Sheet 4

Exercise 1 (Scheffé’s Lemma).

Suppose that $(X_n,n\ge 1)$ is a sequences of random variables in ${ℒ}^1(\mathrm{\Omega },ℱ,\mathbb{P})$ such that ${lim}_{n\to \mathrm{\infty }}{X}_n=X_{\mathrm{\infty }}$ almost surely. Show that the following statements are equivalent:

1. $X_n\to X_{\mathrm{\infty }}$ in ${ℒ}^1$, as $n\to \mathrm{\infty }$;

2. , as $n\to \mathrm{\infty }$.

Hint.

Apply Fatou’s Lemma to $\mathrm{|}{X}_n\mathrm{|}\mathrm{+}\mathrm{|}{X}_{\mathrm{\infty }}\mathrm{|}\mathrm{-}\mathrm{|}{X}_n\mathrm{-}{X}_{\mathrm{\infty }}\mathrm{|}\mathrm{\ge }\mathrm{0}$.

Solution.

(i) $\Rightarrow$ (ii) Follows from the reverse triangle inequality

$\left||x|-|y|\right|\le |x-y|,$ (1)

and Jensen’s inequality

$\left|𝔼\left[|X_n|\right]-𝔼\left[|X_{\mathrm{\infty }}|\right]\right|$ $=\left|𝔼\left[|X_n|-|X_{\mathrm{\infty }}|\right]\right|$

$\stackrel{\text{Jensen}}{\le }𝔼\left[\left||X_n|-|X_{\mathrm{\infty }}|\right|\right]$

$\stackrel{(\mathit{\text{<a href="\#S0.E1" title="(1) ‣ Exercise 1 (Scheffé’s Lemma). ‣ Sheet 4" class="ltx\_ref ltx\_font\_italic" style="font-size:70\%;"><span class="ltx\_text ltx\_ref\_tag">1</span></a>}})}{\le }𝔼\left[|X_n-X_{\mathrm{\infty }}|\right]\to 0,(n\to \mathrm{\infty }).$

(ii) $\Rightarrow$ (i): By Fatou’s Lemma we have

$2𝔼[|X_{\mathrm{\infty }}|]$ $=𝔼[\underset{n}{lim\; inf}|X_n|+|X_{\mathrm{\infty }}|-|X_n-X_{\mathrm{\infty }}|]$

$\stackrel{\text{Fatou}}{\le }\underset{n\to \mathrm{\infty }}{lim\; inf}𝔼[|X_n|+|X_{\mathrm{\infty }}|-|X_n-X_{\mathrm{\infty }}|]$

$=2𝔼[|X_{\mathrm{\infty }}|]-\underset{n\to \mathrm{\infty }}{lim\; sup}𝔼[|X_n-X_{\mathrm{\infty }}|]$

This implies

$\underset{n\to \mathrm{\infty }}{lim\; sup}𝔼[|X_n-X_{\mathrm{\infty }}|]\le 0,$

and therefore convergence. ∎

Exercise 1’ (Generalized Scheffé’s Lemma - Optional).

Assume that ${\mathrm{(}{X}_n\mathrm{)}}_{n\mathrm{\in }\mathbb{N}}\mathrm{\subset }{ℒ}^{\mathrm{1}}$ converges in probability to $X_{\mathrm{\infty }}\mathrm{\in }{ℒ}^{\mathrm{1}}$. Then the following statements are equivalent:

1. , as $n\to \mathrm{\infty }$.

2. For all $ϵ>0$ we have ${lim\; sup}_{n\to \mathrm{\infty }}𝔼[|X_n|{\mathrm{𝟙}}_{|X_{\mathrm{\infty }}-X_n|>ϵ}]\le ϵ$

3. $\{X_{\mathrm{\infty }},X_1,X_2,\mathrm{\dots }\}$ are uniformly integrable

4. $X_n\to X_{\mathrm{\infty }}$ in ${ℒ}^1$, as $n\to \mathrm{\infty }$;

Hint.

