text:
\begin{document}

\title{On the excursions of reflected local time \\ processes and stochastic fluid queues}

\authorone[Heriot-Watt University]{Takis Konstantopoulos}
\vspace{-0.8cm}
\addressone{
School of Mathematical Sciences, Heriot-Watt University, Edinburgh, EH14 4AS, UK}
\authortwo[University of Bath]{Andreas E. Kyprianou}
\addresstwo{Department of Mathematical Sciences, University of Bath, Claverton Down, Bath, BA2 7AY, UK}

\vspace{-0.8cm}
\authorthree[ \AA bo Akademi
University]{Paavo Salminen}
\addressthree{Department of Mathematics, \AA bo Akademi
University, Turku, FIN-20500, Finland}
\renewcommand{\thefootnote}{\arabic{footnote}}

\begin{abstract}
This paper extends previous work by the authors.
We consider the local time process of a strong Markov process, add negative
drift, and reflect it \`a la Skorokhod. The resulting process is used to model a fluid queue.  We derive an expression
for the joint law of the duration of an excursion, the maximum
value of the process on it, and the time distance between
successive excursions. We work with a properly constructed
stationary version of the process.
Examples are also given in the paper.

\keywords{L\'evy process, local time, Skorokhod reflection, stationary
process}

\ams{60G51, 60G10}{90B15}

\end{abstract}
\section{Introduction}
\label{Intro}
Consider a stationary strong Markov process $X=(X_t, t \in \R)$,
defined on some filtered probability space
$(\Omega, \FF, P, (\FF_t, t \in \R))$,
with values in $\R_+$, a.s.\ \cadlag paths, and adapted to $(\FF_t)$.

In this paper, the local time $L$ of the process $X$ at $x=0$
is considered an $(\FF_t)$-adapted stationary random measure
that regenerates jointly with $X$ at every (stopping)
time that $X$ hits $0$. More precisely:
\begin{itemize}\em
\item[{\bf (A1)}]
$L$ assigns a nonnegative random
variable $L(B,\omega)$ to each $B \in \BB(\R)$ such that $L(\cdot, \omega)$
is a Radon measure for each $\omega \in \Omega$.
\item[{\bf (A2)}]
For any a.s.\ finite $(\FF_t)$-stopping time $T$ at which $X_T=0$,
the process $((X_{T+t}, L(T, T+t)), t \ge 0)$ is independent of $\FF_T$.
\end{itemize}
We take the broader perspective with regard to the process $L$ and we allow for the case that it is a local time of an irregular point (in which case $L$ has discontinuous paths) as well as the case that $0$ is a sticky point
(in which case $L$ is absolutely continuous with respect to the Lebesgue measure with density $c\1(X_t = 0)$ for some $c>0$).
We refer to \cite[Chap.\ IV]{BER} (in particular Corollary 6), \cite[Chap.\ 6]{K}, and \cite[\S V.3]{BG}
for further discussion.

For each $s \in \R$ define the inverse local time
process with respect to $t $ by
\begin{equation}
\label{Linv}
L_{s; u}^{-1} := \inf\{ t>0:~ L[s, s+t] > u \},
\quad u \ge 0.
\end{equation}

What is important is that, owing to this definition, the inverse
of the cumulative local time is a L\'evy process in the following sense:
\begin{lemma}
\label{basic}
If $L$ is continuous then for every a.s.\ finite $(\FF_t)$-stopping time
$T$ such that $X_T=0$, the process
$(L^{-1}_{T;u}, u \ge 0)$ is a subordinator
with $L^{-1}_{T;0}=0$.
\end{lemma}
If $L$ is not continuous, that is to say if $0$ is an irregular point for $X$, then this Lemma is taken as an
additional requirement to the definition of $L$. This is easily arranged by choosing $L$ to be a modification of
the counting process on $\ZZZ$, the discrete set of times that $X$ visits $0$, so that the inverse is a subordinator.
To do this, we assign, to each element of $\ZZZ$,
an i.i.d.,  unit-mean exponentially distributed weight. Then  let
the local time on an interval $I$ to be the sum of all the weights of
the points of $\ZZZ$ in $I$.

We summarise this as an assumption, in addition to (A1)-(A2) above:
\begin{enumerate}
\em
\item[\bf (A3)]
If $L$ is discontinuous then we require that
for every a.s.\ finite $(\FF_t)$-stopping time
$T$ such that $X_T=0$, the process
$(L^{-1}_{T;u}, u \ge 0)$ is a subordinator.
\end{enumerate}
We will also need the following assumption:
\begin{enumerate}
\em
\item[\bf (A4)]
The stationary random measure $L$ has finite rate not
exceeding $1$, i.e.
\[
EL(0,t) = \mu\,t,
\]
where $0<\mu<1.$
\end{enumerate}

Then, as in \cite{MNS}, \cite{KoSa}, \cite{Si}, and \cite{KKSS},  we define a
stationary process $Q=(Q_t, t \in\mathbb{R})$ by
\begin{equation}
\label{Q}
Q_t = \sup_{-\infty < s \le t} \left\{ L(s,t]-(t-s)\right\},
\quad t \in \R.
\end{equation}
Furthermore, $Q$ is ergodic (its invariant $\sigma$-field is trivial.)
Notice that $Q$ also satisfies, pathwise,
\begin{align}
Q_t &= Q_s + L(s,t]-(t-s)  - \inf_{s \le r \le t}
\big( Q_s + L(s,r]-(r-s) \big) \wedge 0,
\nonumber
\\
&= \sup_{s \le r \le t} (L(r,t]-(t-r)) \vee
(Q_s + L(s,t]-(t-s))
\label{sss}
\end{align}
for all $-\infty < s < t < \infty$.
It is worth recalling \cite{KKSS} that if we consider \eqref{sss}
as a fixed point equation for $Q$ then process defined by \eqref{Q}
is the unique stationary and ergodic solution of \eqref{sss}.
A typical sample path of $Q$ is depicted in Figure \ref{typical} below.
It consists of isolated excursions away from zero (also called
``busy periods''), followed by intervals of time at which $Q$ stays
at zero (called ``idle periods''). In this respect, the process $Q$ is thought of as the workload in a stochastic fluid queue. Amongst other things
in \cite{KoSa}, \cite{Si}, and \cite{KKSS}, expressions are  derived
for the marginal distribution of $Q$ and the Laplace transform of the duration of a typical
idle and busy periods.

In this paper, we shall derive an expression for
the joint law of three random variables: the duration of a busy period,
the duration of an idle period, and the maximum of $Q$ over a busy period.

It is assumed, throughout, that $(\Omega, \FF, P)$ is endowed with
a $P$-preserving measurable flow $\theta_t:\Omega\to \Omega$, $t \in \R$,
with a measurable inverse $\theta_t^{-1} = \theta_{-t}$.
All stationary random processes and measures can be constructed on
$\Omega$ in such a way that the flow commutes with the natural shift, e.g.,
$Q_t (\theta_s \comp \omega) = Q_{t+s}(\omega)$,
and $L(B, \theta_s\omega) = L(B+s, \omega)$, for all
$s,t \in \R$, Borel sets $B \subset \R$, and $\omega \in \Omega$.
The flow will be explicitly used  in
Section \ref{cycleform} to obtain distributions conditional on
observing a positive (or a zero) value of $Q_0$.

\section{A closer look at the reflected process}\label{closerlook}
Consider now any a.s.\ finite  $(\FF_t)$-stopping time $T$,
such that $X_T=0$.
Then $(L^{-1}_{T; t}, ~ t \ge 0)$ is a subordinator starting from zero
(owing to Lemma \ref{basic} or assumption (A3)) with law that does not
depend on $T$. It turns out that the process of interest is $\Lambda_T = \{\Lambda_{T,t}: t\geq 0\}$, where
\begin{equation}
\label{LLL}
\Lambda_{T;t} = t - L^{-1}_{T; t} , \quad t\geq 0.
\end{equation}
Note that, irrespective of $T$, the process $\Lambda_T$ has
the law of the same bounded variation, spectrally negative L\'evy process which
is issued from the origin at time zero.
By (A4), $L$ has rate $\mu < 1$;
hence $E \Lambda_{T;1} = 1-\frac{1}{\mu}< 0$.  Since $L^{-1}_{T;t}$ is a subordinator, it has
a well-defined, possibly nonzero, drift.

If this drift is larger than or equal to unity then $-\Lambda_T$
is a subordinator and, as it will turn out, this is a trivial case.

We therefore  assume in the sequel that the drift of $L^{-1}_T$ is less
than unity or, equivalently, that
\begin{enumerate}
\em
\item[\bf (A5)]
The drift $\delta_\Lambda$ of the process $\Lambda$ defined by
\eqref{LLL}
is strictly positive.
\end{enumerate}

Under this assumption,
the point $0$ is irregular for $(-\infty,0)$ for $\Lambda_T$
(this follows as a standard results for bounded variation spectrally
negative L\'evy processes, see Bertoin \cite[Chap.\ VII]{BER}.)

In addition, under (A5), it is clear that the time taken for $\Lambda_T$ to
first enter $(-\infty, 0)$ is almost surely strictly positive.
It will be shown below (Lemma \ref{BI}) that this implies that the
excursions of the process $Q$, i.e.~the busy periods,
have strictly positive Lebesgue length with probability one.
\begin{figure}[h]
\begin{center}
\includegraphics[width=10cm]{figlambda}
\caption{\em The construction of the process $(\Lambda_{T;t},~t \ge 0)$
and related processes, assuming that $T=0$. Note that $\Lambda$ may have
countably many jumps on finite intervals.}
\label{lambda}
\end{center}
\end{figure}
It can be intuitively seen, via a geometric argument
involving the reflection of the space-time path of
$\Lambda_T$ about the diagonal (see Figure \ref{lambda}), that the time taken for $\Lambda_T$ to
first enter $(-\infty, 0)$ is almost surely equal to the length of
the excursion of $Q$ started at time $T$.

In this light, note also that $\Lambda_T$ cannot creep downwards
because it is spectrally negative with paths of bounded variation
(cf. Bertoin \cite[Chap.\ VII]{BER}).
Hence the overshoot at first passage of $\Lambda_T$ into $(-\infty, 0)$
is almost surely strictly positive. It will turn out (Lemma \ref{basic})
that this overshoot agrees with the idle period following the aforementioned
excursion of $Q$.

