^{s-1}}{(s-1)!}\frac{d^{s-1}}{db^{s-1}}\left(\frac{2\pi i}{e^{2\pi iab}-e^{2\pi i(a-1) b}}\right).$$
I would be pleased to learn if this is a known result for this special function, or a reference to a derivation in the literature.

REPLY [1 votes]: Let me suggest a more pedestrian approach than the elegant solution of Christian Remling but which is arguably easier to come up with: contour integration.
To get the sum over $\mathbb{Z}$ from a contour integral we need a function which has poles at all integers. And thankfully we know such a function - it is $\frac{1}{\sin(\pi z)}$ ! More precisely, the function
$$f(z)=\frac{\pi e^{iz(\alpha - \pi)}}{(z+w)\sin(\pi z)}$$
has residue $\frac{e^{in\alpha}}{w+n}$ at the point $n$ (I switched $z$ from the OP to $w$ because I want $z$ be a free complex variable). Note that $-\pi$ in the exponent is because residues of $\frac{1}{\sin(\pi z)}$ alternate between $\frac{1}{\pi}$ and $\frac{-1}{\pi}$.
Now let us consider a circle of radius $R$ centred at the origin with, say, $R = N+\frac{1}{2}$ for integer $N$ just to be far from the poles. Since $-\pi < \alpha - \pi < \pi$ we can see that the integral tends to $0$ (exponent $e^{iz(\alpha - \pi)}$ is being killed by the growth of $\sin (\pi z)$) and therefore sum of the residues is zero. On the other hand there's only one more residue that we have: $z = -w$ and the value at it gives us precisely the result.