# Exercise - Martingale - Dominated convergence

here's an exercise I was able to solve completely, however I need some help concerning the justifications why I am allowed to use dominated convergence in the second point. Here is the complete statement of the problem:

Let $(X_n)_{n\geq 1}$ be an i.i.d. sequence of random variables with zero means and common variance $\sigma^2$, i.e., $\mathbb{E}[X_n]=0$ and $\sigma^2=var[X_n]<\infty$.

• If $M_n=X_1+...+X_n$, then show that $(M_n^2-n\sigma^2)$ is a martingale.
• If $T$ is an integrable stopping time, show that $$var\left[\sum_{n=1}^TX_i\right]=\sigma^2\mathbb{E}[T].$$

So here is what I did: T is integrable, so we know that $T<\infty \quad a.s.$. Since $T$ is a stopping time, the stopped process $(M_{n\land T } ^2-(n\land T)\sigma^2)$ is also a martingale, so that we have $\mathbb{E}\left[M_{n\land T } ^2\right]=\mathbb{E}\left[(n\land T))\right]\sigma^2$. By Beppo-Levi $$\mathbb{E}\left[(n\land T))\right] \longrightarrow \mathbb{E}\left[T\right].$$

Then I have problems for this:

1. I want to show $\mathbb{E}\left[M_{n\land T } ^2\right] \longrightarrow \mathbb{E}\left[M_T^2\right]$ by dominated convergence, but I do not know how to dominate $M_{n\land T } ^2$ by something integrable. Of course I have to use that T is integrable, but I am not able to show that I can apply dominated convergence.
2. Moreover I have to show $\mathbb{E}[M_T]=0$, because then $var\left[\sum_{n=1}^TX_i\right]=\mathbb{E}\left[M_{T } ^2\right]$. Again I would like to use the fact that $M_{T\land n}$ is a martingale, so for each $n$, I have $\mathbb{E}[M_{T\land n}]=\mathbb{E}[M_0]=0$. Then, I want to use again dominated convergence in order to show $\mathbb{E}\left[M_{n\land T}\right] \longrightarrow \mathbb{E}\left[T\right]$, but for this I have to dominate $|M_{n\land T}|$ by something integrable.

So I want to know if anybody can help me justifying why/if I can apply dominated convergence? Thanks in advance!

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Suggestion: 1. $M_{n\wedge T}^2=M_n^2\mathbb{1}_{\{T>n\}}+M_T^2\mathbb{1}_{\{T\leq n\}}$ and 2. $M_T=\mathbb{1}_{\{T=n\}} \sum_{i=1}^n X_i$ – alexlo Jan 26 '13 at 23:59

Since $(M_n)_{n\geq 0}$ is a martingale we know that $(M_{n \wedge T})_{n \geq 0}$ is a martingale by optional stopping. In particular

$$\forall m \leq n: \mathbb{E}(M_{n \wedge T} \cdot M_{m \wedge T}) = \mathbb{E}(M_{m \wedge T} \cdot \mathbb{E}(M_{n \wedge T}|\mathcal{F}_m))=\mathbb{E}(M_{m \wedge T}^2) \tag{1}$$

by tower property. Using $\mathbb{E}(M_{n \wedge T}^2)=\sigma^2 \cdot \mathbb{E}(n \wedge T)$ we obtain

$$\mathbb{E}((M_{n \wedge T}-M_{m \wedge T})^2) \stackrel{(1)}{=} \mathbb{E}(M_{n \wedge T}^2-M_{m \wedge T}^2)=\sigma^2 \cdot \mathbb{E}(n \wedge T - m \wedge T) \to 0 \qquad (m,n \to \infty)$$

since $T \in L^1$ and $T \wedge n \to T$ a.s. as $n \to \infty$. This means that $(M_{n \wedge T}^2)_{n \geq 0}$ is a $L^2$-Cauchy-sequence, thus convergent (in $L^2$). Since $M_{n \wedge T} \to M_T$ a.s. we conclude $M_{n \wedge T} \to M_T$ in $L^2$.

This solves your first question. The second one follows from the fact that $L^2$-convergence implies $L^1$-convergence.

(I don't see how to apply dominated convergence in this case, because you always need some kind of "uniform boundedness", for example the boundedness of the increments, i.e. $$\mathbb{P}(\sup_n |N_n-N_{n-1}| \leq k)=1$$ for some $k \in \mathbb{R}$.)

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Thank you! Yes, you are right the other method does not work since one can not find a uniform bound which works... Thanks a lot! – Mathoman Jan 27 '13 at 16:36