Sheet 5

Prof. Leif Döring, Felix Benning

^†^†course: Wahrscheinlichkeitstheorie 1^†^†semester: FSS 2022^†^†tutorialDate: 21.03.2022^†^†dueDate: 10:15 in the exercise on Monday 21.03.2022

Exercise 1 (Uniform Integrability).

(i)

(Prop. 6.3.14 (ii) $\Rightarrow$ (i)) $(X_{\alpha},\alpha\in I)$ a family of random variables with $\sup_{\alpha\in I}\mathbb{E}[|X_{\alpha}|]<\infty$ , i.e. bounded in $\mathcal{L}^{1}(\Omega,\mathcal{F},\mathbb{P})$ . Suppose that for every $\epsilon>0$ , there exists $\delta>0$ , such that for all $A\in\mathcal{F}$ with $\mathbb{P}(A)<\delta$ , we have

$\mathbb{E}[|X_{\alpha}|\mathbbm{1}_{A}]<\epsilon,\qquad\forall\alpha\in I.$

Prove that the $X_{\alpha}$ are uniformly integrable.

Solution.

Let $\epsilon>0$ . By the Markov inequality we have

$\displaystyle\mathbb{P}(\lvert X_{\beta}\rvert\geq M)\leq\frac{\mathbb{E}[% \lvert X_{\beta}\rvert]}{M}\leq\frac{\sup_{\alpha\in I}\mathbb{E}[\lvert X_{% \alpha}\rvert]}{M}\coloneqq\frac{c}{M}.$

So for all $M\geq M_{0}:=\lceil\frac{c}{\delta(\epsilon)}\rceil$ and for all $\beta\in I$ , we have

$\displaystyle\mathbb{P}(\lvert X_{\beta}\rvert\geq M)\leq\delta(\epsilon),$

where $\delta(\epsilon)$ is selected such, that by assumption we get

$\displaystyle\mathbb{E}[\lvert X_{\alpha}\rvert\cdot\mathbbm{1}_{\lvert X_{% \beta}\rvert\geq M}]\leq\epsilon\quad\forall\alpha\in I.$

Since $\delta(\epsilon)$ does not depend on $\beta$ , it holds for all $\beta\in I$ . So in particular we have

$\displaystyle\sup_{\alpha\in I}\mathbb{E}[\lvert X_{\alpha}\rvert\cdot\mathbbm% {1}_{\lvert X_{\alpha}\rvert\geq M}]\leq\epsilon\quad\forall M\geq M_{0}.$

As $\epsilon>0$ was arbitrary, this is by definition uniform integrability. ∎
(ii)

Prove that a sequence of identically distributed random variables $\{X_{i},i\in\mathbb{N}\}$ with $\mathbb{E}[|X_{1}|]<\infty$ is uniformly integrable.

Solution.

Since $\mathbb{E}[|X_{1}|]<\infty$ , we deduce by the DCT

$\displaystyle\lim_{M\to\infty}\mathbb{E}[|X_{1}|\mathbbm{1}_{|X_{1}|>M}]=% \mathbb{E}[|X_{1}|\underbrace{\lim_{M\to\infty}\mathbbm{1}_{|X_{1}|>M}}_{=0}]=0.$

Therefore for every $\epsilon>0$ , there exists $M>0$ , such that

$\mathbb{E}[|X_{1}|\mathbbm{1}_{|X_{1}|>M}]<\epsilon.$

The sequence is identically distributed, therefore, for every $i\in\mathbb{N}$ , we conclude that

$\sup_{i\in\mathbb{N}}\mathbb{E}[|X_{i}|\mathbbm{1}_{|X_{i}|>M}]=\mathbb{E}[|X_% {1}|\mathbbm{1}_{|X_{1}|>M}]<\epsilon.\qed$
(iii)

Let $(X_{i},i\in\mathbb{N})$ be a sequence of i.i.d. random variables with $\mathbb{E}[|X_{1}|]<\infty$ . Let $S_{n}=\sum_{i=1}^{n}X_{i}$ for every $n\in\mathbb{N}$ . Then the family $\{\frac{1}{n}S_{n},n\in\mathbb{N}\}$ is uniformly integrable.

Solution.