Prove them in order. For (i) $\mathrm{\Rightarrow }$ (ii) you may want to use

$𝔼[|X_n|{\mathrm{𝟙}}_{|X_{\mathrm{\infty }}-X_n|>ϵ}]$

$=𝔼[|X_n|]-𝔼[|X_n|{\mathrm{𝟙}}_{|X_{\mathrm{\infty }}-X_n|\le ϵ}]$

$=\underset{\mathit{\text{Similar total mass}}}{\underbrace{𝔼[|X_n|]-𝔼[|X_{\mathrm{\infty }}|]}}+𝔼[(|X_{\mathrm{\infty }}|-|X_n|){\mathrm{𝟙}}_{|X_{\mathrm{\infty }}-X_n|\le ϵ}]+\underset{\mathit{\text{Convergence in probability}}}{\underbrace{𝔼[|X_{\mathrm{\infty }}|{\mathrm{𝟙}}_{|X_{\mathrm{\infty }}-X_n|>ϵ}]}}$

For (ii) $\mathrm{\Rightarrow }$ (iii) you might want to ponder how $X_n$ can be larger than some $M$ when $X_{\mathrm{\infty }}$ is smaller than $M\mathrm{-}ϵ$.

Solution.

(i) $\Rightarrow$ (ii): Fix $ϵ>0$ and choose any $\eta >0$. Then there exists $n_0\in \mathbb{N}$ such that

$𝔼[|X_n|]-𝔼[|X_{\mathrm{\infty }}|]\le \frac{\eta }{2}\mathit{\hspace{1em}}\forall n\ge n_0.$ (2)

Since $\{X_{\mathrm{\infty }}\}$ is uniformly integrable, there exists $\delta (\eta )>0$ such that

Since $X_n\to X_{\mathrm{\infty }}$ in probability, there therefore exists $n_1\in \mathbb{N}$ such that for all $n\ge n_1$

$\mathbb{P}(|X_{\mathrm{\infty }}-X_n|>ϵ)\le \delta (\eta )⟹𝔼[|X_{\mathrm{\infty }}|{\mathrm{𝟙}}_{|X_{\mathrm{\infty }}-X_n|>ϵ}]\le \frac{\eta }{2}$ (3)

Now we plug things together. We have for $n\ge \max \{n_0,n_1\}$

$𝔼[|X_n|{\mathrm{𝟙}}_{|X_{\mathrm{\infty }}-X_n|>ϵ}]$ $=𝔼[|X_n|]-𝔼[|X_n|{\mathrm{𝟙}}_{|X_{\mathrm{\infty }}-X_n|\le ϵ}]$

This implies the claim. The first sum exploits the fact that the mass of the random variables has to be similar. So we can not concentrate mass on a small event such as $X_n$ being far from $X_{\mathrm{\infty }}$. Which is what examples of convergence in probability without $L^1$ convergence exploit.

(ii) $\Rightarrow$ (iii): We want to show

$\underset{M\to \mathrm{\infty }}{lim}\underset{n}{sup}𝔼[|X_n|{\mathrm{𝟙}}_{|X_n|>M}]=0.$

So fix some $\delta >0$. We select $ϵ:=\frac{\delta }{4}$ and then select $n_0\in \mathbb{N}$ such that for all $n\ge n_0$ we have

$𝔼[|X_n|{\mathrm{𝟙}}_{|X_{\mathrm{\infty }}-X_n|>ϵ}]\le ϵ+\frac{\delta }{4}.$ (4)

As $\{X_1,\mathrm{\dots },X_{n_0}\}$ is a finite set it is uniformly integrable, and there exists $M_0$ such that

$\underset{0\le n\le n_0}{sup}𝔼[|X_n|{\mathrm{𝟙}}_{|X_n|>M}]\le \delta \mathit{\hspace{1em}}\forall M\ge M_0.$ (5)

Similarly there exists $M_1$, such that

$𝔼[|X_{\mathrm{\infty }}|{\mathrm{𝟙}}_{|X_{\mathrm{\infty }}|>M-ϵ}]\le \frac{\delta }{4}\mathit{\hspace{1em}}\forall M\ge M_1$ (6)

To put things together, note that we always have

${\mathrm{𝟙}}_{|X_n|>M}\le {\mathrm{𝟙}}_{|X_{\mathrm{\infty }}-X_n|>ϵ}+{\mathrm{𝟙}}_{|X_{\mathrm{\infty }}|>M-ϵ}{\mathrm{𝟙}}_{|X_{\mathrm{\infty }}-X_n|\le ϵ}.$ (7)

This implies for all $M\ge \max \{M_0,M_1\}$

$\underset{n\in \mathbb{N}}{sup}𝔼[|X_n|{\mathrm{𝟙}}_{|X_n|>M}]$

(iii) $\Rightarrow$ (iv): Fix some $ϵ>0$, then due to uniform integrability there exists some $\delta >0$, such that implies for all $n$

