TITLE: The positivity of matrix with one parameter changing
QUESTION [2 upvotes]: Set
$$D=\begin{bmatrix}0\\d_{1}&0\\d_{2}&d_{1}&0\\\vdots&\vdots&\ddots&\ddots\\d_{n}&d_{n-1}&\cdots&d_{1}&0\end{bmatrix}$$
and
$$W(\alpha)=(I-\alpha D)^{-1}-\frac{1}{2}I$$
$$\Sigma(\alpha)=W(\alpha)+W^{*}(\alpha)$$
where $\alpha$ is a real number in $[0,1]$, $*$ represents conjugate transpose.
The question is:
suppose $$\Sigma(1)>0$$, if we can think $$\Sigma(\alpha)>0$$ with $\alpha $ changing from $0$ to $1$ ? Thanks a lot
I know that $W(\alpha)$ is also a lower triangular matrix with diagonal elements $\frac{1}{2}$.

REPLY [1 votes]: Let $C=(I-\alpha D)$. Then
\begin{align}
\Sigma(\alpha)
&= (C^\ast)^{-1} + C^{-1} - I\\
&= (C^\ast)^{-1}(C + C^\ast - C^\ast C)C^{-1}\\
&= (C^\ast)^{-1}\left[I-(C^\ast-I)(C-I)\right]C^{-1}\\
&= (C^\ast)^{-1}(I-\alpha^2D^\ast D)C^{-1}\\
\end{align}
Since $\Sigma(1)\succ0$, we must have $I\succ D^\ast D$. Hence $I-\alpha^2D^\ast D$ and in turn $\Sigma(\alpha)$ are positive definite for every $-1\le\alpha\le1$.


TITLE: Find limits of value/derivatives defining a polynomial at 2 points to bound it in between
QUESTION [0 upvotes]: The following properties define the polynomial $p(x)$ uniquely:
$\text{deg}(p(x))=7\\p(-1)=y_1,\ p'(-1)=d_{1,1},\ p''(-1)=d_{2,1},\ p'''(-1)=d_{3,1}\\
p(1)=y_2,\ \ \ \ p'(1)=d_{1,2},\ \ \ \ p''(1)=d_{2,2},\ \ \ \ p'''(1)=d_{3,2}$
The parameters $y_1,\ d_{1,1},\ d_{2,1},\ d_{3,1},\ y_2,\ d_{1,2},\ d_{2,2},\ d_{3,2}$ are real and chosen in such a way that the implication $-1\leq x\leq 1\Rightarrow-1\leq p(x)\leq1$ is true.
For each parameter, I need to find all real numbers that, when replacing the parameter,  make the implication $-1\leq x\leq 1\Rightarrow-1\leq p(x)\leq1$ true.
There is always one solution, because, for instance, if I replace $y_1$ with a real number $r$, one solution is $r=y_1$. Moreover, the complete solution is an interval, $a\leq r\leq b$, because all points with $x\in[-1,1]$ will move the same way, when a parameter is modified monotonously.
How should this be done numerically? (I don't think there is an general exact solution)

REPLY [0 votes]: You could start with the convex combinations of the Chebyshev polynomials $T_n(x)=\cos(n\arccos(x))$. More general than the constants functions and also than just the convex combinations of the monomials.

More precisely, the convex combinations of $T_n$ and $-T_n$, so after collecting the coefficients for positive and negative versions,
$$\sum_{n=0}^7c_kT_n(x)\text{ with }\sum_{k=0}^7|c_k|\le 1.$$


TITLE: Adjoints to cofree modules tensor?
QUESTION [2 upvotes]: If $M$ is a cofree $R$-module and $A,B$ are arbitrary $R$-modules then, is there a left adjoint to the functor $M\otimes_R -$,  i.e. is there an endofunctor $F$ on $_R \mathrm{Mod}$ such that $\mathrm{Hom}_R(F(A),B) \cong \mathrm{Hom}_R(A,M\otimes_R B)$?

REPLY [1 votes]: $M \otimes_R -$ is almost never continuous and therefore has no left adjoint.


TITLE: Density Function of $F\left(Y_1\right)+F\left(Y_2\right)$
QUESTION [0 upvotes]: I came across a question which stated:

Suppose $Y_1,Y_2,...$ are independently and identically distributed random variables with common distribution function $F$. Then find the pdf of $$U=F\left(Y_1\right)+F\left(Y_2\right)$$

From what I understood of the question, I proceeded as follows (which, I assume, is not correct as I couldn't understand the question properly):
I tried to find the CDF of $U$ and I wrote this
$$\begin{align}P\left(U\leq u\right)&=P\left(F\left(Y_1\right)+F\left(Y_2\right)\leq u\right)\\
&=\int_0^1P\left(F\left(Y_1\right)+y_2\leq u\right)\cdot P\left(F\left(Y_2\right)\leq y_2\right)\cdot dy_2\\
&=\int_0^1\left(F\left(Y_1\right)\leq u-y_2\right)\cdot\left(F\left(Y_2\right)\leq y_2\right)\cdot dy_2
\end{align}$$
I cannot go any further and am confused as to how this problem can be done. Any suggestions and hints will be very much helpful.

Some thoughts crept into my mind as I was writing this down here:
Suppose I take another random variable (say $X$) such that I can write the above integral as
$$\int_0^1\left(P\left(X\leq Y_1\right)\leq u-y_2\right)\cdot\left(P\left(X\leq Y_2\right)\leq y_2\right)\cdot dy_2$$
But I still cannot figure out anything...