Using (ii), $(X_{i},i\in\mathbb{N})$ is uniformly integrable. By Proposition 6.3.14 (Theorem 6.24 in Klenke), for every $\epsilon>0$ , there exists $\delta>0$ (dependent on $\epsilon$ but not on $i$ ), such that for every measurable set $A$ with $\mathbb{P}(A)<\delta$ ,

$\sup_{i\in\mathbb{N}}\mathbb{E}[|X_{i}|\mathbf{1}_{A}]<\epsilon.$

Then for every $n\in\mathbb{N}$ ,

$\mathbb{E}\bigg{[}\frac{|S_{n}|}{n}\mathbbm{1}_{A}\bigg{]}\leq\frac{1}{n}\sum_% {i=1}^{n}\mathbb{E}[|X_{i}|\mathbf{1}_{A}]<\epsilon.$

We also have $(S_{n}/n,n\in\mathbb{N})$ is bounded in $L^{1}$ :

$\sup_{n\in\mathbb{N}}\mathbb{E}\bigg{[}\frac{|S_{n}|}{n}\bigg{]}\leq\sup_{n\in% \mathbb{N}}\frac{1}{n}\mathbb{E}\left[\sum_{i=1}^{n}|X_{i}|\right]=\mathbb{E}[% |X_{1}|]<\infty.$

Then by Proposition 6.3.14 (Theorem 6.24 in Klenke), $(S_{n}/n,n\in\mathbb{N})$ is uniformly integrable. ∎

Exercise 2 (Not Uniformly Integrable).

Let $(U_{n},n\geq 1)$ be i.i.d. Bernoulli random variables with $\mathbb{P}(U_{n}=1)=p\in(0,1)$ and $\mathbb{P}(U_{n}=0)=1-p$ . Set $\tau:=\inf\{n\geq 1\colon U_{n}=1\}$ . Define $X_{n}:=(1-p)^{-n}\mathbbm{1}_{\tau>n}$ .

(i)

Show that $(X_{n})$ is a martingale (choose a suitable filtration).

Solution.

Choose $\mathcal{F}_{n}=\sigma(U_{1},\dots,U_{n})$ . Because of

\displaystyle\{\tau>n\}=\{U_{1}=0,\dots,U_{n}=0\}\in\mathcal{F}_{n},

$X_{n}$ is adapted, and we have

	$\displaystyle\mathbb{E}[X_{n}\,\|\,\mathcal{F}_{n-1}]$	$\displaystyle=(1-p)^{-n}\mathbb{E}[\mathbbm{1}_{\tau>n}\mathbbm{1}_{\tau>n-1}% \,\|\,\mathcal{F}_{n-1}]$
		$\displaystyle=(1-p)^{-n}\mathbb{E}[\mathbbm{1}_{\tau>n-1}\mathbbm{1}_{U_{n}=0}% \,\|\,\mathcal{F}_{n-1}]$
		$\displaystyle=(1-p)^{-n}\cdot\mathbbm{1}_{\tau>n-1}\cdot\mathbb{P}(U_{n}=0)$
		$\displaystyle=(1-p)^{-n}\cdot\mathbbm{1}_{\tau>n-1}$
		$\displaystyle=X_{n-1}.$

By induction and positivity of $X_{n}$ we also have $\mathbb{E}[|X_{n}|]=\mathbb{E}[X_{n}]=\mathbb{E}[X_{0}]=1<\infty$ . So $X_{n}$ is a martingale. ∎

(ii)

Prove that $X_{n}\to 0$ almost surely

Solution.

As $X_{n}$ is a martingale, we have

$\displaystyle 1=\mathbb{E}[X_{n}]=(1-p)^{-n}\mathbb{E}[\mathbbm{1}_{\tau>n}],$

so we get

$\displaystyle\sum_{n\in\mathbb{N}_{0}}\mathbb{P}(\mathbbm{1}_{\tau>n}>\epsilon% )=\sum_{n\in\mathbb{N}_{0}}\mathbb{P}(\tau>n)=\sum_{n\in\mathbb{N}_{0}}(1-p)^{% n}<\infty.$

By Borell-Cantelli we therefore have

$\displaystyle\mathbb{P}(\mathbbm{1}_{\tau>n}\to 0)$ $\displaystyle=1-\mathbb{P}(\exists\epsilon>0,\forall n_{0}\in\mathbb{N}\exists n% \geq n_{0}:\mathbbm{1}_{\tau>n}>\epsilon)$

$\displaystyle=1-\mathbb{P}\left(\bigcup_{\epsilon>0}\bigcap_{n_{0}\in\mathbb{N% }}\bigcup_{n\geq n_{0}}:\{\mathbbm{1}_{\tau>n}>\epsilon\}\right)$

