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I've got a theorem (without proof) that say:

If $(S_n)$ is a martingale refer to $(X_n)$, then $\mathbb E[S_n]=\mathbb E[S_1]$.

I don't really understand why. Is there an intuitive why to see it ?

idm
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1 Answers1

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$$\mathbb{E}\left[S_n\right]=\mathbb{E}\left[\mathbb{E}\left[S_n\mid X_{n-1}\right]\right]=\mathbb{E}\left[S_{n-1}\right]=\mathbb{E}\left[\mathbb{E}\left[S_{n-1}\mid X_{n-2}\right]\right]$$ $$=\mathbb{E}\left[S_{n-2}\right]=\ldots=\mathbb{E}\left[\mathbb{E}\left[S_{2}\mid X_{1}\right]\right]=\mathbb{E}\left[S_{1}\right].$$

To show the first equality, we use the definition of the conditional expectation : for all random variable $Y$ which is $\sigma\left(X_1,\ldots X_{n-1}\right)$-measurable and such that $\mathbb{E}\left[S_nY\right]$ exists, we have $$\mathbb{E}\left[S_nY\right]=\mathbb{E}\left[\mathbb{E}\left[S_n\mid X_1,\ldots X_{n-1}\right]Y\right]$$ and for $Y=1$, we get $$\mathbb{E}\left[S_n\right]=\mathbb{E}\left[\mathbb{E}\left[S_n\mid X_1,\ldots X_{n-1}\right]\right].$$ Then we use the martingale property $\mathbb{E}\left[S_n\mid X_1,\ldots X_{n-1}\right]=S_{n-1}$. As $1$ is constant, it is measurable for any $\left(\sigma\left(X_1,\ldots,X_k\right)\right)_{k\geq1}$ and then the above argument holds for each $k\geq1$.

Nicolas
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  • I don't understand where come from the first and the second equality. For the second one, for me, $$S_{n-1}=\mathbb E[S_{n}\mid X_{1},...,X_{n-1}].$$ Could you explain please ? – idm May 04 '15 at 13:04
  • Sure, I edit my post in a few seconds. – Nicolas May 04 '15 at 13:04
  • Ok first the first one. But for the second, to me, the martingale property is $$\mathbb E[S_n\mid X_{1},...,X_{n-1}]=S_{n-1}$$ and not $$\mathbb E[S_n\mid X_{n-1}]=S_{n-1}.$$ – idm May 04 '15 at 13:13
  • No problem, the above argument holds if you put $\sigma\left(X_1,\ldots,X_{n-1}\right)$ instead of $\sigma\left(X_{n-1}\right)$ because $1$ still will be measurable. In general, it is better to choose $\left(X_n\right)$ such that $\sigma\left(X_n\right)\subset\sigma\left(X_{n+1}\right)$ (to do this, you can take $\mathcal{F}n=\sigma\left(X_1,\ldots,X_n\right)$ and then $\mathcal{F}_n\subset\mathcal{F}{n+1}$). – Nicolas May 04 '15 at 13:17
  • When $X_n$ are random variables, it is best to use $\sigma(X_1,\dots,X_n)$ as your sigma-fields, since $\sigma(X_n) \subseteq \sigma(X_{n+1})$ may not be true. On the other hand, if $X_n$ are sigma-fields, the original argument is fine. (We do not know what $X_n$ is, or what "refer" means in the question...) – GEdgar May 04 '15 at 14:05
  • Totally agree. But it is often assumed that $\sigma(X_n)\subset\sigma(X_{n+1})$, that's why I guessed it first. To me,$X_n$ are random variables and $\mathbb{E}\left[S_n\mid X_n\right]$ means $\mathbb{E}\left[S_n\mid \sigma(X_1,\ldots,X_n)\right]$. – Nicolas May 04 '15 at 14:08
  • So just to be sure, if I replace $\mathbb E[S_n\mid X_{n-1}]$ by $\mathbb E[S_n\mid X_{n-1},...,X_1]$ it still work, right ? – idm May 04 '15 at 17:13
  • Yes, it works because $1$ is in any $\sigma$-algebra. I've edited my post anyway. If you have any other questions, don't hesitate! – Nicolas May 04 '15 at 17:18