Second Equality of Wald [closed]

Hi,

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 $F = (F_i)$ be a filtration such as $X$ is adapted. We suppose that $X_i$ and $F_{i-1}$ are independant 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 is there a mistake in the beginning ?

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.

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Also on math.SE: math.stackexchange.com/q/153247/5363 – Theo Buehler Jun 3 at 12:47
This forum is not for exercises. You can certainly get helped in stats.stackexchange.com. – Valerio Capraro Jun 3 at 12:48
No the right place for the question, voting to close. – Igor Rivin Jun 3 at 12:48
I didn't know that this forum was inappropriate for exercices. Thanks for the information – Merli Jun 3 at 12:56
Is there a way to delete the post by myself ? – Merli Jun 3 at 12:58