Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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!

share|cite|improve this question
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
up vote 1 down vote accepted

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 $k \in \mathbb{R}$.)

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.