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I'm stuck with an martingales exercise here:

$$\lim_{n\to\infty}\int_0^1\int_0^1\cdots\int_0^1\sin\left(\frac{x_1+x_2+\cdots+x_n}{n}\right)dx_1dx_2\cdots dx_n$$

I tried to do it without martingales and it work fine for me. But I can't see how to apply the definition of Martingales here.

Can someone please explain me what I exactly have to do? Thank's!

Pedro
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Eva Leon
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  • How did you solve the question? Actually, the strong law of large numbers is the easiest way to solve this question, I think. – saz Jul 02 '15 at 06:00
  • I don't see how it is possible to compute this directly. What trick did you use? – shalop Jul 02 '15 at 06:09

1 Answers1

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Let $X_1,X_2,X_3,...$ be a sequence of i.i.d. random variables which are distributed uniformly over $[0,1]$.

Then we know that $$\int_0^1\int_0^1\cdots\int_0^1\sin\left(\frac{x_1+x_2+\cdots+x_n}{n}\right)dx_1dx_2\cdots dx_n \;\;= \;\;E \bigg[\sin \bigg( \frac{X_1+...+X_n}{n} \bigg) \bigg]$$

By the law of large numbers, we know that $\frac{X_1+...+X_n}{n}$ converges almost surely to $1/2$. Thus, by continuity of the sine function, we know that $\sin \big( \frac{X_1+...+X_n}{n} \big)$ converges almost surely to $\sin(1/2)$.

Hence by using the Dominated convergence theorem as well as the fact that sine is bounded, we see that $$ \lim_{n \to \infty} \int_0^1\int_0^1\cdots\int_0^1\sin\left(\frac{x_1+x_2+\cdots+x_n}{n}\right)dx_1dx_2\cdots dx_n$$ $$=\lim_{n \to \infty} E \bigg[\sin \bigg( \frac{X_1+...+X_n}{n} \bigg) \bigg] = E \bigg[ \lim_{n \to \infty} \;\sin \bigg( \frac{X_1+...+X_n}{n} \bigg) \bigg] $$$$ = E\big[\sin(1/2)\big]=\sin(1/2)$$

I apologize in advance if there is a mistake somewhere in my calculations.

shalop
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