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Show $M_n=X_1+X_2+...+X_n-n\theta$ is a martingale w.r.t ${X_n}$, given that $X_i$ are i.i.d. random variables with $\mathbb{E}[X_i]=\theta$

this is what I've done: $$\mathbb{E}[M_{n+1}|X_{\le n}]=\mathbb{E}[X_1+...+X_n+X_{n+i}-(n+1)\theta|X_{\le n}]$$ $$\mathbb{E}[X_1+X_2+...+X_n-n\theta|X_{\le n}]+\mathbb{E}[X_{n+1}-\theta|X_{\le n}]$$ $$M_n+\theta-\theta=M_n$$

is this correct?

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    $\begingroup$ I can understand your step and it is correct, but the notation/steps used looks quite weird to me. Also for a more rigorous answer you will want to state that the expectation is finite, although that is obvious. $\endgroup$ – BGM Apr 13 '16 at 7:17
  • $\begingroup$ Also consider the Doob-Meyer decomposition of $S_n := \sum_{k=1}^n X_k$. $\endgroup$ – Math1000 Apr 15 '16 at 12:53

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