Upon a request, I'm uploading a picture of the question here:

REPLY [1 votes]: Let $V:=F(Y)$, where $F$ is the distribution of $Y$. Then $V$ is a random variable taking values in $(0,1)$. For simplicity, assuming that $F$ is strictly monotone and continuous
$P[V\leq v]=P[F(Y)\leq v]=P[Y\leq F^{-1}(v)]=F(F^{-1}(v))=v$
For more general distriutions, you may use the quantile function $Q(q):=\inf\{x:F(x)\geq q\}$ which works like an inverse for $F$.


TITLE: Question about solution to showing $\mathbb{P}(\lim_{s\to t} X_s \;\text{exists})=0$ if $X_t$ are independent $N(0,t)$.
QUESTION [1 upvotes]: This is a problem from Rene Schilling's Brownian Motion.
Let $(X_t)_{t\ge 0}$ be a stochastic process such that $X_t$ are independent  $N(0,t)$ random variables. Show that $\mathbb{P}(\lim_{s\to t} X_s \;\text{exists})=0$ for every $t>0$.
Below is the solution to the problem. In the solution, I'm not sure how we get a uniform q by step 1. So in step 1  we find out that for any sequence $t_n \to t$, we have $P(\lim X_{t_n}\; \text{exists}) <1$. But $P(\lim X_{t_n}\; \text{exists})$  depends on the particular sequence $t_n$ chosen. So how do we get a uniform $q \in (0,1)$ that is $q > P(\lim X_{t_n}\; \text{exists})$ for all $t_n$?

REPLY [2 votes]: I think there are numerous problems in the solution, though the idea is somewhat clear.
Consider the following proof instead.
Take $t>0$, and fix a sequence $t_n\to t$ such that it is strictly decreasing, so that, in particular, each $t_n > t$. Consider the event $$E:=\{\omega\mid X_{t_n}(\omega)\mbox{ converges}\}.$$ Fix $\epsilon > 0$. Consider further the event $$E_n:=\{\omega\mid \forall m > n: |X_{t_m}(\omega)-X_{t_n}(\omega)|<\epsilon\}.$$ Since for every $m > n$, $t<t_m<t_n$, $X_{t_m}$ is independent of $X_{t_n}$ having variance at least $t$, and the normal distribution $N(0,t_m)$ is unimodal centered at $0$, hence, $$\begin{align*}\mathbb{P}\{|X_{t_m}-X_{t_n}|<\epsilon\} & \le\mathbb{P}\{|X_{t_m}|<\epsilon\} \\ & \le \mathbb{P}\{|N(0,t)|<\epsilon\} \\ & =c \\ & <1\mbox{ (as }t>0\mbox{)}.\end{align*}$$ From here, you conclude that $$\mathbb{P}\{E_n\}=0,$$and, since $$E\subset \cup_n E_n,$$we have $$\mathbb{P}\{E\}=0.$$ Then, the result follows, as $$\{\omega\mid \exists\lim X_s(\omega)\mbox{ as }s\to t\}\subset E.$$
P.S. There was a question, how to show that $\mathbb{P}\{E_n\}=0$, because $X_{t_m}-X_{t_n}$ are not independent.
Well, why this is true, it does not really matter in this case. The inequalities showing that $\mathbb{P}\{|X_{t_m}-X_{t_n}|<\epsilon\}\le c<1$ mean that this holds conditional on any value of $X_{t_n}$, and, hence, on any event.
Formally, define $$E_{n,k}=\{\omega\mid |X_{t_m}-X_{t_n}|<\epsilon\mbox{ for }m=n+1,\dotsc,n+k\}.$$ Further, define $$F_{n,k}=\{\omega\mid |X_{t_{n+k}}-X_{t_n}|<\epsilon\}.$$ Then, $$\begin{align*}\mathbb{P}\{E_{n,k+1}\} & =\mathbb{P}\{E_{n,k}\cap F_{n,k+1}\} \\ & =\mathbb{P}\{F_{n,k+1}\mid E_{n,k}\}\mathbb{P}\{E_{n,k}\} \\ & \le c \cdot \mathbb{P}\{E_{n,k}\}\end{align*}$$by the above, and $$E_n=\cap_k E_{n,k}.$$


TITLE: Is motion really relative?
QUESTION [2 upvotes]: I had a question that had been bothering me for some time. I tried to come up with an explanation but I wanted to know whether it is right.

Imagine two different observers say A & B floating in space in their corresponding frames say P&Q with nothing else surrounding them. A observes that B is accelerating away from him and B observes that A is accelerating away from him. I feel that this situation is completely symmetrical for both A & B and thus if the motion is actually relative, then the observations of both A & B must be similar. However, then how is it possible that only one of them say, A, happens to observe a ball in his hand following Newton's laws of motion showing that he is in an inertial frame of reference whereas a ball in the hands of B seems to be acted upon by pseudoforces?
  Basically, my question is what is the reason for the seemingly asymmetry between them ?

REPLY [3 votes]: Motion is relative but acceleration is absolute. You can know if you are being accelerated without any reference to the outside world. You will feel it in your internal organs and if you carry out an experiment you will observe pseudo forces, even if you don't look outside your space ship.
If a body A feels none of these effects and sees that another body B is being accelerated ´relative´ to it, then A is in an inertial frame (but it is unknown whether it is at rest, since this is relative) and B must be accelerating. This is not symmetrical, since B will feel an acceleration.