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?

  • 2
    $\begingroup$ 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 ${\rm E}[S_{n+1}\mid\mathcal{F}_n]=S_n$. $\endgroup$ Jun 13 '13 at 8:56
  • 1
    $\begingroup$ 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 $E(Z_{n+1}|F_n)=0$ ? $\endgroup$ Jun 13 '13 at 9:46

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 $\mathcal{F}_n$.

  • $\begingroup$ I forgot about first two properties. Thank you for help. $\endgroup$ Jun 13 '13 at 11:38

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.