The above analysis implies that, on finite intervals of time, $Q$ has finitely many excursions (busy periods)
separated by positive-length idle periods.
Denote by
\[
\cdots < g(-1) < g(0) < g(1) < g(2) < \cdots
\]
the beginnings of the idle periods and by
\[
\cdots < d(-1) < d(0) < d(1) < d(2) < \cdots
\]
their ends, see Figure \ref{typical}.
We choose the indexing so that $g(0) \le 0 < g(1)$.
Let $N_g$ (respectively, $N_d$) be the point process with points
$\{g(n): n \in \Z\}$ (respectively, $\{d(n): n \in \Z\}$).
As $Q$ is a stationary process,
$N_g$ and $N_d$ are jointly stationary with finite, nonzero, intensity
\cite{KKSS} denoted by $\lambda$ (an expression for which
is given by \eqref{commonrate} and is derived in \S \ref{Rates} below).

\begin{figure}[h]

\begin{center}
\includegraphics[width=\columnwidth]{figpath}

\end{center}
\caption{\em The definition of $g(n)$ and $d(n)$. By convention,
the origin of time is between $g(0)$ and $g(1)$, under the original
measure $P$. Under $P_d$, the origin of time is at $d(0)$. Under $P_g$,
the origin of time is at $g(0)$. The random variable $Q^*$ is the maximum
deviation from $0$ of $Q$ within the typical busy period.}
\label{typical}
\end{figure}
Corresponding to point processes $N_g$, $N_d$ we have the
Palm probabilities $P_g$, $P_d$, respectively.
Let us consider $Q$ under the measure $P_d$. Then
$P_d(d(0)=0)=1$, i.e.\ the origin of time is placed at
the beginning of a busy period.
By the strong Markov property, the ``cycles''
\[
\CC_n:= \left\{Q_t:~ d(n) \le t < d(n+1)\right\}, \quad n \in \Z,
\]
are i.i.d.\ under measure $P_d$.
In particular,
the pairs of random variables
\[
\big(g(n+1)-d(n),~ d(n+1)-g(n+1)\big), \quad n \in \Z,
\]
are i.i.d.\ under $P_d$.
Consider the triple
\begin{equation}
\label{triple}
(B, I, Q^*) := \left(g(1)-d(0),~ d(1)-g(1),~ \sup_{d(0) < t < g(1)} Q_t \right),
\end{equation}
which is a function of $\CC_0$.
We are primarily interested in the $P_d$-law of $(B,I,Q^*)$
Since, under $P_d$, the origin of time is placed at $d(0)$,
we interpret $B, I,Q^*$ as the typical busy period, the typical
idle period, and the maximum value of $Q$ over a typical
busy period, respectively.

The next lemma is proved in \cite{KKSS}:
\begin{lemma}
\label{Dd}
Let $D =\inf\{t >0:~ X_t =0\}$ and $d=\inf\{t >0:~ Q_t >0\}$.
Then $d=D$ a.s.\ on $\{Q =0\}$.
\end{lemma}

We now obtain an alternative expression for $B= g(1) -d(0)$
and $I = d(1)-g(1)$
in terms of the inverse local time.
\begin{lemma} We have that
\label{BI}
\begin{align}
B& = g(1)-d(0) = \inf\{u>0: ~ L^{-1}_{d(0);u} > u \},
\label{B}
\\
B+I &= d(1)-d(0) =  L^{-1}_{d(0); g(1)-d(0)}.
\label{I}
\end{align}
\end{lemma}
\begin{proof}
Since $d(0)$ is the end of an idle period (and the beginning
of a busy period), we have $Q_{d(0)-}=0$.
Using then expression (\ref{sss}) we obtain
\[
Q_t =  L[d(0), t] - (t-d(0)), \quad d(0) \le t < g(1),
\]
which gives
\[
B= g(1) -d(0)
= \inf\{t > 0:~ L[d(0), d(0)+t] =t\}.
\]
Consider now $L^{-1}_{d(0);u}$, defined by \eqref{Linv}.
By Lemma \ref{Dd}, $d(0)$ is a point of increase of the function
$t \mapsto L[d(0),d(0)+t]$.
Hence $g(1) > d(0)$.
Also, when $L[d(0),d(0)+t]-t$ decreases, it does so continuously.
Therefore,
\[
B = \inf\{t > 0:~ L[d(0),d(0)+t]<t\}.
\]
Notice also that, for all $t,x > 0$,
\[
L[d(0),d(0)+t] < x \iff t< L^{-1}_{d(0); x-},
\]
where
$L^{-1}_{d(0); x-} = \lim_{\epsilon \downarrow 0}L^{-1}_{d(0); x-\epsilon}$.
It follows that,
\begin{align*}
B &= \inf\{t >0:~ t< L^{-1}_{d(0); t-}\}
\\
&= \inf\{t >0:~ L^{-1}_{d(0); t} > t\},
\end{align*}
by the right continuity of $t \mapsto L^{-1}_{d(0); t}$.
To prove the expression for $B+I$, notice that $L$ does
not charge the interval $[g(1), d(1))$ because, by definition,

$Q$ is zero for all $t$ in this interval.
\QED
\end{proof}

Henceforth it will be convenient to work with the process $\Lambda= (\Lambda_t, t\geq 0)$ where
\[
\Lambda_t := t - L^{-1}_{d(0);t}, \quad t \ge 0.
\]
Note also that $d(0)$ is an $(\FF_t)$-stopping time at which $X$ takes the value 0 and hence in
our earlier notation $\Lambda_t = \Lambda_{d(0);t}$.

{From} the expression (\ref{B}), and as discussed in the introduction of  Section \ref{closerlook}, we see that $B$ is simply the first
time at which $\Lambda$ enters $(-\infty, 0)$,
\begin{equation}
\label{Bnew}
B= \inf\{t>0:~ \Lambda_t < 0\},
\end{equation}
which is necessarily strictly positive thanks to the irregularity of $0$ for $(-\infty,0)$ for $\Lambda$.
{From} \eqref{B} and \eqref{I} we see that
\begin{equation}
\label{Inew}
I= L^{-1}_{d(0);g(1)}-B = L^{-1}_{d(0);B} -B
= - \Lambda_B,
\end{equation}
i.e.\ $I$ is, in absolute value, equal to the value of $\Lambda$ at the first time it becomes
negative. Again we recall from the discussion at the beginning of Section \ref{closerlook}
that  $\Lambda$ cannot creep downwards and hence
$
 I >0
$
almost surely.

Consider now the random variable $Q^*=\sup_{d(0) < t < g(1)} Q_t$.
If we define
\begin{equation}
\label{tau}
\tau_x := \inf\{t>0:~ \Lambda_t > x\}
= \inf\{t>0:~ \Lambda_t = x\}
\end{equation}
we immediately see that
\begin{equation}
\label{Mnew}
\{Q^*< x\} = \{ B< \tau_x \}.
\end{equation}

\section{The triple law}

Recall that $P_d$ is the Palm probability with respect to the point
process $\{d(n), n \in \Z\}$.
The function
\[
H(\alpha, \beta, x)
= E_d\left[e^{-\alpha B - \beta I} \1(Q^* \le x) \right]
\]
characterizes the joint law of the triple $(B,I,Q^*)$ under $P_d$.
Since $P_d(d(0)=0)=1$, we have that
\begin{equation}
\label{LAMBDA}
\text{$\Lambda_t = t-L^{-1}_{0;t}$,
with $\Lambda_0 =0$, $P_d$-a.s.}
\end{equation}
Recalling the expressions \eqref{Bnew}, \eqref{Inew} and \eqref{Mnew}
for $B$, $I$ and $Q^*$, respectively, we write
\begin{equation}
\label{H}
H(\alpha, \beta, x)
= E_d\left[e^{-\alpha B +\beta\Lambda_B } \1(B \le \tau_x)\right].
\end{equation}

Since our primary object is the process $\Lambda$ defined in \eqref{LAMBDA},
and in view of \eqref{LLL} and \eqref{H}, it makes sense to consider
the process on its canonical probability space and denote its law
by $\P_0$.
Then
\begin{equation}
\label{HH}
H(\alpha, \beta, x)
= \E_0 \left[e^{-\alpha B +\beta \Lambda_B} \1(B < \tau_x)\right].
\end{equation}
The latter function may now be expressed in terms of so-called
scale functions for spectrally negative L\'evy processes.
To define the latter, let
\[
\psi_\Lambda(\theta): = \log \E_0  e^{\theta \Lambda_1}, \, \theta\geq 0,
\]
 be the Laplace exponent of $\Lambda$ under $\mathbb{P}_0 $.
Then the, so-called, $q$-scale function for
$(\Lambda, \mathbb{P}_0 )$,
denoted by $W^{(q)}(x)$,
satisfies $W^{(q)}(x) = 0$ for $x<0$ and
on $[0,\infty)$ it is the unique continuous
(right continuous at the origin) monotone increasing function whose Laplace
transform is given by
\begin{equation}
\label{alpha}
\int_0 ^\infty e^{-\theta x} W^{(q)}(x) dx =
\frac{1}{\psi_\Lambda(\theta)-q}, \quad \text{for } \beta>\Phi_\Lambda(q),
\end{equation}
where
\[
 \Phi_\Lambda(q) = \sup\{\theta\geq 0 : \psi_\Lambda(\theta) = q\}
\]
is the right inverse of $\psi_\Lambda$. (See for example the discussion in Chapter 9 of \cite{K}).

\begin{theorem}
Let $\Lambda$ be the process defined by \eqref{LAMBDA}, $B$ its first entry time to
$(-\infty, 0)$ as in \eqref{Bnew}, and $\tau_x$ the first hitting time of $\{x\}$ as
in \eqref{tau}.
For $\alpha, \beta, x\geq 0$ we have
\begin{eqnarray}
\label{Hformula}
\nonumber
H(\alpha, \beta, x)
&=& \E_0 \left[e^{-\alpha B +\beta \Lambda_B} \1(B< \tau_x)\right]\\
&=& 1 - \frac{1}{\delta_\Lambda}
~
\frac{1 + (\alpha-\psi_\Lambda(\beta))
\int_0 ^x e^{-\beta y} W^{(\alpha)}(y) dy}
{e^{-\beta x} W^{(\alpha)}(x)}.
\end{eqnarray}
\end{theorem}
\begin{proof}
Let $\GG_t:= \sigma(\Lambda_s, s \le t)$ and define,
for all $\beta \ge 0$, the exponential $(\GG_t)$-martingale
\[

M^\beta_t := e^{\beta \Lambda_t - \psi_\Lambda(\beta)t}, \quad t \ge 0.
\]
Let, on the canonical space of $\Lambda$, $\P^\beta_0$ be a probability
measure, absolutely continuous with respect to $\P_0$ on $\GG_t$ for
each $t$, with Radon-Nikod\'ym derivative
\[
\left. \frac{d\P_0 ^\beta}{d\P_0 } \right|_{\GG_t} := M^\beta_t.
\]
Notice that $\Lambda$ is still a L\'evy process under $\P^\beta_0$

with Laplace exponent
\begin{equation}
\label{psil}
\psi_\Lambda^\beta(\theta) =\log \E^\beta_0 e^{\theta \Lambda_1}
= \psi_\Lambda(\beta+\theta)-\psi_\Lambda(\beta).
\end{equation}
It is straightforward to check from the above formula that, under $\P^\beta_0$,
$\Lambda$ is spectrally negative, with bounded variation paths
and drift coefficient equal to $\delta_\Lambda$.
Since on the stopped $\sigma$-field $\GG_B$ we have
$(d\P_0^\beta/d\P_0)\big|_{\GG_B} = M^\beta_B$,
we may substitute
\[
e^{\beta \Lambda_B} = M^\beta_B e^{\psi_\Lambda(\beta) B}
\]
in the equation \eqref{HH} for $H$ to obtain
\begin{align*}
H(\alpha, \beta, x)
= \E_0  \left[M^\beta_B e^{\psi_\Lambda(\beta) B} e^{-\alpha B}
\1(B < \tau_x) \right]
= \E_0 ^\beta \left[ e^{-(\alpha-\psi_\Lambda(\beta)) B}
\1(B < \tau_x) \right].
\end{align*}
Let
\[
q := \alpha-\psi_\Lambda(\beta),
\]
and assume that $q \ge 0$.
It follows from \cite[Thm,\ 8.1(iii)]{K} that
\begin{equation}
\label{Hformula2}
H(\alpha, \beta, x)
=\E_0 ^\beta \left[ e^{-qB} \1(B < \tau_x) \right]
= Z^{(q)}_\beta (0) - Z^{(q)}_\beta (x) \frac{W^{(q)}_\beta (0)}{W^{(q)}_\beta (x)},
\end{equation}
where $W^{(q)}_\beta$ is the $q$-scale function for $(\Lambda, \P^\beta_0)$
and $Z^{(q)}_\beta $ is given by
\[
Z^{(q)}_\beta (x) = 1+q \int_0 ^x W^{(q)}_\beta (t) dt.
\]
It is easy to see \cite[Lemma 8.4]{K}
that the Laplace transform of $W^{(q)}_\beta(\cdot)$
is the Laplace transform of $W^{(q)}(\cdot)$ shifted by $\beta$
and this ensures that

\begin{equation}
 W^{(q)}_\beta (x) = e^{-\beta x} W^{(\alpha)}(x).
\label{Wq}
 \end{equation}
Moreover, since $\Lambda$ still has drift coefficient $\delta_\Lambda$ under $\P^\beta_0$,
\cite[Lemma 8.6]{K} tells us that, irrespective of the value of $q$ and $\beta$,
$W^{(q)}_\beta(0) = 1/\delta_\Lambda$.
Putting the pieces together, this gives us the desired expression for $\alpha \geq\psi_\Lambda(\beta)$.
However \cite[Lemma 8.3]{K}, since $W^{(q)}(x)$ is analytic in $q$,
the  condition on $\alpha$ can be relaxed to $\alpha\geq 0$
by using a straightforward analytic extension argument.
\QED
\end{proof}

