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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?

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  • $\begingroup$ 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
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    $\begingroup$ 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
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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 $Y_k$.

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