$\displaystyle=1-\lim_{\epsilon\to 0}\underbrace{\mathbb{P}(\limsup_{n}\{% \mathbbm{1}_{\tau>n}>\epsilon\})}_{=0\quad\text{(BC)}}.$

$\mathbbm{1}_{\tau>n}\to 0$ almost surely implies there exists $N(\omega)$ large enough such that $\mathbbm{1}_{\tau>n}=0$ for all $n>N(\omega)$ . So we can conclude that

$\displaystyle X_{n}(\omega)=(1-p)^{-n}\cdot\mathbbm{1}_{\tau>n}(\omega)=0$

for all $n>N(\omega)$ . Thus, $X_{n}\to 0$ a.s. ∎
(iii)

Prove that $(X_{n},n\geq 1)$ is not uniformly integrable.

Solution.

If $(X_{n})_{n\in\mathbb{N}}$ was u.i., then $X_{n}\to X_{\infty}$ in $\mathcal{L}^{1}$ and $\mathbb{E}[X_{\infty}]=1$ by Theorem 6.3.16. This is however a contradiction to $X_{\infty}=0$ a.s. ∎

Exercise 3 (Tail $\sigma$ -Algebra).

(i)

$\mathcal{A}$ is trivial if an only if all $(\mathcal{A},\mathcal{B}(\mathbb{R}^{d}))$ -measurable functions are constant

Solution.

Assume that $\mathcal{A}$ is trivial and let $f:\Omega\to\mathbb{R}^{d}$ be $(\mathcal{A},\mathcal{B}(\mathbb{R}^{d}))$ -measurable, then there exists $y\in f(A)$ and therefore

$\displaystyle\emptyset\neq f^{-1}(\{y\})\in\mathcal{A}=\{\emptyset,\Omega\}$

therefore $f^{-1}(y)=\Omega$ which implies $f(\omega)=y$ for all $\omega\in\Omega$ .

Assume on the other hand that, that $\mathcal{A}$ is not trivial, then there exists some set $A\in\mathcal{A}$ with $\emptyset\neq A\neq\Omega$ . Then $\mathbbm{1}_{A}$ is measurable but not constant. ∎

Let $X_{1},X_{2},\dots$ be a sequence of random variables with $\tau_{n}=\sigma(X_{n},\dots)$ and tail $\sigma$ -algebra $\tau=\bigcap_{n}\tau_{n}$ .

(ii)

Prove that $\liminf_{n}X_{n}$ and $\limsup_{n}X_{n}$ are $\tau$ measurable.

Solution.

It is sufficient to prove measurability on a generator. In particular it is enough if the preimages of the sets $(-\infty,\lambda]$ for all $\lambda\in\mathbb{R}$ are measurable. Now for every $N\in\mathbb{N}$ we have

$\displaystyle\{\liminf_{n}X_{n}\leq\lambda\}$ $\displaystyle=\{\lim_{n\to\infty}\inf_{k\geq n}X_{k}\leq\lambda\}=\bigcap_{n=1% }^{\infty}\bigcup_{k\geq n}\{X_{k}\leq\lambda\}$

$\displaystyle=\bigcap_{n=N}^{\infty}\underbrace{\bigcup_{k\geq n}\{X_{k}\leq% \lambda\}}_{\in\tau_{n}\subseteq\tau_{N}}\in\tau_{N}$

therefore the set is in $\tau$ . Replacing $\inf$ with $\sup$ and unions with intersections and vice versa proves the same result for the $\limsup$ . ∎
(iii)

Prove that

$\displaystyle C:=\{\omega\in\Omega:\lim_{n\to\infty}X_{n}\ \text{exists}\}\in\tau$

Solution.

As

$\displaystyle Y:=\limsup_{n\to\infty}X_{n}-\liminf_{n\to\infty}X_{n}$

is $\tau$ -measurable by (ii), we have

$\displaystyle C$ $\displaystyle=\{\liminf_{n}X_{n}=\limsup_{n}X_{n}\}=Y^{-1}(0)\in\tau.\qed$

Exercise 4 (Gambler’s ruin).

Let $(X_{n})_{n\geq 1}$ be a sequence of i.i.d. random variables with $\mathbb{P}(X_{1}=1)=1-\mathbb{P}(X_{1}=-1)=p$ for some $p\in(0,1)$ , $p\neq 1/2$ . Let $a,b\in\mathbb{N}$ with $0<a<b$ . Define $S_{0}\coloneqq a$ and for every $n\geq 1$ , $S_{n}\coloneqq S_{n-1}+X_{n}$ . Finally, define the following stopping time:

\displaystyle T\coloneqq\inf\{n\geq 0:S_{n}=0\text{ or }S_{n}=b\}.