In view of \eqref{Bnew}, \eqref{Inew}, \eqref{Mnew}, and \eqref{H},
we get the following corollary.
\begin{corollary}[Joint law of typical $B$, $I$ and $Q^*$]
Assume that (A1)--(A5) hold. Then
the joint law of the length $B$ of a typical busy period,
the length $I$ of a typical idle period, and the maximum
$Q^*$ of $Q$ over the typical busy period is expressed by
the formula
\begin{equation}
\label{triplelaw}
E_d[e^{-\alpha B - \beta I} \1(Q^* \le x)]
=
1 - \frac{1}{\delta_\Lambda}
~
\frac{1 + (\alpha-\psi_\Lambda(\beta))
\int_0 ^x e^{-\beta y} W^{(\alpha)}(y) dy}
{e^{-\beta x} W^{(\alpha)}(x)}
\end{equation}
where $\alpha, \beta, x\geq 0$.
\end{corollary}

\section{Marginal distributions}
Clearly, formula \eqref{triplelaw} can be used to extract more detailed information
about typical  behaviour of $Q$.

Let us first derive the distribution (Laplace transform) of the
pair $(B,I)$ under the measure $P_d$. We have
\[
E_d [e^{-\alpha B} e^{-\beta I}]
= \E_0  [ e^{-\alpha B} e^{\beta \Lambda_B}]
= \lim_{x \to \infty} H(\alpha, \beta, x).
\]
To derive the limit, let us temporarily assume that $q = \alpha- \psi_\Lambda(\beta)>0$ and $\beta\geq 0$. Consider \eqref{Hformula} in the
form \eqref{Hformula2}
and use the limiting result
\[
\lim_{x \to \infty}
\frac{Z^{(q)}_\beta (x)}{W^{(q)}_\beta (x)}
= \frac{q}{\Phi^\beta_\Lambda(q)},
\]
from \cite[Exercise 8.5]{K}
where the function $\Phi^\beta_\Lambda$ is the right inverse of $\psi_\Lambda^\beta$.
That is to say
\begin{align*}
\Phi_\Lambda^\beta(q) &= \sup\{\theta \geq 0 : \psi_\Lambda^\beta(\theta) = q\}\\
&=\sup\{\theta \geq 0 : \psi_\Lambda^\beta(\theta) = \alpha - \psi_\Lambda(\beta)\}\\
&=\sup\{\theta \geq 0 : \psi_\Lambda(\theta+\beta) = \alpha \}\\
&= \Phi_\Lambda(\alpha)-\beta.
\end{align*}
This gives
\begin{equation}
E_d [e^{-\alpha B} e^{-\beta I}]
=
1-\frac{1}{\delta_\Lambda}
\frac{\alpha-\psi_\Lambda(\beta)}{\Phi_\Lambda(\alpha) -\beta}.
\label{no-max}
\end{equation}
To remove the restriction that $\alpha> \psi_\Lambda(\beta)$ in (\ref{no-max}) and replace it instead by just $\alpha\geq 0$,
one may again proceed with an argument involving analytical extension taking care to note for the case that $\alpha  =\psi_\Lambda(\beta)$,
\[
 \lim_{|\alpha - \psi_\Lambda(\beta)|\rightarrow 0} \frac{\alpha-\psi_\Lambda(\beta)}{\Phi_\Lambda(\alpha) -\beta}=
\lim_{|\alpha - \psi_\Lambda(\beta)|\rightarrow 0} \frac{\psi^\beta_\Lambda(\Phi_\Lambda(\alpha) -\beta)}{\Phi_\Lambda(\alpha) -\beta}
= {\psi_\Lambda^{\beta}}'(0+) = \psi_\Lambda'(\beta).
\]

\subsection{Busy period}
Letting $\beta=0$ in (\ref{no-max}), we find the $P_d$-law of $B$. That is to say,
\[
E_d[e^{-\alpha B}] = 1-\frac{1}{\delta_\Lambda}
\frac{\alpha}{\Phi_\Lambda(\alpha)}.
\]
This formula is consistent with the result of
\cite[Prop.\ 8]{KKSS} and, moreover,
we see that the mean duration of the busy period is given by
\begin{equation}
\label{busym}
E_d[B] = \frac{1}{\delta_\Lambda \Phi_\Lambda(0)}.
\end{equation}

\subsection{Idle period}

To find the $P_d$-law of $I$ we need to set $\alpha=0$.
Recall however from the beginning of Section \ref{closerlook} that $E_d(\Lambda_1) <0$. This implies that $\Phi_\Lambda(0)>0$ and hence
we have
\begin{equation}
\label{typicalI}
E_d[e^{-\beta I}] = 1-\frac{1}{\delta_\Lambda}
\frac{\psi_\Lambda(\beta)}{\beta-\Phi_\Lambda(0)}.
\end{equation}
It follows that
the mean idle period is thus equal to
\begin{equation}
E_d[I] = \frac{-\psi_\Lambda'(0+)}{\delta_\Lambda \Phi_\Lambda(0)},
\label{meanidle}
\end{equation}
where $\psi_\Lambda'(0+) = E_d(\Lambda_1)<0$.

\subsection{Rates}
\label{Rates}
A cycle of the process $Q$ is defined as the interval from the beginning of
a busy period until the beginning of the next busy period.
We therefore have
\begin{equation}
\label{meancycle}
\text{mean cycle length = } E_d[B+I] = \frac{1-\psi_\Lambda'(0+)}
{\delta_\Lambda \Phi_\Lambda(0)}.
\end{equation}
We can express the common rate, $\lambda$, of $N_g$ and $N_d$ as
the inverse of the mean cycle length:
\begin{equation}
\label{commonrate}
\lambda := E N_d(0,1) = E N_g(0,1) = \frac{1}{ E_d[B+I]}
= \frac{\delta_\Lambda \Phi_\Lambda(0)}{1-\psi_\Lambda'(0+)}.
\end{equation}

\subsection{The maximum over a busy period}
We now derive the $P_d$-distribution of $Q^*$. Letting $\alpha=\beta=0$ in
\eqref{Hformula} we obtain
\[
P_d(Q^* \le x) = 1 -\frac{1}{\delta_\Lambda W(x)},
\]
where $W(x) \equiv W^{(0)}(x)$ is defined through its
Laplace transform
\begin{equation}
\label{Wlap}
\int_0 ^\infty e^{-\theta x} W(x) dx
= \frac{1}{\psi_\Lambda(\theta)}, \quad \text{for } \theta> \Phi_\Lambda(0).
\end{equation}
An immediate observation is that $\lim_{x \to 0} P_d (Q^* \le x) =0$,
since

$W(0) = \lim_{\theta \to 0} \theta/\psi_\Lambda(\theta)=1/\delta_\Lambda$.
So under $P_d$, the random variable $Q^*$
has no atom at zero--which is, of course, expected.

We now show that $Q^*$ has exponential tail under $P_d$ and derive
the precise asymptotics.

To do this, let
\[
\beta^*:= \Phi_\Lambda(0).
\]
Then \eqref{Wlap} gives
that the Laplace transform of $x \mapsto e^{-\beta^* x} W(x)$ is
$\theta \mapsto 1/\psi_\Lambda(\beta^*+\theta)$.
{From} the final value theorem for Laplace transforms,
\[
\lim_{x \to \infty} e^{-\beta^* x} W(x) = \lim_{\theta \to 0}
\frac{\theta}{\psi_\Lambda(\beta^*+\theta)}
= \frac{1}{{\psi_\Lambda}'(\beta^*)},
\]
where we used the fact that $\psi_\Lambda(\beta^*)=0$.
It follows that
\[
 P_d (Q^* > x)\sim \frac{\psi_\Lambda'(\Phi_\Lambda(0))}{\delta_\Lambda} e^{-\Phi_\Lambda(0)x}
\]
as $x\rightarrow\infty$.

\section{Cycle formulae}
\label{cycleform}
We now show how the use of cycle formulae of Palm calculus
enable us to find (Proposition \ref{Iobs} below) the joint law of
the endpoints of an idle period conditional on the event that
the idle period contains the origin of time.
Also (Proposition \ref{Bobs} below) we characterise the joint law of the
endpoints of a busy period, together with the maximum of $Q$ over this
busy period, conditional on the event that
the busy period contains the origin of time.

It is well-known that if $(\Omega, \FF, P)$ is endowed with
a $P$-preserving flow $(\theta_t, t \ge 0)$ (see end of Section \ref{Intro})
then for any random measure $M$ with finite intensity $\lambda_M$,
and any point process $N$ with finite intensity $\lambda_N$
such that $M(B, \theta_t\omega) = M(B+t, \omega)$,
$N(B, \theta_t\omega) = N(B+t, \omega)$, for $t \in \R$, $B$ Borel subset of $\R$,
and $\omega \in \Omega$, and any nonnegative measurable $Z :\Omega \to \R$, we have
\begin{equation}
\label{exch}
\lambda_M E_M[Z]
=
\lambda_N E_N \int_{T_k}^{T_{k+1}} Z \comp \theta_{t} ~ M(dt),
\end{equation}
where $P_M$, $E_M$ (respectively, $P_N$, $E_N$) denotes Palm probability
and expectation with respect to $M$ (respectively, $N$), $T_0$ is
the first atom of $N$ which is $\le 0$, and $T_k, T_{k+1}$
are any two successive atoms of $N$.

