Sheet 2

Prof. Leif Döring, Felix Benning

^†^†course: Wahrscheinlichkeitstheorie 1^†^†semester: FSS 2022^†^†tutorialDate: 28.02.2022^†^†dueDate: 10:15 in the exercise on Monday 28.02.2022

Exercise 1.

Calculate the following conditional expectations and determine the conditional distributions.

(i)

$\mathbb{E}\left[(Y-X)^{+}|X\right]$ , where $X, Y$ are independent uniform random variables on $[0,1]$ .

Solution.

Using our kernel representation we get

$\displaystyle\mathbb{E}[(Y-X)^{+}\,|\,X]$ $\displaystyle=\int(y-X)^{+}\mathbb{P}(Y\in dy\,|\,X)$

$\displaystyle\stackrel{{\scriptstyle\text{independent}}}{{=}}\int(y-X)^{+}% \mathbb{P}(Y\in dy)$

$\displaystyle=\int_{0}^{1}(y-X)^{+}dy=\int_{X}^{1}y-Xdy=\int_{0}^{1-X}sds$

$\displaystyle=\tfrac{1}{2}(1-X)^{2}\qed$

(ii)

$\mathbb{E}\left[X|Y\right]$ , where $(X,Y)$ has a joint distribution with density:

f(x,y)=4y(x-y)e^{-(x+y)}\mathbbm{1}_{0<y<x}.

Solution.

We know that

\displaystyle\mathbb{E}[X\,|\,Y]=\int x\mathbb{P}(X\in dx\,|\,Y)=\int xf(x\,|% \,Y)dx

with

\displaystyle f(x\,|\,y)=\frac{f(x,y)}{f_{Y}(y)},

where

	$\displaystyle f_{Y}(y)$	$\displaystyle=\int f(x,y)dx=\int 4y(x-y)e^{-(x+y)}\mathbbm{1}_{0<y<x}dx=\int_{% y}^{\infty}4y(x-y)e^{-(x+y)}dx$
		$\displaystyle\stackrel{{\scriptstyle z=x-y}}{{=}}\int_{0}^{\infty}4yze^{-(z+2y% )}dz=4ye^{-2y}\underbrace{\int_{0}^{\infty}ze^{-z}dz}_{=\Gamma(2)=1!}$
		$\displaystyle=4ye^{-2y}.$

Using

	$\displaystyle\int xf(x,y)dx$	$\displaystyle=\int_{y}^{\infty}4xy(x-y)e^{-(x+y)}dx=\int_{0}^{\infty}4(z+y)yze% ^{-(z+2y)}dz$
		$\displaystyle=4ye^{-2y}\int_{0}^{\infty}(z+y)ze^{-z}dz=4ye^{-2y}(\Gamma(3)+y% \Gamma(2))$
		$\displaystyle=4ye^{-2y}(2+y)$

we therefore get

\displaystyle\mathbb{E}[X\,|\,Y]

\displaystyle=\int xf(x\,|\,Y)dx=\frac{\int xf(x,Y)dx}{f_{Y}(Y)}=\frac{4Ye^{-2% Y}(2+Y)}{4Ye^{-2Y}}=2+Y.\qed

(iii)

$\mathbb{E}\left[X|X+Y\right]$ , where $X, Y$ are independent Poisson random variables with parameters $\lambda$ and $\mu$ respectively.

Solution.

Since $X$ and $Y$ are both Poission we are in the discrete setting. So first, we want to compute the conditional probability:

\displaystyle\mathbb{P}(X=x\,|\,X+Y=z)=\frac{\mathbb{P}(X=x,\>X+Y=z)}{\mathbb{% P}(X+Y=z)}.

Let us calculate each part separately. Starting with the nominator, we get:

	$\displaystyle\mathbb{P}(X=k,\>X+Y=z)$	$\displaystyle=\mathbb{P}(X=x,\>Y=z-x)$
		$\displaystyle=\frac{\lambda^{x}e^{-\lambda}}{x!}\cdot\frac{\mu^{z-x}e^{-\mu}}{% (z-x)!}\mathbbm{1}_{z\geq x}$

For the denominator we can use the results from Stochastics 1, that $X+Y\sim\text{Poi}(\lambda+\mu)$ . In case, you have not seen it so far, the calculation below shows the claim from Sto1:

	$\displaystyle\mathbb{P}(X+Y=z)$	$\displaystyle=\sum_{x\in\mathbb{N}_{0}}\mathbb{P}(X=x,\>Y=z-x)$
		$\displaystyle=\sum_{x=0}^{z}\frac{\lambda^{x}e^{-\lambda}}{x!}\cdot\frac{\mu^{% z-x}e^{-\mu}}{(z-x)!}$
		$\displaystyle=\frac{e^{-(\lambda+\mu)}}{z!}\underbrace{\sum_{x=0}^{z}\frac{z!}% {x!(z-x)!}\mu^{z-x}\lambda^{x}}_{=(\mu+\lambda)^{z}}$
		$\displaystyle=\frac{(\mu+\lambda)^{z}}{z!}e^{-(\lambda+\mu)}$

Now we can calculate the conditional expectation $\mathbb{E}\left[X|X+Y\right]$ :

	$\displaystyle\mathbb{E}\left[X\|X+Y\right]$	$\displaystyle=\sum_{x\in\mathbb{N}_{0}}x\cdot\mathbb{P}(X=x\,\|\,X+Y=z)$
		$\displaystyle=\sum_{x=0}^{z}x\frac{\frac{\lambda^{x}e^{-\lambda}}{x!}\frac{\mu% ^{z-x}e^{-\mu}}{(z-x)!}}{\frac{(\mu+\lambda)^{z}}{z!}e^{-(\lambda+\mu)}}$
		$\displaystyle=\frac{\lambda}{(\mu+\lambda)^{z}}\sum_{x=0}^{z}\left(\frac{d}{d% \lambda}\lambda^{x}\right)\mu^{z-x}\frac{z!}{x!(z-x)!}$
		$\displaystyle=\frac{\lambda}{(\mu+\lambda)^{z}}\frac{d}{d\lambda}\underbrace{% \sum_{x=0}^{z}\lambda^{x}\mu^{z-x}\binom{z}{x}}_{=(\lambda+\mu)^{z}}$
		$\displaystyle=\frac{\lambda\cdot z}{\mu+\lambda}$

Thus, we can conclude that $\mathbb{E}\left[X|X+Y\right]=\frac{\lambda}{\lambda+\mu}(X+Y)$ . ∎

Exercise 2 ((Sub-/Super-)Martingales).

(i)

The (sub-/super-)martingale property holds over several time steps, i.e.

$\displaystyle\mathbb{E}[X_{m}\,|\,\mathcal{F}_{n}]\begin{cases}=X_{n}&\text{% martingale}\\ \leq X_{n}&\text{supermartingale}\\ \geq X_{n}&\text{submartingale}\end{cases}\quad\forall m\geq n$

Solution.

We use induction over $m$ with induction start $n$ , so assuming the statement holds for $m$ we have for $m+1$

$\displaystyle\mathbb{E}[X_{m+1}\,|\,\mathcal{F}_{n}]$ $\displaystyle=\mathbb{E}[\mathbb{E}[X_{m+1}\,|\,\mathcal{F}_{m}]\,|\,\mathcal{% F}_{n}]\lesseqgtr\mathbb{E}[X_{m}\,|\,\mathcal{F}_{n}]\lesseqgtr X_{n}.\qed$
(ii)

Expectation increase/decrease/stay-constant, i.e.

$\displaystyle\mathbb{E}[X_{m}]\begin{cases}=\mathbb{E}[X_{n}]&\text{martingale% }\\ \leq\mathbb{E}[X_{n}]&\text{supermartingale}\\ \geq\mathbb{E}[X_{n}]&\text{submartingale}\end{cases}\quad\forall m\geq n$

Solution.

Simply taking the expectation over the previous result yields the claim. ∎
(iii)

$X$ is a supermartingale iff $-X$ is a submartingale.

Solution.

Adapted and $\mathbb{E}[|X_{n}|]<\infty$ follows from both statements. So let $X$ be a supermartingale, then we have

$\displaystyle\mathbb{E}[-X_{n+1}\,|\,\mathcal{F}_{n}]=-\underbrace{\mathbb{E}[% X_{n+1}\,|\,\mathcal{F}_{n}]}_{\leq X_{n}}\geq-X_{n}$

therefore $-X$ is a submartingale. The opposite direction is analogous. ∎

Exercise 3 (Markov Inequality for Conditional Expectation).

Prove that for two random variables $X, Y$ and an increasing $h:[0,\infty)\to[0,\infty)$ we have for any $\epsilon>0$

\displaystyle\mathbb{P}(|X|\geq\epsilon\,|\,Y)\leq\frac{\mathbb{E}[h(|X|)\,|\,% Y]}{h(\epsilon)}\quad\text{a.s.}

Remark.