We consider the filtration generated by $(X_{n})_{n\geq 1}$ .

(i)

Show that $\mathbb{E}[T]<\infty$ .

Solution.

We have for every $n\geq 0$ , $\mathbb{P}(T\leq n+b|\mathcal{F}_{n})>\inf(p^{b},(1-p)^{b})>0$ . We deduce from Sheet 5, Exercise 5 that $\mathbb{E}[T]<\infty$ . ∎
(ii)

Consider for every $n\geq 0$ ,

$\displaystyle M_{n}\coloneqq\bigg{(}\frac{1-p}{p}\bigg{)}^{S_{n}}\quad\text{% and}\quad N_{n}\coloneqq S_{n}-n(2p-1).$

Prove that $(M_{n})_{n\geq 0}$ and $(N_{n})_{n\geq 0}$ are martingales.

Solution.

$(S_{n})_{n\geq 0}$ is adapted to the filtration $(\mathcal{F}_{n})_{n\geq 0}$ generated by $(X_{n})_{n\geq 0}$ and then so are $(M_{n})_{n\geq 0}$ and $(N_{n})_{n\geq 0}$ . They are integrable: clear for $(N_{n})_{n\geq 0}$ and comes from $|S_{n}|\leq a+n$ for every $n\geq 0$ for $(M_{n})_{n\geq 0}$ . Then for every $n\geq 0$ ,

$\displaystyle\mathbb{E}[M_{n+1}|\mathcal{F}_{n}]=M_{n}\mathbb{E}\bigg{[}\bigg{% (}\frac{1-p}{p}\bigg{)}^{X_{n+1}}\bigg{]}=M_{n}\bigg{[}p\frac{1-p}{p}+(1-p)% \frac{p}{1-p}\bigg{]}=M_{n},$

and

$\displaystyle\mathbb{E}[N_{n+1}-N_{n}|\mathcal{F}_{n}]$ $\displaystyle=\mathbb{E}[X_{n+1}]-(2p+1)=[p-(1-p)]-(2p-1)=0.\qed$
(iii)

Deduce the values of $\mathbb{P}(S_{T}=0)$ and $\mathbb{P}(S_{T}=b)$ .

Solution.

We apply optional stopping Theorem to obtain that $(M_{n\wedge T})_{n\geq 0}$ is a martingale. Since $T<\infty$ a.s., we have $\lim_{n\to\infty}M_{n\wedge T}=M_{T}$ a.s. We also notice that this process is bounded:

$|M_{n\wedge T}|\leq\max\left(\bigg{(}\frac{1-p}{p}\bigg{)}^{b},1\right),\quad n% \in\mathbb{N}.$

By dominated convergence,

$\mathbb{E}[M_{T}]=\lim_{n\to\infty}\mathbb{E}[M_{n\wedge T}]=\mathbb{E}[M_{0}].$

That is,

$\displaystyle\bigg{(}\frac{1-p}{p}\bigg{)}^{a}=\mathbb{E}\bigg{[}\bigg{(}\frac% {1-p}{p}\bigg{)}^{S_{0}}\bigg{]}=\mathbb{E}\bigg{[}\bigg{(}\frac{1-p}{p}\bigg{% )}^{S_{T}}\bigg{]}=\mathbb{P}(S_{T}=0)+\bigg{(}\frac{1-p}{p}\bigg{)}^{b}% \mathbb{P}(S_{T}=b).$ (1)

Since $\mathbb{P}(S_{T}=b)=1-\mathbb{P}(S_{T}=0)$ , we get

$\displaystyle\mathbb{P}(S_{T}=b)$ $\displaystyle=\frac{\displaystyle\bigg{(}\frac{1-p}{p}\bigg{)}^{a}-1}{% \displaystyle\bigg{(}\frac{1-p}{p}\bigg{)}^{b}-1},\quad\text{and}\quad\mathbb{% P}(S_{T}=0)=\frac{\displaystyle\bigg{(}\frac{1-p}{p}\bigg{)}^{b}-\bigg{(}\frac% {1-p}{p}\bigg{)}^{a}}{\displaystyle\bigg{(}\frac{1-p}{p}\bigg{)}^{b}-1}.\qed$
(iv)

Compute the value of $\mathbb{E}[T]$ .

Solution.