$𝔼[|X_n|{\mathrm{𝟙}}_A]\le ϵ\mathit{\hspace{1em}}\text{and}\mathit{\hspace{1em}}𝔼[|X_{\mathrm{\infty }}|{\mathrm{𝟙}}_A]\le ϵ.$ (8)

Now we choose $n_0\in \mathbb{N}$ such that

$\mathbb{P}(|X_{\mathrm{\infty }}-X_n|>ϵ)\le \delta \mathit{\hspace{1em}}\forall n\ge n_0.$

With (8) this implies for all $n\ge n_0$

$𝔼[|X_{\mathrm{\infty }}-X_n|]\le \underset{\le ϵ}{\underbrace{𝔼[|X_{\mathrm{\infty }}-X_n|{\mathrm{𝟙}}_{|X_{\mathrm{\infty }}-X_n|\le ϵ}]}}+\underset{\le ϵ}{\underbrace{𝔼[|X_{\mathrm{\infty }}|{\mathrm{𝟙}}_{|X_{\mathrm{\infty }}-X_n|>ϵ}]}}+\underset{\le ϵ}{\underbrace{𝔼[|X_n|{\mathrm{𝟙}}_{|X_{\mathrm{\infty }}-X_n|>ϵ}]}}$

(iv) $\Rightarrow$ (i): This finally follows from the reverse triangle inequality

$\left||x|-|y|\right|\le |x-y|,$ (9)

and Jensen’s inequality

$\left|𝔼\left[|X_n|\right]-𝔼\left[|X_{\mathrm{\infty }}|\right]\right|$ $=\left|𝔼\left[|X_n|-|X_{\mathrm{\infty }}|\right]\right|$

$\stackrel{\text{Jensen}}{\le }𝔼\left[\left||X_n|-|X_{\mathrm{\infty }}|\right|\right]$

$\stackrel{(\mathit{\text{<a href="\#S0.E9" title="(9) ‣ Exercise 1’ (Generalized Scheffé’s Lemma - Optional). ‣ Sheet 4" class="ltx\_ref ltx\_font\_italic" style="font-size:70\%;"><span class="ltx\_text ltx\_ref\_tag">9</span></a>}})}{\le }𝔼\left[|X_n-X_{\mathrm{\infty }}|\right]\to 0,(n\to \mathrm{\infty }).\mathit{∎}$

Exercise 2 (Lemma 6.3.5).

Let $(X_n)$ be an $({ℱ}_n)$-submartingale and $S,T$ be bounded stopping times such that $S\le T$ a.s. Prove that, $𝔼[X_S]\le 𝔼[X_T]$.

Hint.

Consider the stochastic integral of $H\mathrm{=}\mathrm{(}{H}_n\mathrm{)}$ against $X$ with .

Solution.

Since $S,T$ are stopping times, we have that

is ${ℱ}_{n-1}$ measurable and therefore that the process $H=(H_n)$ is previsible. The stochastic integral

${(H\cdot X)}_n$

$=X_{T\wedge n}-X_{S\wedge n}.$

is a submartingale since $H_n$ is positive. Since $S,T$ are bounded, there exists $N\in \mathbb{N}$ such that

$𝔼[X_T]-𝔼[X_S]$ $=𝔼[X_{T\wedge N}-X_{S\wedge N}]=𝔼[{(H\cdot X)}_N]\ge 𝔼[{(H\cdot X)}_0]=0.\mathit{∎}$

Exercise 3 (Uniform Integrability).

1. (Example 6.3.12 (i)) Prove that every finite set of integrable random variables is uniformly integrable.

   Solution.

   The first property follows by the fact that the set of random variables is integrable. The second property follows from

   $\underset{M\to \mathrm{\infty }}{lim}\left(\underset{\alpha \in I}{sup}𝔼[|X_{\alpha }|\cdot {\mathrm{𝟙}}_{|X_{\alpha }|\ge M}]\right)$ $\le \underset{M\to \mathrm{\infty }}{lim}\left({\displaystyle \sum _{\alpha \in I}}𝔼[|X_{\alpha }|\cdot {\mathrm{𝟙}}_{|X_{\alpha }|\ge M}]\right)$

   $={\displaystyle \sum _{\alpha \in I}}\underset{=0\mathit{\hspace{1em}}\text{(DCT)}}{\underbrace{\underset{M\to \mathrm{\infty }}{lim}𝔼[|X_{\alpha }|\cdot {\mathrm{𝟙}}_{|X_{\alpha }|\ge M}]}},$

   where we use that limits can be interchanged with finite sums. ∎

2. If $X_{\alpha }\sim {\delta }_{x_{\alpha }}$, then the set ${(X_{\alpha })}_{\alpha \in I}$ is uniformly bounded if and only if $S=\{x_{\alpha }:\alpha \in I\}\subseteq ℝ$ is uniformly bounded.