The next result can be found for some special cases in \cite{KoSa} (diffusions), \cite{KKSS} (L\'evy processes), and  in \cite{Si} the general expression is derived. Here we offer a new proof in the general case based on (\ref{exch}).

\begin{proposition}[Joint law of endpoints of idle period]
\label{Iobs}
Assume that (A1)--(A5) hold.
Then, conditional on $Q_0=0$, the left end-point, $g(0)$, and right
end-point, $d(0)$, of the idle period containing $t=0$ have
joint Laplace transform given by
\[
E[ e^{-\alpha d(0)+\beta g(0)} \mid Q_0 =0]
= \frac{\Phi_\Lambda(0)}{-\psi'_\Lambda(0+)} \cdot \frac{1}{\alpha-\beta}
\bigg(
\frac{\psi_\Lambda(\alpha)}{\alpha- \Phi_\Lambda(0)}
-
\frac{\psi_\Lambda(\beta)}{\beta- \Phi_\Lambda(0)}
\bigg) ,
\]
for non-negative $\alpha$ and $\beta$ ($\alpha\not=\beta$).
\end{proposition}
\begin{proof}
Let $M_I$ be the restriction of the Lebesgue measure on the idle
periods:
\[
M_I(A) = \int_A \1(Q_t=0) dt, \quad A \in \BB(\R).
\]
Then $E_{M_I}[Z] = E[Z | Q_0=0]$ for all nonnegative random variables $Z$.
Apply \eqref{exch} with $M=M_I$, $N=N_d$, and $Z=e^{-\alpha d(0)+\beta d(0)}$:
\[
\lambda_{M_I} E_{M_I}[e^{-\alpha d(0)+\beta g(0)}]
=
\lambda E_{d} \int_{d(-1)}^{d(0)}
e^{-\alpha d(0) \comp \theta_t
+\beta g(0)\comp \theta_t}  ~ M_I(dt).
\]
Here $\lambda$ is the rate of $N_d$ and is given by \eqref{commonrate}.
The rate $\lambda_{M_I}$ is given by
\[
\lambda_{M_I} = \frac{E_d[I]}{E_d[B+I]} .
\]
Hence
\[
\frac{\lambda}{\lambda_{M_I}} = \frac{1}{E_d[I]}
= \frac{\delta_\Lambda \Phi_\Lambda(0)}{-\psi'_\Lambda(0+)},
\]
where we used \eqref{meanidle} and \eqref{meancycle}.
Now, $P_d(d(0)=0)=1$. To compute the integral above,
note that $M_I$ is zero on the interval $(d(-1), g(0))$,
and that, for $g(0) \le t \le 0$, we have $d(0)\comp \theta_t = -t$,
and $g(0)\comp\theta_t = g(0)-t$. So the integral above equals
\[
\int_{g(0)}^0 e^{(\alpha-\beta)t-\alpha g(1)}~ dt
= \frac{e^{\beta g(0)} - e^{\alpha g(0)}}{\alpha-\beta}.
\]
Combining the above we obtain
\[
E[ e^{-\alpha d(0)+\beta g(0)} \mid Q_0 =0]
=
\frac{\Phi_\Lambda(0)}{-\psi'_\Lambda(0+)}
\cdot
\frac{E_d[e^{\beta g(0)}]-E_d[e^{\alpha g(0)}]}{\alpha-\beta}.
\]
Since $E_d[e^{\beta g(0)}] = E_d[e^{-\beta I}]$, the result is
obtained by using \eqref{typicalI}.
\QED
\end{proof}

\begin{proposition}[Joint law of endpoints of busy period and maximum over it]
\label{Bobs}
Assume that (A1)--(A5) hold.
Then, conditional on $Q_0>0$, the left end-point, $d(0)$, and right
end-point, $g(1)$, of the busy  period containing $t=0$, together
with the maximum of $Q_s$ for $s$ ranging over this busy period have
a joint law which is characterised by
\begin{eqnarray}
\label{max00}
\nonumber
\lefteqn{
E [e^{-\alpha g(1)+\beta d(0)} \1(Q^* \le x)\mid Q_0 >0]
}
\\[2mm]
&=& \frac{\Phi_\Lambda(0)}{\alpha-\beta}

\bigg(
\frac{1+\alpha\int_0^x W^{(\alpha)}(y)dy}{W^{(\alpha)}(x)}
-
\frac{1+\beta\int_0^x W^{(\beta)}(y)dy}{W^{(\beta)}(x)}
\bigg),
\end{eqnarray}
for  non-negative $\alpha$ and $\beta$ ($\alpha\not=\beta$).
\end{proposition}
\begin{proof}
Let $M_B$ be the restriction of the Lebesgue measure on the busy periods:
\[
M_B(A) = \int_A \1(Q_t>0) dt, \quad A \in \BB(\R).
\]

Then $E_{M_B}[Z] = E[Z|Q_0 > 0]$ for all random variables $Z \ge 0$.
Apply \eqref{exch}:
\begin{align}
\lefteqn{
\lambda_{M_B} E_{M_B} [ e^{-\alpha g(1)+\beta d(0)} \1(Q^* \le x)]
}
\nonumber \\
&= \lambda E_d \int_{d(0)}^{d(1)}
e^{-\alpha g(1)\comp\theta_t +\beta d(0)\comp\theta_t }
\1(Q^*\comp\theta_t  \le x) ~M_B(dt)
\nonumber \\
&= \lambda E_d \int_0^{g(1)}
e^{-\alpha (g(1)-t) - \beta t} \1(Q^* \le x) ~ dt
\nonumber \\
&= \lambda E_d
\bigg[
\1(Q^* \le x) ~ e^{-\alpha g(1)} ~
\frac{e^{(\alpha-\beta)g(1)}-1}{\alpha-\beta}
\bigg]
\nonumber \\
&= \frac{\lambda}{\alpha-\beta}
\bigg(
E_d[e^{-\beta g(1)} \1(Q^* \le x) ]
-
E_d[e^{-\alpha g(1)} \1(Q^* \le x) ]
\bigg)
\nonumber \\
&= \frac{\lambda}{\alpha-\beta}
(H(\beta,0,x) - H(\alpha, 0,x)),
\label{argument}
\end{align}
where $H(\alpha,\beta,x)$ is the right-hand side of
\eqref{triplelaw}. Using \eqref{commonrate},
\eqref{meancycle} and \eqref{busym}, we have
\[
\frac{\lambda}{\lambda_{M_B}} = \frac{1}{E_d[B]}
=
\delta_\Lambda \Phi_\Lambda(0).
\]
Combining the above we obtain the announced formula.
\QED
\end{proof}

Proposition \ref{Bobs} yields the next corollary which recovers a result obtained in
\cite{Si} using different methods (for special cases, see \cite{KoSa} and \cite{KKSS}).
Clearly, Corollary \ref{cor0} could also be proved analogously as Proposition \ref{Iobs}.
\begin{corollary}
\label{cor0}
Assume that (A1)--(A5) hold.
Then, conditional on $Q_0>0$, the left end-point, $d(0)$, and right
end-point, $g(1)$, of the busy  period containing $t=0$ have joint Laplace transform given by
\[
E [e^{-\alpha g(1)+\beta d(0)} \mid Q_0 >0]
=
\frac{\Phi_\Lambda(0)}{\alpha-\beta}
\cdot
\bigg(
\frac{\alpha}{\Phi_\Lambda(\alpha)}
-
\frac{\beta}{\Phi_\Lambda(\beta)}
\bigg),
\]
for non-negative $\alpha$ and $\beta$ ($\alpha\not=\beta$).
\end{corollary}
\begin{proof}
The argument proceeds as in the proof Proposition \ref{Bobs}
by omitting the factor $\1(Q^* \le x)$, i.e.\ by formally
replacing $x$ with $+\infty$. The last line of \eqref{argument}
will give $\frac{\lambda}{\alpha-\beta}
(H(\beta,0,\infty) - H(\alpha, 0,\infty))$,
where $H(\alpha,\beta,\infty)$ is given by the right-hand side
of \eqref{no-max}.
\QED
\end{proof}

\begin{corollary}
Assume that (A1)--(A5) hold.
Then, conditional on $Q_0>0$,
the maximum of $Q$ over the  busy  period containing $t=0$
has distribution
\begin{equation}
\label{max1}
P(Q^* \le x \mid Q_0 >0) = \Phi_{\Lambda}(0)\frac{W(x)\int_0^xW(y)dy - \int_0^x W(x-y)W(y)dy}{W(x)^2}
\end{equation}

for $x\geq 0$.
\end{corollary}

\begin{proof} Letting $\alpha,\beta\to 0$ in (\ref{max00}) yields

\begin{equation*}
P(Q^* \le x \mid Q_0 >0)
= \Phi_{\Lambda}(0)\,\lim_{\alpha \to 0} \frac{\partial \hat H}{\partial\alpha}(\alpha,0,x),
\end{equation*}
where
\[
\hat H(\alpha, 0, x)
= 1 -
\frac{1+\alpha \int_0^x W^{(\alpha)}(y) dy}{ W^{(\alpha)}(x)}.
\]
Next recall that for each $x>0$, $W^{(\alpha)}(x)$ is an entire function in the variable $\alpha$ and in particular
\[
W^{(\alpha)}(x) = \sum_{k\geq 0} \alpha^k W^{*(k+1)}(x)
\]
where $W^{*(k+1)}(x)$ is the $(k+1)$-th convolution of $W$ (cf. Bertoin \cite{B97}). From this one easily deduces that
$$
\frac{\partial}{\partial\alpha}W^{(\alpha)}(x)|_{\alpha=0}=\int_0^x W(y)W(x-y)dy.
$$
The result now follows from straightforward differentiation.
\QED
\end{proof}

\section{Example:
Local time storage from reflected Brownian motion with negative drift}
\label{example1}
Let $X = \{X_t, t \in \R\}$ be a reflected Brownian motion
with drift $-c<0$  in stationary state
living on $I=[0, \infty),$ and let $P_0$ denote the probability
measure associated with $X$ when initiated from 0 at time 0.
Its local time (at 0) for $s<t$ is given by
\begin{equation}
         L(s,t]:= \lim_{\varepsilon \searrow 0}
         \frac{1}{2\varepsilon}
         \int_s^t \1_{[0,\varepsilon)}
         (X_u) \,du.
         \label{eqloc}
    \end{equation}
Let $Q$ be the stationary process defined as in  (\ref{Q}):

$$
Q_t:=\sup_{s \leq t}\{L(s,t]-(t-s)\}.
$$
This particular example of fluid queues was introduced and analysed in

\cite{{MNS}} and further
studied in
\cite{KoSa} and

\cite{KKSS}.

Recall that $E_0L(0,1]=c,$ and, hence $Q$
is well-defined if and only if $0<c<\nolinebreak 1$.

Here we make this example more complete by finding the
$\alpha$-scale function associated with process
$\Lambda_t:=t-L^{-1}_t, t\geq 0,$ where
$$
L^{-1}_t:=\inf\{s: L(0,s] > t\}, \quad t \geq 0.
$$
is the inverse local time process.
As seen from formulae (\ref{max1}) and (\ref{Hformula}),
the $\alpha$-scale function is the key ingredient needed for computing
the distribution of the maximum of $Q$ over a busy period
and related random variables.