The proof works the same when one conditions on a sigma algebra instead of a random variable $Y$ .

Solution.

Simply using monotonicity of conditional expectation we get

	$\displaystyle\mathbb{P}(\|X\|\geq\epsilon\,\|\,Y)$	$\displaystyle=\mathbb{E}[\mathbbm{1}_{\|X\|\geq\epsilon}\,\|\,Y]\leq\mathbb{E}[% \mathbbm{1}_{h(\|X\|)\geq h(\epsilon)}\,\|\,Y]\leq\mathbb{E}\left[\frac{h(\|X\|)}{h% (\epsilon)}\mathbbm{1}_{h(\|X\|)\geq h(\epsilon)}\,\|\,Y\right]$
		$\displaystyle\leq\frac{\mathbb{E}[h(\|X\|)\,\|\,Y]}{h(\epsilon)}.\qed$

Exercise 4.

Let $X, Y$ be independent with standard normal distribution $\mathcal{N}(0,1)$ . Show that, for any $z\in\mathbb{R}$ , the conditional distribution $\mathbb{P}(X\in\cdot\,|\,X+Y=z)$ is $\mathcal{N}(z/2,1/2)$ .

Solution.

We know that

\displaystyle\begin{pmatrix}X\\ X+Y\end{pmatrix}=\underbrace{\begin{pmatrix}1&0\\ 1&1\end{pmatrix}}_{=:A}\begin{pmatrix}X\\ Y\end{pmatrix}\sim\mathcal{N}(0,\Sigma),

with

\displaystyle\Sigma=AA^{T}=\begin{pmatrix}1&1\\ 1&2\end{pmatrix}.

With

\displaystyle\det(\Sigma)=2-1=1\quad\text{and}\quad\Sigma^{-1}=\begin{pmatrix}% 2&-1\\ -1&1\end{pmatrix}

we can write the density of this vector explicitly

	$\displaystyle f(x,z)$	$\displaystyle=\det(2\pi\Sigma)^{-\tfrac{1}{2}}\exp\left(-\tfrac{1}{2}\begin{% pmatrix}x&z\end{pmatrix}\Sigma^{-1}\begin{pmatrix}x\\ z\end{pmatrix}\right)$
		$\displaystyle=[(2\pi)^{2}\det(\Sigma)]^{-\tfrac{1}{2}}\exp\left(-\tfrac{1}{2}% \begin{pmatrix}2x-z&-x+z\end{pmatrix}\begin{pmatrix}x\\ z\end{pmatrix}\right)$
		$\displaystyle=\tfrac{1}{\|2\pi\|}\exp(-\tfrac{1}{2}(2x^{2}-zx-xz+z^{2})).$

Lastly we have $X+Y\sim\mathcal{N}(0,2)$ and therefore the marginal density is

\displaystyle f_{X+Y}(z)=\tfrac{1}{\sqrt{4\pi}}\exp(-\tfrac{z^{2}}{4}).

This allows us to calculate the conditional density

	$\displaystyle f_{X\|X+Y}(x\|z)$	$\displaystyle=\frac{f(x,z)}{f_{X+Y}(z)}=\frac{\sqrt{4\pi}}{2\pi}\exp(-\tfrac{1% }{2}(2x^{2}-2zx+z^{2})+\tfrac{z^{2}}{4})$
		$\displaystyle=\frac{1}{\sqrt{\pi}}\exp(-\tfrac{1}{2}(2x^{2}-2zx+\tfrac{1}{2}z^% {2}))$
		$\displaystyle=\frac{1}{\sqrt{\pi}}\exp\left(-(x^{2}-2\tfrac{z}{2}x+\tfrac{1}{4% }z^{2})\right)$
		$\displaystyle=\frac{1}{\sqrt{\pi}}\exp\left(-(x-\tfrac{z}{2})^{2}\right).$

Which finally implies $\mathbb{P}(X\in\cdot\,|\,X+Y)\sim\mathcal{N}(\tfrac{z}{2},\tfrac{1}{2})$ . ∎

Exercise 5 (Stopping Times).