   Solution.

   Assume that $S$ is uniformly bounded. Then there exists some $M_0>0$ such that for all $x\in S$. Therefore we have for all $M\ge M_0$ that ${\mathrm{𝟙}}_{|X_{\alpha }|\ge M}=0$ a.s. for all $\alpha \in I$, which implies

   $\underset{\alpha \in I}{sup}𝔼[|X_{\alpha }|\cdot {\mathrm{𝟙}}_{|X_{\alpha }|\ge M}]=0\mathit{\hspace{1em}\hspace{1em}}\forall M\ge M_0$

   In particular the limit in $M$ is zero. Assume on the other hand that $S$ is not uniformly bounded. Then for every $M\in ℝ$ there exists some $\beta \in I$ with $|x_{\beta }|>M\vee 1$, implying

   $\underset{\alpha \in I}{sup}𝔼[|X_{\alpha }|\cdot {\mathrm{𝟙}}_{|X_{\alpha }|\ge M}]\ge 𝔼[|X_{\beta }|\cdot {\mathrm{𝟙}}_{|X_{\beta }|\ge M}]=|x_{\beta }|\ge 1.$

   Therefore the limes superior in $M$ is at least $1$, so we know that ${(X_{\alpha })}_{\alpha \in I}$ is not uniformly bounded. ∎

Exercise 4 (Uniqueness of Limits in Probability).

1. (Triangle Inequality of Probability) Prove that for $X,Y,Z$ random variables on some metric space with metric $d$ we have

   $\mathrm{\{}d(X,Y)\ge ϵ\}\subseteq \{d(X,Z)\ge \frac{ϵ}{2}\}\cup \{d(Z,Y)\ge \frac{ϵ}{2}\}$

   Solution.

   Assume $\omega$ is in the set on the left, then due to

   $ϵ\le d(X,Y)(\omega )\le d(X,Z)(\omega )+d(Z,Y)(\omega )\le 2\max \{d(X,Z)(\omega ),d(Z,Y)(\omega )\}$

   we have either $\frac{ϵ}{2}\le d(X,Z)(\omega )$ or $\frac{ϵ}{2}\le d(Z,Y)(\omega )$, therefore $\omega$ has to be in the union of sets on the right. ∎

Let us now assume that $X_n\stackrel{P}{\to }X$ and $X_n\stackrel{P}{\to }Y$

1. Prove that $P(d(X,Y)\ge ϵ)=0$ for all $ϵ>0$.

   Hint.

   Recall the proof of uniqueness for deterministic sequences. Apply a similar trick.

   Solution.

   As the triangle inequality of probability with $Z=X_n$, holds for all $n$ it also holds for the limit, therefore

   $\mathbb{P}(d(X,Y)\ge ϵ)$ $\le \underset{n\to \mathrm{\infty }}{lim\; sup}\mathbb{P}(\{d(X,X_n)\ge \frac{ϵ}{2}\}\cup \{d(X_n,Y)\ge \frac{ϵ}{2}\})$

   $\le \underset{n\to \mathrm{\infty }}{lim}\mathbb{P}(d(X,X_n)\ge ϵ)+\mathbb{P}(d(X_n,Y)\ge ϵ)$

   $=0\mathit{∎}$

2. Conclude that $X=Y$ a.s.

   Solution.

   We can calculate the complementing event by

   $\mathbb{P}(X\ne Y)$ $=\mathbb{P}\left({\displaystyle \bigcap _{ϵ>0}}\{d(X,Y)\ge ϵ\}\right)=\underset{ϵ\to 0}{lim}\mathbb{P}(d(X,Y)\ge ϵ)=0.\mathit{∎}$

Exercise 5 (Bound on Stopping Times).