To begin with, we recall some basic formulae. When normalising as in (\ref{eqloc}),
see
\cite[pg.\ 214]{itomckean74}, \cite[pg.\ 22]{BS}  it holds that
\begin{eqnarray}
\label{levy}
 &&
\nonumber
 E_0 (\exp \{-\theta L^{-1}_t\}) =
\exp \left\{ -t\int_0^\infty(1-{\rm e}^{-\theta u})\frac{1}{\sqrt{2\pi
    u^3}}\, {\rm e}^{-c^2u/2}\,du\right\}
\\
&&\hskip3.0cm
=\exp \left\{ -\frac{t}{G_\theta(0,0)} \right\},
\end{eqnarray}
where
$$
G_\theta(0,0)
:= \frac{1}{\sqrt{2 \theta+ c^2}-c}
$$
is the resolvent kernel (Green kernel) of $X$ at $(0,0)$;
see \cite[pg.\ 129]{BS}.
Consequently, we have
\begin{equation*}
   E_0 \left(\exp\{\theta \Lambda_t\}\right) =
   \exp\left \{t\left(\theta-\frac{1}{G_\theta(0,0)}\right) \right\}
   = \exp \{t \psi_{\Lambda }(\theta) \},
   \end{equation*}
where
$$
\psi_{\Lambda }(\theta):= \theta-\sqrt{2 \theta+c^2} + c, \quad \theta \geq 0.
$$

Recall (cf. (\ref{alpha})) that the $\alpha$-scale function ($\alpha\geq 0$) associated with $\Lambda$
is defined for $x\geq 0$ via
\begin{equation}
\label{s1}
\int_0^\infty {\rm e}^{-\theta x}W^{(\alpha)}(x)dx=\frac 1{ \psi_{\Lambda }(\theta)-\alpha};
\end{equation}
 for $x<0$ we set $W^{(\alpha)}(x)=0.$ The 0-scale function is called
simply the scale function and denoted $W.$ For the next proposition introduce
$$
{\rm Erfc}(x):=\frac 2{\sqrt{\pi}}\,\int_x^\infty {\rm e}^{-t^2}dt,
$$
and notice that ${\rm Erfc}(0)=1,\ {\rm Erfc}(+\infty)=0,$ and ${\rm Erfc}(-\infty)=2.$
\begin{proposition}
\label{scale1}
The $\alpha$-scale function $W^{(\alpha)}$ of $\Lambda$ is for $x\geq 0$ given by
\begin{eqnarray}
\label{scale-alpha}
&&\hskip-1cm
\nonumber
W^{(\alpha)}(x)=\frac {{\rm e}^{-c^2 x/2}}{\lambda_1-\lambda_2}
\Big(\lambda_1{\rm  e}^{\lambda_1^2\, x/2}
\,{\rm Erfc}(-\lambda_1\sqrt{x/2})
\\
&&
\hskip5cm
 -\lambda_2{\rm  e}^{\lambda_2^2\, x/2}
\,{\rm Erfc}(-\lambda_2\sqrt{x/2})\Big),
\end{eqnarray}
where
\begin{equation}
\label{roots}
\lambda_{1}:= 1+\sqrt{(1-c)^2+2\alpha},\quad
\lambda_{2}:= 1-\sqrt{(1-c)^2+2\alpha}
\end{equation}
In particular,
\begin{eqnarray}
\label{scalefor}
&&\hskip-1cm
\nonumber
W(x)=\frac {{\rm e}^{-c^2 x/2}}{2(1-c)}\Big((2-c)\,{\rm e}^{(2-c)^2 x/2}
\,{\rm Erfc}(-(2-c)\sqrt{x/2})
\\
&&\hskip5cm
-c\,{\rm e}^{\,c^2x/2}\,{\rm
  Erfc}(-c\sqrt{x/2})\Big),
   \end{eqnarray}
   and $W(0)=1.$
\end{proposition}
\begin{proof} From (\ref{s1})  we have
\begin{equation}
\label{lap20}
\int_0^\infty {\rm e}^{-\theta x}\,W^{(\alpha)}(x)dx= \frac 1{\theta-\sqrt{2 \theta+c^2} + c-\alpha}.
\end{equation}
To invert this Laplace transform, introduce $\lambda:= 2\theta+c^2.$
With this notation,
\begin{eqnarray}
\label{lap21}
&&\hskip-1cm
\nonumber
\frac 1{\theta-\sqrt{2 \theta+c^2} + c-\alpha}=\frac
2{\lambda-2\sqrt{\lambda} +2(c-\alpha)-c^2}
\\
&&
\hskip2.4cm
=\frac 2{(\sqrt{\lambda}-\lambda_1)(\sqrt{\lambda}-\lambda_2)}
\\
&&
\nonumber
\hskip2.4cm
=\frac2{\lambda_1-\lambda_2}\left(\frac
1{\sqrt{\lambda}-\lambda_1}-\frac 1{\sqrt{\lambda}-\lambda_2}\right),
 \end{eqnarray}
where $\lambda_{1,2}$ are the roots of the equation
$z^2-2z+2(c-\alpha)-c^2=0,$ i.e., as in (\ref{roots}).
Next, recall the following Laplace inversion formula
(cf.\ Erd\'elyi \cite[pg.\ 233]{erdelyi54})
\begin{equation}
\label{L0}
{\mathcal L}^{-1}\left(\frac 1{\sqrt{\lambda}+\beta}\right)=\frac
1{\sqrt{\pi x}}-\beta\,{\rm e}^{\beta^2 x}\,{\rm Erfc}(\beta\sqrt{x})
\end{equation}
valid for $\lambda-\beta^2>0.$ Since
\begin{eqnarray*}
&&\hskip-1cm
\int_0^\infty {\rm e}^{-\theta x}W^{(\alpha)}(x)dx=\int_0^\infty {\rm
  e}^{-\lambda y} {\rm
  e}^{c^2\, y}\,W^{(\alpha)}(2y)2dy
 \end{eqnarray*}
we obtain using (\ref{L0})
$$
2{\rm  e}^{c^2\, y}\,W^{(\alpha)}(2y)= \frac 2{\lambda_1-\lambda_2}\left(\lambda_1{\rm  e}^{\lambda_1^2\, y}
\,{\rm Erfc}(-\lambda_1\sqrt{y}) -\lambda_2{\rm  e}^{\lambda_2^2\, y}
\,{\rm Erfc}(-\lambda_2\sqrt{y})\right),
$$
which is formula (\ref{scale-alpha}). In particular, when $\alpha=0$ it holds $\lambda_1=2-c$ and
$\lambda_2=c$ yielding formula (\ref{scalefor}).
\QED
\end{proof}

Using the scale function $W$ and the fact
$$
\Phi_\Lambda(0)=\sup\{\theta>0\,:\, \psi_\Lambda(\theta)=0\}=2(1-c)
$$
formula (\ref{max1}) yields
the distribution of the maximum $Q^*$ over an observed busy period
(i.e.\ over a busy period containing the origin of time).

\begin{proposition}
Let $0 < c < 1$.
The distribution of the maximum $Q^*$ over an
observed  busy period of a reflected Brownian motion
(with drift $-c$) local time storage is given by
\begin{equation}
\label{L12}
P(Q^*\leq x\mid Q_0 > 0)
=2(1-c)\frac{\int_0^x\,W(y)\left( W(x)-W(x-y)\right)\,dy}
{W^2(x)},
\end{equation}
where the scale function $W$ is given by (\ref{scalefor}).
\end{proposition}

We plot the derivative of \eqref{L12} for $c=1/2$ in Figure \ref{fstar} below.
\begin{figure}[h]
\begin{center}
\includegraphics[width=6cm]{figf05}
\caption{\em The density of $Q^*$ conditional on $\{Q_0>0\}$ for
the example corresponding to Brownian motion with drift $-c=-1/2$.
}
\label{fstar}
\end{center}
\end{figure}

We recall some formulas from \cite{KoSa}. First
\begin{eqnarray} \label{busybrown}
 &&E \left( e^{\theta d(0) -\beta g(1)}\,|\, Q_0>0 \right)
\hspace{0.2cm} = \frac{8(1-c)}{\sqrt{2 \theta+(1-c)^2}+ \sqrt{2 \beta+(1-c)^2}}
 \nonumber  \\
 & &  \hspace{1.5cm} \times
 \frac{1}{(\sqrt{2 \theta+(1-c)^2}+1+c)(\sqrt{2 \beta+(1-c)^2}+1+c)} \nonumber \\
 & & \hspace{0.2cm} =:F(\theta, \beta; 1-c),
\end{eqnarray}

and
\begin{equation}
\label{idlebrown}
\E \left(e^{\theta g(0) -\beta d(0)}\,|\, Q_0=0 \right) =
F(\theta, \beta; c).
\end{equation}
Setting $\beta = \theta$ in the right-hand side
of (\ref{busybrown}) and (\ref{idlebrown}), respectively,
 we get
\begin{equation} \label{lapvb}
  E \left( e^{-\theta (g(1) -d(0))} \,|\, Q_0>0 \right) =
  \frac{4 (1-c)}{\sqrt{2 \theta +(1-c)^2} (\sqrt{2 \theta +(1-c)^2}+1+c)^2},
\end{equation}
and
\begin{equation} \label{lapvi}
  E \left( e^{-\theta (d(0) -g(0))} \,|\, Q_0=0 \right) =
  \frac{4 c}{\sqrt{2 \theta +c^2} (\sqrt{2 \theta +c^2}+2-c)^2},
\end{equation}
Taking the inverse Laplace transform of $(\ref{lapvb})$ (cf.
Erd\'elyi \cite[pg.\ 234]{erdelyi54})
we obtain the density of the length of the busy period $g(1) - d(0)$,
given $Q_0>0$, as
$$
f_{g-b} (v) = 2(1-c){\rm e}^{-(1-c)^2v/2}\left( \sqrt{2v/\pi}-  (1+c) v
e^{(1+c)^2v/2}{\rm Erfc}((1+c)\sqrt{v/2})\right)
$$
Note that the density of the length of the idle period $d(0) - g(0)$,
given that $Q_0=0$, is obtained from $f_{g-b}(v)$ by substituting $c$ for $1-c$.
In Figure \ref{fgb} we have ploted $f_{g-b} (v)$ for three different
values of $c$.
\begin{figure}[h]
\begin{center}
\includegraphics[width=7cm]{figfgb}
\caption{\em The density of length of the busy period, given that $Q_0>0$
for three different values of $c$.}
\label{fgb}
\end{center}
\end{figure}
We notice also that the mean busy period length has a simple expression:
\[
E[g(1)-d(0) \mid Q_0 > 0] = \frac{2-c}{(1-c)^2}.
\]
The joint density of $d(0)$ and $g(1)$ is given by
  \begin{eqnarray*}
  && \P(-d(0) \in dx, g(1) \in dy \,|\, Q_0>0)=
   2(1-c){\rm e}^{-(1-c)^2(x+y)/2}
\\
&&
\hskip1cm
\times\left( \sqrt{2/(\pi(x+y))}-  (1+c)
e^{(1+c)^2(x+y)/2}{\rm Erfc}((1+c)\sqrt{(x+y)/2})\right)
\end{eqnarray*}
and, again, the density for $(g(0),d(0))$ is obtained by substituting $c$
for $1-c.$

Next we find the density of $g(1)$ (recall that $-d(0)$ is identical
in law with $g(1)$)  by inverting the Laplace transform
(obtained from (\ref{busybrown}) by choosing $\theta=0$):
\begin{eqnarray}
\label{busyd}
\nonumber
 &&\hskip-1cm
E \left(e^{-\beta g(1)}\,|\, Q_0>0 \right)
\\
&&

