Let $X_1,\dotsc,X_n$ be independent, zero mean random variables and define $Y_k = \alpha^{n-k}X_k$.

Is $\{Z_k\}$ with $Z_k = \sum_{i=1}^k Y_i = \sum_{i=1}^k \alpha^{n-i}X_i$ a martingale? I suppose not, but I know the sum of independent random variables is a martingale. Aren't the $Y_k$ independent?

  • $\begingroup$ The $Y_k$ are indeed independent, so it should be a martingale, should it not? Why are you saying no? $\endgroup$ – Shalop Mar 23 '15 at 15:24
  • 2
    $\begingroup$ Sum of independent random variables is a martingale if, and only if, $EX_k = 0$. $\endgroup$ – Rodrigo Ribeiro Mar 23 '15 at 15:26
  • $\begingroup$ Please show what you have tried. $\endgroup$ – Ben Derrett Mar 23 '15 at 15:26
  • $\begingroup$ @Rodrigo: Yes, you're correct. Sorry about that. $\endgroup$ – Shalop Mar 23 '15 at 15:29

Let $X_1,..,X_n$ independent random variables and $\mathcal{F_k} = \sigma (X_1,..,X_k)$ the natural filtration. So, clearly, each $X_i$ is measurable with respect to $F_k$ for $i \le k$. And $X_{k+1}$ is independent of $\mathcal{F_k}$. With this in mind you have, denoting $S_k = X_1+X_1 + X_2 + ... + X_k$

$$E[S_{k+1} | \mathcal{F_k}] = S_k+E[X_{k+1}] =S_k \iff E[X_k] = 0 $$ for all $k$.

The same is true to your $Y_k$.


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.