Recall that the process $(S_{n\wedge T})_{n\geq 0}$ is bounded by $b$ . For every $n\geq 0$ , we then get

$\displaystyle|S_{n\wedge T}-(2p-1)n\wedge T|\leq b+(2p-1)T\in L^{1}.$

Then from dominated convergence theorem

$\displaystyle a=\mathbb{E}[N_{0}]=\mathbb{E}[N_{n\wedge T}]=\mathbb{E}[S_{n% \wedge T}-(2p-1)n\wedge T]\to\mathbb{E}[S_{T}-(2p-1)T].$

Finally

$\displaystyle\mathbb{E}[T]=\frac{\mathbb{E}[S_{T}]-a}{2p-1}$ $\displaystyle=\frac{b}{2p-1}\frac{\displaystyle\bigg{(}\frac{1-p}{p}\bigg{)}^{% a}-1}{\displaystyle\bigg{(}\frac{1-p}{p}\bigg{)}^{b}-1}-\frac{a}{2p-1}.\qed$

(v)

Let $\tau_{0}\coloneqq\inf\{n\geq 0:S_{n}=0\}$ and $\tau_{b}\coloneqq\inf\{n\geq 0:S_{n}=b\}$ . Compute the value of $\mathbb{P}(\tau_{0}<\infty)$ .

Hint.

Consider $\mathbb{P}(\tau_{0}<\tau_{b})$ and let $b\to\infty$ .

Solution.

Due to

\displaystyle T=\inf\{n\geq 0:S_{n}=0\text{ or }S_{n}=b\}=\tau_{0}\wedge\tau_{% b}.

we can apply (1) to have

	$\displaystyle\bigg{(}\frac{1-p}{p}\bigg{)}^{a}$	$\displaystyle=\mathbb{P}(S_{T}=0)+\bigg{(}\frac{1-p}{p}\bigg{)}^{b}\mathbb{P}(% S_{T}=b)$
		$\displaystyle=\mathbb{P}(\tau_{0}<\tau_{b})+\bigg{(}\frac{1-p}{p}\bigg{)}^{b}% \mathbb{P}(\tau_{0}>\tau_{b}).$		(2)

Case 1:

If $p>1/2$ , then $\frac{1-p}{p}<1$ . Letting $b\to\infty$ , we have

$\displaystyle\bigg{(}\frac{1-p}{p}\bigg{)}^{a}$ $\displaystyle=\lim_{b\to\infty}\Big{[}\mathbb{P}(\tau_{0}<\tau_{b})+% \underbrace{\bigg{(}\frac{1-p}{p}\bigg{)}^{b}}_{\to 0}\mathbb{P}(\tau_{0}>\tau% _{b})\Big{]}$

$\displaystyle=\lim_{b\to\infty}\mathbb{P}(\tau_{0}<\tau_{b})$

$\displaystyle\stackrel{{\scriptstyle\tau_{b}<\tau_{b+1}}}{{=}}\mathbb{P}\bigg{% (}\bigcup_{b=1}^{\infty}\{\tau_{0}<\tau_{b}\}\bigg{)}\geq\mathbb{P}(\tau_{0}<\infty)$

Where the inequality follows from $\tau_{b}\geq b-a$ , so $\tau_{b}\to\infty$ as $b\to\infty$ , which implies

$\displaystyle\{\tau_{0}<\infty\}\subseteq\bigcup_{b=1}^{\infty}\{\tau_{0}<\tau% _{b}\}.$

As $\tau_{0}\wedge\tau_{b}=T<\infty$ almost surely, and the countable union of their complements is still a zero set, we have

$\displaystyle\mathbb{P}\bigg{(}\bigcap_{b=1}^{\infty}\{\tau_{0}\wedge\tau_{b}<% \infty\}\bigg{)}=1$

As intersections with probability one sets do not change the probability, we have

$\bigg{(}\frac{1-p}{p}\bigg{)}^{a}=\mathbb{P}\bigg{(}\bigcup_{b=1}^{\infty}\{% \tau_{0}<\tau_{b}\}\cap\bigcap_{b=1}^{\infty}\{\tau_{0}\wedge\tau_{b}<\infty\}% \bigg{)}\leq\mathbb{P}(\tau_{0}<\infty).$

Because if $\omega$ is in the set in the middle, there exists $b$ such that

$\displaystyle\omega\in\{\tau_{0}<\tau_{b}\}\cap\{\tau_{0}\wedge\tau_{b}<\infty% \}\subseteq\{\tau_{0}<\infty\}.$