= \frac{4(1-c)}{\left(\sqrt{2 \beta+(1-c)^2}+ 1-c\right)
\left(\sqrt{2 \beta+(1-c)^2}+1+c\right)}.
 \end{eqnarray}
Letting  $\lambda:=2 \beta+(1-c)^2$ we rewrite (\ref{busyd})
as
\begin{eqnarray}
\label{busyd2}
\nonumber
 &&\hskip-1cm
E \left(e^{-\beta g(1)}\,|\, Q_0>0 \right)
= \frac{2(1-c)}{c}\left(\frac{1}{\sqrt{\lambda}+ 1-c}-\frac{1}{\sqrt{\lambda}+ 1+c}\right).
 \end{eqnarray}
{From} (\ref{L0})

\begin{align}
\lefteqn{
{\mathcal L}^{-1}\left(\frac{1}{\sqrt{\lambda}+ 1-c}
-\frac{1}{\sqrt{\lambda}+
  1+c}\right)}
\nonumber \\
& =(1+c)\,{\rm e}^{(1+c)^2 x}\,{\rm Erfc}\left((1+c)\sqrt{x}\right)
-(1-c)\,{\rm e}^{(1-c)^2 x}\,{\rm Erfc}\left((1-c)\sqrt{x}\right).
\label{L01}
\end{align}
Consequently,
\begin{multline}
2 {\rm e}^{-(1-c)^2x}f_{g(1)}(2x)
= \frac{2(1-c)}{c}
\Big((1+c)\,{\rm e}^{(1+c)^2 x}\,{\rm Erfc}((1+c)\sqrt{x})
\\
-(1-c)\,{\rm e}^{(1-c)^2 x}\,{\rm Erfc}((1-c)\sqrt{x})\Big),
\label{den2}
\end{multline}
where $f_{g(1)}$ denotes the density of $g(1)$ conditioned on $\{Q_0>0\}. $
From (\ref{den2}) we obtain
\begin{multline}
 f_{g(1)}(x)=\frac{(1-c)\,{\rm e}^{-(1-c)^2x/2}}{c} ~
\Big((1+c)\,{\rm e}^{(1+c)^2 x/2}\,{\rm Erfc}\left((1+c)\sqrt{x/2}\right)
\\
-(1-c)\,{\rm e}^{(1-c)^2 x/2}\,{\rm Erfc}\left((1-c)\sqrt{x/2}\right)\Big).
\label{den3}
\end{multline}
Moreover, the density  $f_{d(0)}$ of $d(0)$ conditional on
$\{Q_0=0\}$ is obtained from (\ref{den3}) by
substituting $c$ instead of $1-c:$
\begin{multline}
f_{d(0)}(x)=\frac{c\,{\rm e}^{-c^2x/2}}{1-c}~
\Big((2-c)\,{\rm e}^{(2-c)^2 x/2}\,{\rm Erfc}\left((2-c)\sqrt{x/2}\right)
\\
-c\,{\rm e}^{c^2 x/2}\,{\rm Erfc}\left(c\sqrt{x/2}\right)\Big).
\label{den4}
\end{multline}
It is striking how similar formulae (\ref{scalefor}) and (\ref{den4}) are.

\begin{remark}
The scale function formulae (\ref{scale-alpha}) and (\ref{scalefor}) are clearly valid for all $c\geq 0.$  In case $c=0$ the process $\{L(0,t]\,;\, t\geq 0\}$ is a version of the Brownian local time, and the $\alpha$-scale function $ W_0^{(\alpha)} $ of the corresponding process $\Lambda$ is given by
\begin{multline*}
W_0^{(\alpha)}(x)=\frac{{\rm e}^{(1+\alpha)x}}{2\sqrt{1+2\alpha}}\Big(
(1+\sqrt{1+2\alpha})\,{\rm e}^{x\sqrt{1+2\alpha}}\,{\rm Erfc}\left(-(1+\sqrt{1+2\alpha})\,\sqrt{x/2}\right)
\\
-(1-\sqrt{1+2\alpha})\,{\rm e}^{-x\sqrt{1+2\alpha}}\,{\rm Erfc}\left(-(1-\sqrt{1+2\alpha})\,\sqrt{x/2}\right)
\Big).
\end{multline*}
In particular,
\begin{equation}
\label{rbm1}
W_0(x)={\rm e}^{2x}\,{\rm Erfc}(-\sqrt{2x}).
\end{equation}
In case $c=1$ it holds
$\lambda_{1,2}=1\pm\sqrt{2\alpha}$ and for $\alpha\not= 0$ formula (\ref{scalefor}) can be used directly. For the 0-scale function we need to take the limit as $c\to 1$ in (\ref{scalefor}):
\begin{equation}
\label{p111}
W_1(x)=(1+x)\,{\rm Erfc}(-\sqrt{x/2})+\sqrt{\frac{2x}{\pi}}\,{\rm e}^{-x/2}.
\end{equation}
\end{remark}
\begin{remark}
Here we display some formulae for
Laplace transforms apparent from above and point out a misprint in Erd\'elyi et al. \cite{erdelyi54}.

First,
from (\ref{lap20}),  (\ref{lap21}), and (\ref{p111})
we have the following Laplace inversion
formula valid for $\lambda>1$:
\begin{equation}
\label{p1}
{\mathcal L}^{-1}\left(\frac 1{(\sqrt{\lambda}-1)^2}\right)=
(1+2x)\,{\rm e}^x\,{\rm Erfc}(-\sqrt{x})+\frac{2\sqrt x}{\sqrt\pi},
\end{equation}
and this can be ``extended''  (for $a>0$) to
\begin{equation}
\label{p2}
{\mathcal L}^{-1}\left(\frac 1{(\sqrt{\lambda}-a)^2}\right)=
(1+2a^2x)\,{\rm e}^{a^2x}\,{\rm Erfc}(-a\sqrt{x})+\frac{2a\sqrt x}{\sqrt\pi}.
\end{equation}
Furthermore, it can be checked that (\ref{p2}) is valid for all $a<0$
by evaluating the Laplace transform of the right-hand side.
This can be done
term by term  by using, e.g., Erd\'elyi et al. \cite[pp.\ 137, 177]{erdelyi54}
(well-known formulae):
$$
{\mathcal
  L}\left(\sqrt{x}\right)=\frac{\sqrt{\pi}}2\,\lambda^{-3/2},
$$
$$
{\mathcal L}\left({\rm e}^{a^2 x}\,{\rm
  Erfc}(a\sqrt{x})\right)=\lambda^{-1/2}\,(\lambda^{1/2}+a)^{-1},
$$
and
\begin{eqnarray*}
&&
{\mathcal L}\left(x{\rm e}^{a^2 x}\,{\rm
  Erfc}(a\sqrt{x})\right)=-\frac{\partial}{\partial \lambda}
{\mathcal L}\left({\rm e}^{a^2 x}\,{\rm
  Erfc}(a\sqrt{x})\right)
\\
&&
\hskip3.3cm
=-\frac{\partial}{\partial \lambda}\lambda^{-1/2}\,(\lambda^{1/2}+a)^{-1}
\\
&&
\hskip3.3cm
=\frac 1{2a}\left(\lambda^{-3/2} -\lambda^{-1/2}\,(\lambda^{1/2}+a)^{-2}\right).
\end{eqnarray*}
We remark that formula (10)
in \cite{erdelyi54} p. 234:
\begin{equation}
\label{e1}
{\mathcal L}^{-1}\left(\frac 1{(\sqrt{\lambda}+\sqrt{b})^2}\right)=
1-2\sqrt{bx/\pi}+ (1-2bx)\,{\rm e}^{bx}\left({\rm Erf}(\sqrt{bx})-1\right).
\end{equation}
is {\bf not} correct since it does not coincide with formula
(\ref{p2}) (for $a<0$). Indeed, because
$$
{\rm Erf}(x):=\frac 2{\sqrt{\pi}}\,\int_0^x {\rm e}^{-t^2}dt,
$$
the right-hand side of (\ref{e1}) is zero at zero but the right-hand side of
(\ref{p2}) is 1 at zero.
\end{remark}

\section{Further examples}
In the previous example we derived a local time process from a given Markov process. However, it is also possible to consider examples where just the local time process $L$, or equivalently the subordinator $L^{-1}$, is specified.
Indeed the subordinator that will play the role of
$L^{-1}$ in this example has no drift and has L\'evy measure given by
\[
\Pi(x,\infty) = \frac{ \gamma^\nu}{\Gamma(\nu)}x^{\nu-1}e^{-\gamma x} + \varphi \frac{ \gamma^\nu}{\Gamma(\nu)}\int_x^\infty
y^{\nu-1}e^{-\gamma y} dy,
\]
where the constants $\varphi,  \gamma>0$ and $\nu\in(0,1)$. Note in particular then that $L^{-1}$ is the sum of two independent subordinators, one of which is a compound Poisson process with gamma distributed jumps, the other has infinite activity  and is of the so called tempered-stable type.
Clearly $\Pi$ also describes the L\'evy measure of $-\Lambda$ too.

According to
\cite{HK}, the process $\Lambda$ belongs to the Gaussian Tempered Stable Convolution class and moreover,
\[
\psi_\Lambda(\theta) = (\theta  - \varphi)\left(1 - \left(\frac{\gamma}{\gamma+\theta}\right)^\nu\right)
\]
for $\theta\geq 0$. In particular $\delta_\Lambda =1$ and $\Phi_\Lambda(0) = \varphi$. It is a straightforward exericse to show that
\[
\mathbb{E}(\Lambda_1)  = \psi_\Lambda'(0+) = -\varphi  \frac{\nu}{\gamma}
\]
and this implies that
\[
 \mu = \frac{1}{1+ \varphi\nu/\gamma}<1,
\]
as required.

From \cite{HK} we also know that
\[
W(x) = e^{\varphi x}+ \gamma^\nu e^{\varphi x}\int ^x e^{-(\gamma + \varphi)y} y^{\nu-1}{\rm E}_{\nu,\nu}(\gamma^\nu y^\nu)dy
\]
where
\[
 {\rm E}_{\alpha, \beta}(x) := \sum_{n\geq 0}\frac{z^n}{\Gamma(\alpha n + \beta)}
\]
 is the two parameter Mittag-Leffler function.

We may now deduce from the theory presented earlier that, for example,
\[
P_d(Q^*\leq x) = \frac{1-e^{-\varphi x}+ \gamma^\nu \int ^x e^{-(\gamma + \varphi)y} y^{\nu-1}{\rm E}_{\nu,\nu}(\gamma^\nu y^\nu)dy
}{1+ \gamma^\nu \int ^x e^{-(\gamma + \varphi)y} y^{\nu-1}{\rm E}_{\nu,\nu}(\gamma^\nu y^\nu)dy
}
\]
and
\[
 P_d(Q^*>x)\sim \left(1 - \left(\frac{\gamma}{\gamma+\varphi}\right)^\nu\right)e^{-\varphi x}.
\]
\vskip.5cm
\noindent
{\bf Acknowledgement.} We thank Ilkka Norros for posing the problem for finding the distribution of the maximum and Andrey Borodin for discussions on Laplace transforms.

config: arxiv


TITLE: A question about finite fields with order $p^n$.
QUESTION [1 upvotes]: If we have a finite field $E$ with $p^n$ elements, where $p$ is a prime number. Is it necessary that it is a field extension of a subfield isomorphic to $\Bbb{Z_p}$, and consequently is of characteristic $p$?
I believe it is true that any finite field of characteristic $p$ has order $p^n$, and it is a field extension of a subfield isomorphic to $\Bbb{Z_p}$. However, I'm interested  in the converse.

REPLY [1 votes]: Theorem 2.1 here would be of interest to you.  Also note that any field with $p$ elements is isomorphic to $\mathbb{Z}_p$.


