I'm trying to prove that the sum of a martingale difference sequence $\sum_0^n \epsilon_i\delta M_i$ is a martingale, but I am stuck. $$ \delta M_i = Y_i - E(Y_i|F_{i-1}) $$
If this is a martingale, then $E(\sum_0^n \epsilon_i\delta M_i) = \sum_0^{n-1} \epsilon_i\delta M_i$. But I am finding that $E(\sum_0^n \epsilon_i\delta M_i) = 0$
$$ \sum_0^n \epsilon_i\delta M_i = \epsilon_n(Y_n - E(Y_n|F_{n-1})) + ... \epsilon_1(Y_1 - E(Y_1|F_{1-1})) $$ $$ E(\sum_0^n \epsilon_i\delta M_i) = E\epsilon_n(Y_n - E(Y_n|F_{n-1})) + ... E\epsilon_1(Y_1 - E(Y_1|F_{1-1})) = \\ = \epsilon_n(E(Y_n) - EE(Y_n|F_{n-1})) + ... \epsilon_1(E(Y_1) - EE(Y_1|F_{1-1})) ) \\ = \epsilon_n(E(Y_n) - E(Y_n)) + ... \epsilon_1(E(Y_1) - E(Y_1)) = \epsilon_n(0) + ... + \epsilon_1(0) = 0 $$
Note: the sum is supposed to start at zero... But I feel this is an error, because $Y_0 - Y_{0-1}$ doesn't make sense to me.
What am I doing wrong? Why am I not finding $E(\sum_0^n \epsilon_i\delta M_i) = \sum_0^{n-1} \epsilon_i\delta M_i$
There is another way I can look at this problem, $\delta M_i = Y_i - Y_{i-1}$, but the sum would be: $$ E(\sum_0^n \epsilon_i\delta M_i) = E\epsilon_n(Y_n - Y_{n-1}) + ... + E\epsilon_1(Y_1 - Y_{0}) + E\epsilon_0 Y_0 = \\ = \epsilon_n(Y_{n-1} - Y_{n-2}) + ... + \epsilon_1(Y_0 - Y_{0}) + \epsilon_0 Y_0 $$ The above is the case because $Y$ is a martingale, but... this would work if not for the $\epsilon$ being different. For this to be a martingale I think the leading $\epsilon$ would have to be $\epsilon_{n-1}$ not $\epsilon_n$
You found $0$ because you took the ordinary expectation while you should instead take the conditional expectation with respect to $\mathcal F_{n-1}$. If $\epsilon_i$ is $\mathcal F_{i-1}$-measurable, then the sum up to $n-1$ is $\mathcal F_{n-1}$-measurable and it remains to check that $\mathbb E\left[\epsilon_n\delta M_n\mid \mathcal F_{n-1}\right]=0$.