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In the hypothesis of the martingale central limit theorem, my book says that given a sequence of random variables $X_n$ with the condition that $E(X_n \mid \mathcal F_{n-1}) = 0$, then $S_n = \sum_{i=0}^n X_i$ is a $\mathcal F_n$ martingale.

Now I suppose that the $\left(\mathcal F_n\right)_{n \ge 0}$ is a filtration (thought it's not specified) but apart from that are there any other condition to be assumed on $\mathcal F_n$? Because otherwise, I can't show that $S_n$ is a martingale.

If $S_m$ is measurable wrt to $\mathcal F_m$, then it's easy, since $$E(S_n \mid \mathcal F_m) = E(S_m \mid \mathcal F_m) + E(X_{m+1} + \dots + X_n \mid \mathcal F_m) = S_m + E(X_{m+1} + \dots + X_n \mid \mathcal F_m)$$

And one can use the power property of conditional expectation to show that $E(X_{m+1} + \dots + X_n \mid \mathcal F_m) = 0$

But how to show that $S_m$ is $\mathcal F_m$ measurable?

Should we require $\mathcal F_m = \sigma(\bigcup_{i=0}^m \sigma(X_i))$?

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We have to require that $X_n$ is $\mathcal F_n$-measurable. Otherwise, take $\mathcal F_n=\mathcal F=$ a $\sigma$-algebra independent of the $X_n$'s and suppose that $X_n$ are centered and not degenerated. Then we have $\mathbb E[X_n\mid \mathcal F_{n-1}] =0$ for each $n$ but $X_n$ is not $\mathcal F_n$-measurable.

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  • $\begingroup$ I see, thank you :) So I take it was a forgetfulness of the book? $\endgroup$ – Ant Mar 18 '15 at 13:31

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