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I'm reviewing some old homework in probability theory and I encountered the following exercise. We computed $$\lim_{n\rightarrow\infty} \int_0^1 \cdots \int_0^1 f\left(\frac{x_1 + \ldots + x_n}{n}\right)dx_1\ldots dx_n$$ for any continuous and bounded function $f$.

Since we have to use methods of probability theory we can replace the integral by the expected value

$$E[f\left(\frac{X_1 + \ldots + X_n}{n}\right)]$$

while $X_i$ is uniformly distributed over $[0,1]$.

Afterwards, we can simply apply Lebesgue's dominated convergence theorem, the continuity property of $f$ and law of large numbers to find the limit of $f(1/2)$.

My problem is to remember why we can replace the integral by the expected value $E$.

KCd
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Marc
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1 Answers1

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Because the distribution of $(X_1,\ldots,X_n)$ is the Lebesgue measure on $[0,1]^n$ hence, for every measurable function $g$, $$ \mathbb E(g(X_1,\ldots,X_n))=\iint_{[0,1]^n}g(x_1,\ldots,x_n)\mathrm dx_1\ldots\mathrm dx_n. $$ Now, apply this fact to the function $$ g:(x_1,\ldots,x_n)\mapsto f\left(\frac{x_1+\cdots+x_n}n\right). $$

Did
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