TITLE: How do I solve Schroedinger's equation for sterile neutrinos?
QUESTION [0 upvotes]: When reading Light singlet neutrinos and the primordial nucleosynthesis the following equation is given:
$$\tag{5} i \frac{\partial}{\partial t } \begin{pmatrix} c_e (t) \\ c_x (t) \end{pmatrix}= \frac{\Delta _m }{2}\begin{pmatrix} cos2\theta_m & sin2\theta_m \\ sin2\theta_m & cos 2\theta_m\end{pmatrix} \begin{pmatrix} c_e (t) \\ c_x (t) \end{pmatrix}$$
Given the wave function below:
$$\nu(t) = c_e (t) \nu_e + c_x (t) \nu_x$$
in here $\nu_x$ represents sterile neutrinos and $\nu_e$ represents electron-neutrinos.
The equation above can be solved as to give the probability to find such wavefunction in the state $\nu_x$ at the time $t$:
$$P_{\nu_e \to \nu_x} (t)  = |c_x (t) |^2 = sin^2 2\theta_m  sin^2 \left( \frac{1}{2} \Delta_m t\right) \tag{7}$$
How is (5) solved to give us (7)? How does the $c_e$ term vanish in this integration?
The variable integrated in this case is time, if so, then how does one integrate the $c_e(t)$ and $c_x(t)$ terms that depend on the variable?

My take on it:
$$i \frac{\partial}{ \partial t} c_x = \frac{\Delta m}{2} (sin 2\theta c_e + cos2\theta c_x)$$
$$i \frac{\partial}{ \partial t} c_e = \frac{\Delta m}{2} (cos 2\theta c_e + sin 2\theta c_x)$$
I manipulated the top equation to obtain $c_x $in terms of $c_e $. And rewrote it as :
$$c_e = \frac{2i}{\Delta m sin2 \theta} \frac{\partial}{\partial t} c_x - tan^{-1} 2\theta c_x$$
Tried substituting this back into the bottom equation but I didn't know how to differentiate it.

REPLY [1 votes]: This is not the way to solve two simultaneous coupled differential equations. (To be honest, I'm not completely sure how exactly you managed to get what you say you did!) I won't solve the whole problem, but I'll give you a couple of pointers of how to get to the solution.
The point is that rate of change of the variables $c_x$ depends not only on the value of $c_x$, but also of $c_e$, and so you can't simply "integrate" the left and right hand sides to get anything meaningful. The best thing to do is to find some combination of $c_x$ and $c_e$ such that this pair of coupled equations reduces to two uncoupled differential equations, which you can then integrate. (This is very similar to how the two-masses-three-springs problems are solved.)
The easiest way to solve such equations is using matrices, but it isn't clear to me how much linear algebra you know, and so I'm going to try an approach using only differential equations, but it requires a little bit of mathematical intuition to decouple the equations.
The equations in question that you have are
\begin{equation*}
\begin{aligned}
\frac{\text{d}}{\text{d}t}c_e(t) &= -\frac{i \Delta}{2} \Bigg( \cos(2\theta)\,\, c_e(t) + \sin(2\theta)\,\, c_x(t)\Bigg)\\
\frac{\text{d}}{\text{d}t}c_x(t) &= -\frac{i \Delta}{2} \Bigg( -\sin(2\theta)\,\, c_e(t) + \cos(2\theta)\,\, c_x(t)\Bigg)\\
\end{aligned}
\end{equation*}
where, since the coefficients depend only on time, I have replaced the partial derivative with a total derivative, and I've ignored all the $m$s to make it easier to write. Also, you had an error in your matrix: in the second row, it should be $-\sin(2\theta)$.
Now, the trick is to realise that if you multiply the second equation by $i$ and you add and subtract the equations, you can show that (I'll leave it as an exercise):
\begin{equation*}
\begin{aligned}
\frac{\text{d}}{\text{d}t} \Big( c_e(t) + i\, c_x (t) \Big) &= -\frac{i \Delta}{2} e^{- i 2\theta} \Big(c_e(t) + i\, c_x(t)\Big)\\
\frac{\text{d}}{\text{d}t}\Big( c_e(t) - i\, c_x (t) \Big)  &= -\frac{i \Delta}{2} e^{i 2\theta} \Big(c_e(t) -i\, c_x(t)\Big)\\
\end{aligned}
\end{equation*}
If we define two new variables $c_1 = c_e(t) + i\, c_x (t)$ and $c_2 = c_e(t) - i\, c_x (t)$, our equations magically become decoupled!
\begin{equation*}
\begin{aligned}
\frac{\text{d}}{\text{d}t} c_1(t) &= -\frac{i \Delta}{2} e^{- i 2\theta} c_1(t)\\
\frac{\text{d}}{\text{d}t} c_2(t)  &= -\frac{i \Delta}{2} e^{i 2\theta} c_2(t)\\
\end{aligned}
\end{equation*}
You can now solve these equations for $c_1(t)$ and $c_2(t)$, and from there you can find $c_e(t)$ and $c_x(t)$.
The state of the system at any time $t$ is given by $\nu(t) = \begin{pmatrix}c_e(t)\\c_x(t)\end{pmatrix}$, where the upstairs element is the probability (amplitude) of finding an electron-neutrino and the downstairs element is the probability (amplitude) of finding a "sterile" neutrino. In the problem, you are given that the state originally starts off as an electron neutrino, and so $c_e = 1, c_x = 0$ (which are the initial conditions you will need when you solve the above differential equations). However, at some later time, $c_e(t) \neq 1$ and $c_x(t) \neq 0$, meaning that there is a probability that a measurement will result in a neutrino that isn't an electron-neutrino.
The probability of finding the neutrino as a "sterile" neutrino at some later time is $|c_x(t)|^2$, and so after you find the value of $c_x(t)$, then the probability that you get a "sterile" neutrino at some time $t$, given that you started with an electron neutrino at $t=0$, denoted by $P_{\nu_e \to \nu_x}(t)$, is
$$P_{\nu_e \to \nu_x}(t)=|c_x(t)|^2.$$

EDIT:
Ok this is odd. I don't think my method is wrong, but I don't seem to be able to reproduce the result in Equation (7) given in the paper, and I don't really know why. Looking online, Equation 3.2 here, seems to use a different matrix $$\begin{pmatrix}-\cos{2\theta}&\sin{2\theta}\\ \sin{2\theta}&\cos{2\theta}\end{pmatrix},$$
and using this I get the expected result for the probability. I'm not sure why this is happening. Perhaps the paper you've linked messed up their matrix, or perhaps there's something going on here that's above my pay-grade... I'd appreciate it if anyone had any insights.


TITLE: Show that for every natural number $n$ there exists a unique natural number $m$ such that $n$ lies between the sum of $m$ and $m-1$ numbers
QUESTION [0 upvotes]: How do we show that for every $n$ $\epsilon$ $N$, there exists a unique $m$ $\epsilon$ $N$such that $n$ can be expressed as $$\frac{m(m-1)}{2}<n \le\frac{m(m+1)}{2}$$

REPLY [1 votes]: Here I am assuming $\mathbb{N} = \{ 1,2,...\}$
For $m \in \mathbb{N}$ Set $$I_{m} = (\frac{m(m-1)}{2}, \frac{m(m+1)}{2}].$$
note that for $a<b$
$$1) \text{ }\text{ }\text{ }I_{a}\cap I_{b} = \emptyset$$
as $\frac{b(b-1)}{2} \geq \frac{a(a+1)}{2}$ and the left endpoint of each $I_m$ is not inside $I_{m}$. Furthermore
$$\bigcup_{m=1}^{j}I_{m} = (0, \frac{j(j+1)}{2}] \rightarrow (0, \infty]\text{ as }j\rightarrow \infty$$
thus each $n \in \mathbb{N}$ is in some $I_{p}$ where $p$ must be unique by $1)$


TITLE: Integral of $\sin(ax+b)\cos(ax+b)$
QUESTION [0 upvotes]: $\displaystyle\int \sin(ax+b)\cos(ax+b) \, dx$ is giving me two answers;

$\dfrac{1}{2a}\sin^2(ax+b)+C$
negative of    $\dfrac{1}{4a}\cos2(ax+b) +C$

Are these two results the same? Can you please prove it? Thank you

REPLY [2 votes]: They are same. Just use the formula
$$\cos 2A = 1 - 2 \sin^2 A$$

REPLY [1 votes]: They are the same because you add a fixed constant to an arbitrary constant you should get back an arbitrary constant.


TITLE: Is St. Petersburg a good place for the 2022 Int. Congress of Mathematicians
QUESTION [113 upvotes]: There might be just enough time to pick another location, but I am curious what mathematicians think. Will Ukrainian mathematicians be able to attend a conference in Russia if Russia no longer recognizes their passports?
To be clear: I love Russia, and I am not trying to hurt the feelings of Russian mathematicians or people.
The International Mathematical Union (IMU) have made a decision on moving the ICM to a virtual event, but there is still the (less exciting) decision to make concerning the location of IMU General Assembly.

REPLY [50 votes]: The IMU reached their decision, the ICM will take place entirely virtually except for the IMU general assembly and awards ceremony which will take place in-person outside of Russia. No word on whether Russian government involvement in organizing will continue.  Further details to come.
Terry Tao, who is the chair of the ICM Structure Committee, is inviting discussion of how to best use the virtual format on his blog.
The IMU executive committee has released some further clarifications, including announcing that there will be no Russian government involvement in the ICM.


TITLE: Which of these sums is equal to $\mathbb R^4$?
QUESTION [0 upvotes]: I'm given the following sets:
$$U=\{(0,a,b,a-b): a,b \in \mathbb{R}\} \\ V=\{(x,y,z,w): x=y, z=w\} \\ W=\{(x,y,z,w): x=y\}$$
I'm trying to determine which of the following is equal to $\mathbb R^4$:
$U+V, U+W, V+W$ where $A+B=\{a+b \ , a\in A, b \in B\}$)
Here's my working:
$U+V=\{x, a+y, b+z, a-b+w \quad : a, b \in \mathbb{R}, x=a+y, b+z=a-b+w\}$
$U+W=\{x, a+y, b+z, a-b+w  \quad :a,b \in \mathbb{R}, x=a+y\}=\{x,x, b+z, b+z \}$
$V+W=\{2x,2y,2z,2w \quad : x=y, z=w\}=\{x', x', z', z'\}$
Obviously, $V+W \neq \mathbb R^4$ as $(1,2,3,4) \notin V+W,$ but the other two sums looks very familiar. How would I go about eliminating one of them from being $\mathbb R^4$?

REPLY [1 votes]: You need to have freely changing variables.
$U$ gives you free second and third terms ($a,b$) while $W$ gives you free first, third and fourth terms.
$U+W=\{x, a+y, b+z, a-b+w  \quad :a,b,x,y,z,w \in \mathbb{R}\}=\{x',y', z', w' \quad x',y',z',w' \in \mathbb{R} \}$


TITLE: Is it possible to find an expression for $\frac{d^n}{dx^n}e^{-x^2}$?
QUESTION [3 upvotes]: I am trying to find a general form to derivatives of the function $e^{-x^2}$. I tried to do it by finding a pattern for the first derivatives but with no success.
Any tip is welcome.