Let $T$ be a ${({ℱ}_n)}_{n\in {\mathbb{N}}_0}$-stopping time. Suppose that there exists $N\in \mathbb{N}$ and $\epsilon \in (0,1)$, such that for every $n\in \mathbb{N}$,

$$\mathbb{P}(T\le n+N|{ℱ}_n)>\epsilon ,\text{almost surely}.$$

1. Prove that for each $k\in \mathbb{N}$, we have $\mathbb{P}(T>kN)\le {(1-\epsilon )}^k$.

   Hint.

   $\mathbb{P}(T>kN)=\mathbb{P}(T>kN,T>(k-1)N)$.

   Solution.

   Using induction we have the statement for $k=0$, as ${(1-ϵ)}^0=1$. The induction step is:

   $\mathbb{P}(T>(k+1)N)$ $=𝔼[{\mathrm{𝟙}}_{T>(k+1)N}{\mathrm{𝟙}}_{T>kN}]$

   $=𝔼[𝔼[{\mathrm{𝟙}}_{T>(k+1)N}{\mathrm{𝟙}}_{T>kN}|{ℱ}_n]]$

   $=𝔼[{\mathrm{𝟙}}_{T>kN}\underset{=1-\mathbb{P}(T\le kN+N|{ℱ}_n)\le 1-ϵ}{\underbrace{\mathbb{P}(T>kN+N|{ℱ}_n)}}]$

   $\le (1-ϵ)\mathbb{P}(T>kN)$

   $\stackrel{\text{ind.}}{\le }{(1-ϵ)}^{k+1}.\mathit{∎}$

2. Deduce that .

   Solution.

   As $T$ is positive we have

   $𝔼[T]$

   $={\displaystyle \sum _{k=0}^{\mathrm{\infty }}}{\displaystyle {\int }_{kN}^{(k+1)N}}\underset{\le \mathbb{P}(T>kN)}{\underbrace{\mathbb{P}(T>s)}}𝑑s$

   $\le {\displaystyle \sum _{k=0}^{\mathrm{\infty }}}N\mathbb{P}(T>kN)$

   $\le N{\displaystyle \sum _{k=0}^{\mathrm{\infty }}}{(1-ϵ)}^k\le {\displaystyle \frac{N}{1-ϵ}}\mathit{∎}$

Exercise 6 (Monkey Typewriter Theorem).

At each of times $1,2,\mathrm{\dots }$, a monkey types a capital letter at random, the sequence of letters typed forming an i.i.d. sequence of random variables chosen uniformly amongst the 26 possible capital letters. Let $T$ be the first time by which the monkey has produced the consecutive sequence ABRACADABRA of $11$ letters. We prove that

$𝔼[T]={26}^{11}+{26}^4+26.$

For this, imagine that just before each time $n\in \mathbb{N}$, a new gambler indexed by $n$ arrives on the scene. He bets 1 euro that

the $n$-th letter will be A.

If he loses, he loses all his money and leaves. If he wins, he receives 26 euros, all of which he bets on the event that

the $n+1$-th letter will be B.

If he loses, he loses all his money and leaves. If he wins, he bets his whole current fortune of ${26}^2$ that

the $n+2$-th letter will be R.

and so on through the ABRACADABRA sequence. All gamblers come and play and leave independently of each other. Let $(X_i^k,i\ge 0)$ denote the fortune of the $k$-th gambler just after time $i$: in particular, $X_i^k=1$ for all $i\le k-1$.

1. Show that indeed, .

   Hint.

   See Exercise 5.

   Solution.

   We see that $T$ is a stopping time with respect to ${({ℱ}_n)}_{n\ge 0}$ and $\mathbb{P}(T\le n+11|{ℱ}_n)>{26}^{-11}$ for every $n\ge 0$. Thus by Exercise 5. ∎

For simplicity of notation, we can add infinitely many Z’s after ABRACADABRA. Gamblers then bet on the word ABRACADABRAZZZZZ… We denote by $L_k$ the $k$-th letter of this word for every $k\ge 1$. We also denote by $U_n$ the $n$-th letter typed for every $n\ge 1$ and ${({ℱ}_n)}_{n\ge 0}$ the corresponding filtration.

1. For every $n\ge 1$, let $M_n$ be the total net profit of all gamblers who have participated in the game just after time $n$, i.e.

   $$M_n=\sum _{k=1}^n(X_n^k-1),$$

   and $M_0≔0$. Show that ${(M_n)}_{n\ge 0}$ is a martingale.

