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Let $S_0$, $Z_1$, $Z_2$, $\ldots$ be independent random variables. $S_n=S_0+Z_1+\cdots+Z_n$, $n=0,1,2,\ldots$

$S_n$ is a random walk starting in a random point, $S_0$ I need to find out, when it is a martingale.

Is it enough to show, that $E(S_{n+1}|\mathcal F_n)=S_n$? Does "find out when" regards filtration?

Davide Giraudo
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atomoutside
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    The filtration is probably the natural filtration, i.e. the filtration generated by the sequence $(S_0,Z_1,\ldots,)$. Given that your variables are integrable, then you just have to find a condition that is equivalent to ${\rm E}[S_{n+1}\mid\mathcal{F}_n]=S_n$. – Stefan Hansen Jun 13 '13 at 08:56
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    So I need show that $E(S_{n+1}|F_n)=E(S_0+Z_1+\cdots+Z_n+Z_{n+1}|F_n)=E(S_n|F_n)+E(Z_{n+1}|F_n)=S_n$ and that $E(Z_{n+1}|F_n)=0$ ? – atomoutside Jun 13 '13 at 09:46

1 Answers1

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The definition of a martingale has three items. A sequence $(X_n)_{n\geq 1}$ is said to be martingale with respect to a filtration $(\mathcal{F}_n)_{n\geq 1}$ if

  • $(X_n)_{n\geq 1}$ is adapted to $(\mathcal{F}_n)_{n\geq 1}$, i.e. $X_n$ is $\mathcal{F}_n$-measurable for each $n$,
  • $X_n$ is integrable, i.e. ${\rm E}[|X_n|]<\infty$ for each $n$,
  • and $X_n$ satisfies the martingale condition, i.e. ${\rm E}[X_{n+1}\mid\mathcal{F}_n]=X_n$ for each $n$.

So in order for you to answer the question of when $(S_n)_{n\geq 1}$ is a martingale you need to address the first two bullets first. Let us therefore assume that all variables are integrable, and that the filtration we are working with is indeed the natural filtration, i.e. $$ \mathcal{F}_n=\sigma(S_0,Z_1,\ldots,Z_n),\quad n\geq 1. $$

Then you're correct that you just have to show when ${\rm E}[S_{n+1}\mid\mathcal{F}_n]=S_n$ for all $n$. You correctly calculated that $$ {\rm E}[S_{n+1}\mid\mathcal{F}_n]={\rm E}[S_n\mid\mathcal{F}_n]+{\rm E} [Z_{n+1}\mid\mathcal{F}_n]=S_n+{\rm E}[Z_{n+1}\mid\mathcal{F}_n] $$ and hence ${\rm E}[S_{n+1}\mid\mathcal{F}_n]=S_n$ if and only if ${\rm E}[Z_{n+1}\mid\mathcal{F}_n]=0$. All that is left is to recognize ${\rm E}[Z_{n+1}\mid\mathcal{F}_n]$ as ${\rm E}[Z_{n+1}]$ due to the independence between $Z_{n+1}$ and $\mathcal{F}_n$.

Stefan Hansen
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