REPLY [1 votes]: Set $u(x)=e^{x}$, $v(x)=-x^2$ and use the famous Faà di Bruno's Formula.
$$
\frac{d^{n}}{d x^{n}} u(v(x))=
\sum_{m_1+m_2+\ldots+m_n=n}
\frac{n !}{m_{1} ! m_{2} ! \cdots m_{n} !}
\cdot
u^{\left(m_{1}+\cdots+m_{n}\right)}(v(x))
\cdot
\prod_{j=1}^{n}\left(\frac{v^{(j)}(x)}{j !}\right)^{m_{j}}
$$
For $n=2$ we have
\begin{align}
\frac{d^{2}}{d x^{2}} u(v(x))
=&
\sum_{m_1+m_2=2}
\frac{2!}{m_{1} ! m_{2}!}
\cdot
u^{\left(m_{1}+m_{2}\right)}
\cdot
(v(x))
\cdot
\prod_{j=1}^{2}
\left(\frac{v^{(j)}(x)}{j !}\right)^{m_{j}}
\\
=&
\sum_{m_1+m_2=2}
\frac{2!}{m_{1} ! m_{2}!}
\cdot
u^{\left(m_{1}+m_{2}\right)}
\cdot
(v(x))
\cdot
\left(\frac{v^{(1)}(x)}{1 !}\right)^{m_{1}}
\cdot
\left(\frac{v^{(2)}(x)}{2 !}\right)^{m_{2}}
\end{align}
The partitions of $m_1+m_2=2$ are $(m_1,m_2)=(2,0),(1,1),(0,2)$. Then
\begin{align}
\frac{d^{2}}{d x^{2}} u(v(x))
=&
\frac{2!}{2 ! 0!}
\cdot
u^{\left(2+0\right)}
\cdot
(v(x))
\cdot
\left(\frac{v^{(1)}(x)}{1 !}\right)^{2}
\cdot
\left(\frac{v^{(2)}(x)}{2 !}\right)^{0}
\\
+&
\frac{2!}{1 ! 1!}
\cdot
u^{\left(1+1\right)}
\cdot
(v(x))
\cdot
\left(\frac{v^{(1)}(x)}{1 !}\right)^{1}
\cdot
\left(\frac{v^{(2)}(x)}{2 !}\right)^{1}
\\
+&
\frac{2!}{0 ! 2!}
\cdot
u^{\left(0+2\right)}
\cdot
(v(x))
\cdot
\left(\frac{v^{(1)}(x)}{1 !}\right)^{0}
\cdot
\left(\frac{v^{(2)}(x)}{2 !}\right)^{2}
\end{align}
A most interessant formulation is
Faà di Bruno's determinant formula. If $u$ and $v$ are functions with a sufficient ber of derivatives, and $m \geq 1,$ then
$$
\frac{d^{m}}{d x^{m}} u(v(x))
=
\det
\left(
\begin{array}{cccccc}
\left(\begin{array}{c}
m-1 \\
0
\end{array}\right) u^{\prime} v & \left(\begin{array}{c}
m-1 \\
1
\end{array}\right) u^{\prime \prime} v & \left(\begin{array}{c}
m-1 \\
2
\end{array}\right) u^{\prime \prime \prime} v & \cdots & \left(\begin{array}{c}
m-1 \\
m-2
\end{array}\right) u^{(m-1)} v & \left(\begin{array}{c}
m-1 \\
m-1
\end{array}\right) u^{(m)} v \\
-1 & \left(\begin{array}{c}
m-2 \\
0
\end{array}\right) u^{\prime} v & \left(\begin{array}{c}
m-2 \\
1
\end{array}\right) u^{\prime \prime} v & \cdots & \left(\begin{array}{c}
m-2 \\
m-3
\end{array}\right) u^{(m-2)} v & \left(\begin{array}{c}
m-2 \\
m-2
\end{array}\right) u^{(m-1)} v \\
0 & -1 & \left(\begin{array}{cc}
m-3 \\
0
\end{array}\right) u^{\prime} v & \cdots & \left(\begin{array}{c}
m-3 \\
m-4
\end{array}\right) u^{(m-3)} v & \left(\begin{array}{c}
m-3 \\
m-3
\end{array}\right) u^{(m-2)} v \\
\vdots & \vdots & \vdots & & \vdots & \vdots \\
0 & 0 & 0 & \cdots & \left(\begin{array}{c}
1 \\
0
\end{array}\right) u^{\prime} v & \left(\begin{array}{c}
1 \\
1
\end{array}\right) v^{\prime \prime} v \\
0 & 0 & 0 & \cdots & -1 & \left(\begin{array}{c}
0 \\
0
\end{array}\right) u^{\prime} v
\end{array}
\right)
$$
where $u^{(i)}$ denotes $u^{(i)}(x)$ and $v^{k}$ is to be interpreted as $v^{(k)}(u(x))$


TITLE: Why is angular momentum involved in a spinning ball hitting another?
QUESTION [0 upvotes]: One ball has a certain velocity and is spinning. It hits another ball, and each have a certain friction constant which kind of roughly defines how their surfaces interact.
Now, when I was in high-school and knew less about physics, I could explain the result of this based on the conservation of linear momentum and angular momentum. For linear momentum, this is quite simple and depends on the mass of each ball, for angular momentum there's both the fact that one ball is spinning and the angle at which it hits the other ball (which provides some torque) to take into account as well. The result will be both balls will have a different spin and different velocity but linear momentum and angular momentum are conserved. This can be seen in a way as a consequence of Noether's theorem because of the equivalence of rotated or translated inertial frames.
Now that I know more about physics, this is what happens in my mind and I'm terribly confused why angular momentum is still even a thing:
Both balls are collections of molecules with a certain structure which create intramolecular forces. Regardless of how one ball happened to end up spinning and moving, at the point of contact with the other ball, these molecules will impart linear momentum upon the other ball's molecules, and this linear momentum is caused by the ball spinning and also by the ball's velocity. This linear momentum between molecules needs to be conserved so each will get a share. From that point on, because the balls are squishy, more molecules will become involved and will impart linear momentum, while all the molecules in each ball will still be subject to their intramolecular forces.
Now, the intramolecular bonds in the previously idle ball will start propagating motion through the ball through basically Pauli's principle, and it will basically go at the speed of sound of the material of the ball. This will cause the linear momentum to propagate throughout the ball, but because of the intramolecular forces, they will bend this into a "spin" of the ball as the movement propagates.
As you can see, the final result is explained entirely by linear momentum. There is no such thing as angular momentum involved. Now, I don't want to become the kind of crackpot that says there is no angular momentum. I realise that in another perspective, it could be seen that linear momentum doesn't exist and all of this is just angular momentum being conserved, with a very large radius. However, what I can't see is how both concepts can be fundamental in classical mechanics. I just don't understand anymore how we ever really deal with angular momentum if I can explain everything by molecules imparting linear momentum. Or am I making a mistake?
Note that I have read this post Is Angular Momentum truly fundamental? and it doesn't explain my question.
Thanks for helping me reach deeper understanding,

REPLY [1 votes]: The statements below are from the perspective of rigid body dynamics
You are correct that the impulse exchanged between two bodies is pure momentum when described along a specific direction. With no friction, this is the contact normal, and with friction, it is some skew direction starting from the contact point.
But to an outside observer, you need angular momentum to describe the geometry of the problem. Angular momentum (vector) is analogous to torque (vector). Torque describes the geometry of the loading on a rigid body.

It describes how far an action occurs. It encodes the minimal distance and direction to the line of action. This is a line in 3D space needed to describe either a force (the line of action) or momentum (axis of percussion).

Different observers measure different values because they describe a line in space, which presents itself differently.

You recover the point of action of a force $F$ using list item #3 of this answer and the contact point of an impulse $p$ from item #2
$$\begin{aligned}
  \vec{r} & = \frac{ \vec{\tau} \times \vec{F} }{ \| \vec{F} \|^2} & & \text{point on line of action of force closest to origin}\\
  \vec{r} & = \frac{ \vec{L} \times \vec{p} }{ \| \vec{p} \|^2} & & \text{point on line of action of impulse closest to origin}\\
\end{aligned}$$

It is not a real vector, but a pseudo-vector.

Your question is analogous to asking "do we need torque to describe the loading on a body when only a single force is applied?"
The answer is that you need the force vector, and either a point where the force is going through or the equipollent torque $\vec{\tau} = \vec{r} \times \vec{F}$.
Similarly, for any contact where linear momentum is exchanged between two bodies along a line of action, you need to specify either the point of contact, or the equipollent momentum (angular momentum vector) $\vec{L} = \vec{r} \times \vec{p}$


TITLE: Baby rudin theorem 8.8
QUESTION [2 upvotes]: how he found the two inequalities (56 and the last)?

Thanks!

REPLY [1 votes]: The last inequality is achieved as follows.
$ |Q(re^{i\theta})|$
$=|1+b_k r^k e^{ik\theta}+b_{k+1}r^{k+1}e^{i(k+1)\theta}+ \dots +b_{n}r^{n}e^{in\theta}|$
$\le|1+b_k r^k e^{ik\theta}|+|b_{k+1}r^{k+1}e^{i(k+1)\theta}|+ \dots +|b_{n}r^{n}e^{in\theta}|$
$=1-r^k |b_k|+|b_{k+1}|r^{k+1}+ \dots +|b_{n}|r^{n}$
$=1-r^k$ {$|b_k|-r|b_{k+1}|-\dots-r^{n-k}|b_{n}|$ }


TITLE: Finding the diagonal side of an isosceles triangle
QUESTION [2 upvotes]: Using geometry I would like to find the diagonal length of an isosceles  triangle. I only know that the base of the triangle is 36” long, The lower left angle is 45° and the lower right angle is 90° I don’t know the height. Using geometry how would I find the length of the diagonal leg of the triangle.

REPLY [0 votes]: My assumption is that the "base" is the horizontal. Since one of the angles next to the base is $90^\circ$, then the "height" (vertical) will be the same as the "base" $a=b=36$. The reason is that the triangle is isosceles. Then use Pythagoras' to get the hypotenuse (which will look diagonal in this picture):
$$c=\sqrt{a^2+b^2}=36\sqrt 2$$


TITLE: Prove that if $a_n$ is strictly decreasing, $a_n \rightarrow 0$ and $n_0$ is even, then $ \sum_{n=n_0}^{+ \infty } a_n>0$
QUESTION [1 upvotes]: I have a problem with this task because I don't know how much I must write here.
If   $a_n$ is strictly decreasing and $a_n \rightarrow 0$, then I know that $a_n$ is limited from below by $0$ and all words of $a_n$ are $>0$, because if $a_n<0$ and $a_n$ strictly decreasing, then $a_n \rightarrow - \infty$, but it is contradiction.
If I know that all $a_n$ is $>0$ (because I prove it), then I know that $a_{n_0}+a_{n_1}+a_{n_2}+...= \sum_{n=n_0}^{+ \infty } a_n>0$. However I am afraid that this is not enough evidence. Moreover I don't know for what purpose the information is given that $ a_ {n_0} $ is even.

REPLY [3 votes]: Your assumption that if $\exists n\in \mathbf{N}$ such that $a_n<0$ then that $\{a_n\}$ is strictly decreasing should imply that $a_n\rightarrow -\infty$ is false. Consider $a_n = -1+\frac{1}{n}$ which is strictly decreasing but $a_n\rightarrow -1$ as $n\rightarrow \infty$.
Instead if $a_{n_0}<0$ for some $n_{0}$ then you can conclude that $a_n\not\rightarrow 0$. Do you see how? Consider some neighborhood of $0$ which does not contain any $a_n$ with $n>n_0$.
The fact that $n_0$ is even is immaterial. If $a_n>0$ for every $n$ then
$$\sum_{n=n_0}^{\infty}a_n>0$$