   Hint.

   Show that $M_n^k\mathrm{:=}{X}_n^k\mathrm{-}\mathrm{1}$ is a martingale for every $k$, by defining $M_n^k\mathrm{:=}\mathrm{0}$ for all .

   Solution.

   By definition, ${(M_n)}_{n\ge 0}$ is adapted to the filtration ${({ℱ}_n)}_{n\ge 0}$ and it is also clearly integrable. The net profit of the first gambler $M_n^1:=X_n^1-1$ is as follows:

   $M_0^1=0,M_1^1={\mathrm{𝟚𝟞𝟙}}_{U_1=L_1}-1,M_2^1=26(M_1^1+1){\mathrm{𝟙}}_{U_2=L_2}-1,$

   and more generally, for every $n\ge 0$,

   $M_{n+1}^1=26(M_n^1+1){\mathrm{𝟙}}_{U_{n+1}=L_{n+1}}-1,$

   and then

   $𝔼[M_{n+1}^1|{ℱ}_n]=26(M_n^1+1)\mathbb{P}(U_{n+1}=L_{n+1})-1=M_n^1.$

   So ${(M_n^1)}_{n\ge 0}$ is a martingale with respect to ${({ℱ}_n)}_{n\ge 0}$. More generally, for every $k\ge 1$, $M_n^k=0$ for every and for every $n\ge k$,

   $M_{n+1}^k=26(M_n^k+1){\mathrm{𝟙}}_{U_{n+1}=L_{n+2-k}}-1,$

   and ${(M_n^k)}_{n\ge 0}$ is also a martingale with respect to ${({ℱ}_n)}_{n\ge 0}$. For every $n\ge 1$, $M_n={\sum }_{k=1}^n{M}_n^k$ and therefore ${(M_n)}_{n\ge 0}$ is a martingale with respect to ${({ℱ}_n)}_{n\ge 0}$. ∎

2. Show that $T=inf\{n\ge 0:M_n+n\ge {26}^{11}+{26}^4+26\}$. Conclude that $𝔼[T]={26}^{11}+{26}^4+26$.

   Solution.

   If $T=n$, then all gamblers who started before time $n-10$ (where ABRACADABRA starts) have lost at time $n$ so that $M_n^k=-1$ for every $k\le n-11$. The $(n-10)$-th gambler is the winner: $M_n^{n-10}={26}^{11}-1$. We also have $M_n^{n-3}={26}^4-1$ (because the four first and last letters of ABRACADABRA are the same, viz. ABRA) and for the same reason, $M_n^n=26-1$. Finally, $M_n^k=-1$ for every $k\in \{n-9,\mathrm{\dots },n-4,n-2,n-1\}$. Putting all pieces together,

   $M_n{\mathrm{𝟙}}_{T=n}=({26}^{11}+{26}^4+26-n){\mathrm{𝟙}}_{T=n}.$

   Therefore, almost surely,

   $T\in \{n\ge 0:M_n+n\ge {26}^{11}+{26}^4+26\}.$

   We also see that if , we have

   $M_n={\displaystyle \sum _{k=1}^n}{M}_n^k$

   $\le {\displaystyle \sum _{k=0}^9}{26}^{k+1}-n$

   $=26\cdot \frac{{26}^{10}-1}{26-1}-n\le 26({26}^{10}-1)-n$

   Thus

   $T=inf\{n\ge 0:M_n+n\ge {26}^{11}+{26}^4+26\}.$

   Moreover, since almost surely,

   $M_T+T={\displaystyle \sum _{n=0}^{\mathrm{\infty }}}(M_n+n){\mathrm{𝟙}}_{T=n}={\displaystyle \sum _{n=0}^{\mathrm{\infty }}}({26}^{11}+{26}^4+26){\mathrm{𝟙}}_{T=n}={26}^{11}+{26}^4+26.$

   Now for every $n\ge 0$, $-T\le -(n\wedge T)\le M_{n\wedge T}\le {26}^{11}+{26}^4+26$. Since $T\in L^1$, we derive from the dominated convergence theorem

   $𝔼[M_T]=\underset{n\to \mathrm{\infty }}{lim}𝔼[M_{n\wedge T}]=0.$

   Then, finally,

   $𝔼[T]$ $=𝔼[M_T+T-M_T]=𝔼[M_T+T]-𝔼[M_T]={26}^{11}+{26}^4+26.\mathit{∎}$