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I'm doing an exercice about the second equality of Wald.

Let $(X_i)_{i\ge 1}$ be a sequence of integrable random variables. Let $\mathcal{F} = (\mathcal{F}_i)_{i\ge 1}$ be a filtration such that $X$ is adapted. We suppose that $X_i$ and $F_{i-1}$ are independent for all $i\ge 2$ and that $\mathbb{E}[X_1^2] < \infty$. We put $S_n := X_1 + ... + X_n$. Show that if $T$ is an integrable stopping time such as $T \ge 1$, then $\mathbb{E}[S_T ^2 - T\mathbb{E}[X_1 ^2]]^2 = \sigma ^2 (X_1)\mathbb{E}[T]$.

I would like to do it this way : Show that $Y_n := Z_n^2 - n\sigma (X_1)$ is a martingale, where $Z_n := X_1 + ... + X_n - n\mathbb{E}[X_1]$ is a martingale (this result as been shown in an other exercice).

But I can't show that. We have $\mathbb{E}[Y_{n+1} | F_n] = \mathbb{E}[Z_{n+1} ^2 - (n+1)\sigma ^2 (X_1) | F_n] = \mathbb{E}[Z_{n+1} ^2 | F_n] - (n+1)\sigma ^2 (X_1)$. And then, I tried to develop the square, but I don't get any good result.... Maybe there is a mistake in the beginning ?

Thanks for your help.

P.S. I suppose there is another way to prove that equality, but I want to do it this way.

P.P.S As you can see, English is not my mother tongue, so if you see any mistake, I would be glad to learn how to write it correctly.

Here is the link to the same subject in Mathoverflow : http://mathoverflow.net/questions/98711/second-equality-of-wald

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Sorry, I didn't know that this two sites were related. I added a link to Mathoverflow. –  Merli Jun 3 '12 at 12:53
Thank you. I hope you'll get a good answer soon and a somewhat belated welcome to the site! –  t.b. Jun 3 '12 at 12:56
Thank you very much :) –  Merli Jun 3 '12 at 12:57
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1 Answer

up vote 1 down vote accepted


  • $Z_{n+1}^2=Z_n^2+2Z_n\bar X_{n+1}+\bar X_{n+1}^2$ where $\bar X_{n+1}=X_{n+1}-\mathrm E(X_1)$.
  • $\mathrm E(Z_n^2\mid F_n)=\text{____}$.
  • Since $\bar X_{n+1}$ is independent of $F_n$, $\mathrm E(Z_n\bar X_{n+1}\mid F_n)=\text{____}$ and $\mathrm E(\bar X_{n+1}^2\mid F_n)=\text{____}$.
  • Hence $(Y_n)$ is an $(F_n)$-martingale.

Edit: There are two cases when the conditional expectation of the integrable random variable $U$ conditionally on the sigma-algebra $G$ is simple to determine.

  • If $U$ is $G$-measurable, then $\mathrm E(U\mid G)=U$.
  • If $U$ is independent of $G$, then $\mathrm E(U\mid G)=\mathrm E(U)$.
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Hi did, Thank you for the tips. The first step is ok. Now for the second one, is there any formula to calculate $\mathbb{E}[Z_n ^2 |F_n]$ with a square in it ? Or maybe I just have to develop ? –  Merli Jun 3 '12 at 15:43
I find that $\mathbb{E}[Z_n ^2 |F_n] = Z_n$ –  Merli Jun 3 '12 at 15:45
See Edit. $ $ $ $ –  Did Jun 3 '12 at 15:51
Is this possible that $\mathbb{E}[Z_n \bar{X_{n+1}}] = 0$ ? –  Merli Jun 3 '12 at 15:55
So the exercise is solved with your tips, thank you very much !! :) –  Merli Jun 3 '12 at 16:09
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