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Let the stochastic process $\{X_n\}$ be constructed inductively as follows:

  • $X_0=0$, and
  • for $n\ge 1$, and conditionally on $\mathcal{F}_{n-1}=\sigma(X_0,\ldots,X_{n-1})$, we set
    $\;\,\,\, X_n=nX_{n-1}$ or $0$ with probabilities $1/n$ and $1-1/n$

My problem is to verify that $\{X_n\}$ is a martingale i.e. $E(X_n\mid \mathcal{F}_{n-1})=X_{n-1}$

I guess I can compute $E(X_n\mid\mathcal{F}_{n-1})$ like this: $$ E(X_n\mid\mathcal{F}_{n-1})=nX_{n-1}\cdot \frac{1}{n}+0\cdot(1-\frac{1}{n})=X_{n-1} $$

However, I can not find any theoretical reason letting me do this.

Could you please help? Thank you!

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Which conditional expectations do you know how to compute? –  Did Jan 10 '13 at 0:25
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If $X_0=0$ and $X_n=$ either $0$ or $X_{n-1}$ times some number, then why isn't every term in the sequence equal to $0$? –  Michael Hardy Jan 10 '13 at 0:42
    
@MichaelHardy: Oh, you are right. This is actually an excerpt from a more complicated problem. Sorry I didn't notice that this will bring such a problem which doesn't appear in the original problem. And thank you for your editing. :) –  Roun Jan 10 '13 at 2:20
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1 Answer 1

up vote 1 down vote accepted

Let $Y_i$ be a rv with $P(Y=1)=1/n$ and $P(Y_i=0)=1-1/n$. By assumption suppose that $Y_i$ is an iid sequence of rv and moroever, let $Y_i$ be independent of $X_0,X_1,\ldots,X_{i}$. In other words, $\sigma(Y_i)$ and $F_{i}$ are independent for all $i$. Now write $X_{n+1}=nY_nX_n$. Conditioning on $\mathcal{F}_n$,

$E[X_n\mid\mathcal{F}_{n-1}]=E[nY_{n-1}X_{n-1}\mid\mathcal{F}_{n-1}]=nX_{n-1}E[Y_{n-1}\mid\mathcal{F}_{n-1}]=nX_{n-1}E[Y_{n-1}]=nX_{n-1}(1/n)=X_{n-1}$

where we used the fact that $X_{n-1}$ is $\mathcal{F}_{n-1}$ measurable and independence of $Y_{n-1}$. from $F_{n-1}$.

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