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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 : https://mathoverflow.net/questions/98711/second-equality-of-wald

Community
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Merli
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  • Sorry, I didn't know that this two sites were related. I added a link to Mathoverflow. – Merli Jun 03 '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 03 '12 at 12:56

1 Answers1

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Hints:

  • $Z_{n+1}^2=Z_n^2+2Z_n\bar X_{n+1}+\bar X_{n+1}^2$ 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 $\mathrm E(U\mid G)=\mathrm E(U)$.
Did
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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 03 '12 at 15:43
  • I find that $\mathbb{E}[Z_n ^2 |F_n] = Z_n$ – Merli Jun 03 '12 at 15:45
  • See Edit. $ $ $ $ – Did Jun 03 '12 at 15:51
  • Is this possible that $\mathbb{E}[Z_n \bar{X_{n+1}}] = 0$ ? – Merli Jun 03 '12 at 15:55
  • I fail to see why you ask this but yes, if $U$ and $V$ are independent and $E(U)=0$ then $E(UV)=0$. – Did Jun 03 '12 at 15:56
  • In fact, I just calculate and find that. I find it strange, so I asked :) – Merli Jun 03 '12 at 15:57
  • Beware that $Z_n\bar X_{n+1}$ is not independent on $F_n$ in general, since $\bar X_{n+1}$ is but $Z_n$ is not. – Did Jun 03 '12 at 15:59
  • Okay thanks, that's a remark I have to keep in mind – Merli Jun 03 '12 at 16:00
  • For the last one, I find that $\mathbb{E}[\bar{X}{n+1}^2] = var(X_1)$ : $\mathbb{E}[\bar{X}{n+1}^2] = \mathbb{E}[(X_{n+1} - \mathbb{E}[X_1])^2])$ And since $X_1$ and $X_{n+1}$ have the same distribution – Merli Jun 03 '12 at 16:01
  • Sure, but let me note that your two last questions deal with expectations although you should be concerned with conditional expectations. – Did Jun 03 '12 at 16:03
  • Of course, I just forgot to write it, sorry :( But I write it correctly for my exercise – Merli Jun 03 '12 at 16:04
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    So the exercise is solved with your tips, thank you very much !! :) – Merli Jun 03 '12 at 16:09