(i)

Suppose $T_{1},T_{2},\dots$ are $(\mathcal{F}_{n})$ -stopping times. Prove that $\inf_{k\in\mathbb{N}}T_{k}$ , $\sup_{k\in\mathbb{N}}T_{k}$ , $\liminf_{k\to\infty}T_{k}$ and $\limsup_{k\to\infty}T_{k}$ are stopping times.
Solution.
$S$ is a stopping time, iff $\{S\leq n\}\in\mathcal{F}_{n}\}$ . So we just need to check this definition in every case:
1. (a)
  
  $\{\inf_{k\in\mathbb{N}}T_{k}\leq n\}=\bigcup_{k\in\mathbb{N}}\{T_{k}\leq n\}% \in\mathcal{F}_{n}$
2. (b)
  
  $\{\sup_{k\in\mathbb{N}}T_{k}\leq n\}=\bigcap_{k\in\mathbb{N}}\{T_{k}\leq n\}% \in\mathcal{F}_{n}$
3. (c)
  
  $\{\liminf_{k\in\mathbb{N}}T_{k}\leq n\}=\{\lim_{k}\inf_{j\geq k}T_{j}\leq n\}=% \bigcap_{k\in\mathbb{N}}\{\inf_{j\geq k}T_{j}\leq n\}\in\mathcal{F}_{n}$
4. (d)
  
  $\{\limsup_{k\in\mathbb{N}}T_{k}\leq n\}=\{\lim_{k}\sup_{j\geq k}T_{j}\leq n\}=% \bigcup_{k\in\mathbb{N}}\{\sup_{j\geq k}T_{j}\leq n\}\in\mathcal{F}_{n}$ ∎

(ii)

Suppose $T, S$ are $\mathcal{F}_{n}$ stopping times. Prove $\{S\leq T\}\in\mathcal{F}_{S\wedge T}$ and $\{S=T\}\in\mathcal{F}_{S\wedge T}$ .

Solution.

To prove $A\in\mathcal{F}_{S\wedge T}$ one needs to prove $A\cap\{S\wedge T\leq n\}\in\mathcal{F}_{n}$ , so

	$\displaystyle\{S\leq T\}\cap\{S\wedge T\leq n\}$	$\displaystyle=\left[\{S\leq n,S\leq T\}\cup\{S>n,S\leq T\}\right]\cap\{S\wedge T% \leq n\}$
		$\displaystyle=\left[\bigcup_{k=0}^{n}\{S\leq k\leq T\}\cup\{S>n,S\leq T\}% \right]\cap\{S\wedge T\leq n\}$
		$\displaystyle=\bigcup_{k=0}^{n}\{S\leq k\leq T\}\cup\underbrace{\{S\geq n,S% \leq T,S\wedge T\leq n\}}_{=\emptyset}$
		$\displaystyle=\bigcup_{k=0}^{n}\Big{[}\underbrace{\{S\leq k\}}_{\in\mathcal{F}% _{k}\subseteq\mathcal{F}_{n}}\cap\underbrace{\{k\leq T\}}_{\in\mathcal{F}_{k}% \subseteq\mathcal{F}_{n}}\Big{]}\in\mathcal{F}_{n}$

Similarly we have $\{T\leq S\}\in\mathcal{F}_{S\wedge T}$ , therefore their intersection $\{S=T\}$ is also in $\mathcal{F}_{S\wedge T}$ . ∎

(iii)

(Last Hitting Time is no Stopping Time in General) Consider the stochastic process $X$ defined by $X_{0}=0$ , $X_{1}\sim\mathcal{U}(\{0,1\})$ and $X_{k}=1$ for all $k>1$ . Let $\mathcal{F}_{n}=\sigma(X_{1},\dots,X_{n})$ be the natural filtration generated by $X$ . Show that the last hitting time of $0$

$\displaystyle L:=\sup\{k\geq 0:X_{k}=0\}$

is not a stopping time.

Solution.

We have $\mathcal{F}_{0}=\{\emptyset,\Omega\}$ and

$\displaystyle L=\begin{cases}0&X_{1}=1\\ 1&X_{1}=0\end{cases}$

therefore $L$ is not a stopping time since

$\displaystyle\{L\leq 0\}$ $\displaystyle=\{X_{1}=1\}\notin\mathcal{F}_{0}.\qed$

	$\displaystyle\mathbb{E}[(Y-X)^{+}\,\|\,X]$	$\displaystyle=\int(y-X)^{+}\mathbb{P}(Y\in dy\,\|\,X)$
		$\displaystyle\stackrel{{\scriptstyle\text{independent}}}{{=}}\int(y-X)^{+}% \mathbb{P}(Y\in dy)$
		$\displaystyle=\int_{0}^{1}(y-X)^{+}dy=\int_{X}^{1}y-Xdy=\int_{0}^{1-X}sds$
		$\displaystyle=\tfrac{1}{2}(1-X)^{2}\qed$