Case 2:

If $p<1/2$ , then $\frac{1-p}{p}>1$ . Reordering (2), we get

	$\displaystyle 0$	$\displaystyle=\lim_{b\to\infty}\bigg{(}\frac{1-p}{p}\bigg{)}^{a-b}=\lim_{b\to% \infty}\bigg{(}\frac{1-p}{p}\bigg{)}^{-b}\mathbb{P}(\tau_{0}<\tau_{b})+\mathbb% {P}(\tau_{0}>\tau_{b})$
		$\displaystyle=\lim_{b\to\infty}\mathbb{P}(\tau_{0}>\tau_{b})\stackrel{{% \scriptstyle\tau_{b}<\tau_{b+1}}}{{=}}\mathbb{P}\bigg{(}\bigcap_{b=1}^{\infty}% \{\tau_{0}>\tau_{b}\}\bigg{)}$
		$\displaystyle=1-\mathbb{P}\bigg{(}\bigcup_{b=1}^{\infty}\{\tau_{0}<\tau_{b}\}% \cap\bigcap_{b=1}^{\infty}\{\tau_{0}\wedge\tau_{b}<\infty\}\bigg{)}$
		$\displaystyle\geq 1-\mathbb{P}(\tau_{0}<\infty)$
		$\displaystyle=\mathbb{P}(\tau_{0}=\infty)$

In other words: $\mathbb{P}(\tau_{0}<\infty)=1$ . So in total

\displaystyle\mathbb{P}(\tau_{0}<\infty)

\displaystyle=\begin{cases}1&p<\tfrac{1}{2}\\ \bigg{(}\frac{1-p}{p}\bigg{)}^{a}&p>\tfrac{1}{2}\end{cases}\qed

	$\displaystyle\mathbb{P}(\mathbbm{1}_{\tau>n}\to 0)$	$\displaystyle=1-\mathbb{P}(\exists\epsilon>0,\forall n_{0}\in\mathbb{N}\exists n% \geq n_{0}:\mathbbm{1}_{\tau>n}>\epsilon)$
		$\displaystyle=1-\mathbb{P}\left(\bigcup_{\epsilon>0}\bigcap_{n_{0}\in\mathbb{N% }}\bigcup_{n\geq n_{0}}:\{\mathbbm{1}_{\tau>n}>\epsilon\}\right)$
		$\displaystyle=1-\lim_{\epsilon\to 0}\underbrace{\mathbb{P}(\limsup_{n}\{% \mathbbm{1}_{\tau>n}>\epsilon\})}_{=0\quad\text{(BC)}}.$

	$\displaystyle\{\liminf_{n}X_{n}\leq\lambda\}$	$\displaystyle=\{\lim_{n\to\infty}\inf_{k\geq n}X_{k}\leq\lambda\}=\bigcap_{n=1% }^{\infty}\bigcup_{k\geq n}\{X_{k}\leq\lambda\}$
		$\displaystyle=\bigcap_{n=N}^{\infty}\underbrace{\bigcup_{k\geq n}\{X_{k}\leq% \lambda\}}_{\in\tau_{n}\subseteq\tau_{N}}\in\tau_{N}$

	$\displaystyle\bigg{(}\frac{1-p}{p}\bigg{)}^{a}$	$\displaystyle=\lim_{b\to\infty}\Big{[}\mathbb{P}(\tau_{0}<\tau_{b})+% \underbrace{\bigg{(}\frac{1-p}{p}\bigg{)}^{b}}_{\to 0}\mathbb{P}(\tau_{0}>\tau% _{b})\Big{]}$
		$\displaystyle=\lim_{b\to\infty}\mathbb{P}(\tau_{0}<\tau_{b})$
		$\displaystyle\stackrel{{\scriptstyle\tau_{b}<\tau_{b+1}}}{{=}}\mathbb{P}\bigg{% (}\bigcup_{b=1}^{\infty}\{\tau_{0}<\tau_{b}\}\bigg{)}\geq\mathbb{P}(\tau_{0}<\infty)$

Sheet 5

Exercise 1 (Uniform Integrability).

Solution.

Solution.

Solution.

Exercise 2 (Not Uniformly Integrable).

Solution.

Solution.

Solution.

Exercise 3 (Tail σ-Algebra).

Solution.

Solution.

Solution.

Exercise 4 (Gambler’s ruin).

Solution.

Solution.

Solution.

Solution.

Hint.

Solution.

Exercise 3 (Tail $\sigma$ -Algebra).