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Let f and g be continuous functions on [0,1] satisfying $0\leq f(x)\leq C(g(x))$ for all x and some constant $C>0$. Show that

$\lim_{n\rightarrow\infty}\int _0^1\cdots\int _0^1 \frac{\sum_{i=1}^n f(x_i)}{\sum_{i=1}^n g(x_i)}dx_1\cdots dx_n$

$$\lim_{n\rightarrow\infty}\int _0^1\cdots\int _0^1 \frac{\sum_{i=1}^n f(x_i)}{\sum_{i=1}^n g(x_i)}dx_1\cdots dx_n=\frac{\int_0^1f(x)dx}{\int_0^1g(x)dx}$$

where $\int_0^1g(x)dx\neq 0$

In fact, a more general results is: for i.i.d r.v. X, X_1,X_2,\codts, we have

$$E\frac{\sum_{i=1}^n f(X_i)}{\sum_{i=1}^n g(X_i)}=\frac{E(f(X))}{E(g(X))}$$ where $Eg(X)\neq 0$.

(1) It seems that according to SLLN,

$\lim_{n\rightarrow\infty}\frac{1}{n} \int _0^1 \cdots\int _0^1 \sum_{i=1}^n f(x_i)dx_1\cdots dx_n=\lim_{n\rightarrow\infty}\frac{1}{n} \sum_{i=1}^n\int _0^1 \cdots\int _0^1 f(x_i)dx_1\cdots dx_n\stackrel{a.s.}\rightarrow\int_0^1 f(x) dx$

$\lim_{n\rightarrow\infty}\frac{1}{n} \int _0^1 \cdots\int _0^1 \sum_{i=1}^n g(x_i)dx_1\cdots dx_n=\lim_{n\rightarrow\infty}\frac{1}{n} \sum_{i=1}^n\int _0^1 \cdots\int _0^1 g(x_i)dx_1\cdots dx_n\stackrel{a.s.}\rightarrow\int_0^1 g(x) dx$,

with $\int_0^1 g(x) dx\neq 0$

According to $X_n\stackrel{a.s.}\rightarrow X, Y_n \stackrel{a.s.}{\rightarrow} Y \implies X_n Y_n \stackrel{a.s.}\rightarrow XY$

We can prove

$$\lim_{n\rightarrow\infty}\int _0^1\cdots\int _0^1 \frac{\sum_{i=1}^n f(x_i)}{\sum_{i=1}^n g(x_i)}dx_1\cdots dx_n=\frac{\int_0^1f(x)dx}{\int_0^1g(x)dx}$$

(2) For the second part

assume $du_i$ is the measure for $X_i$

According to SLLN, we have

$\lim_{n\rightarrow\infty}\frac{1}{n} \int _0^1 \cdots\int _0^1 \sum_{i=1}^n f(X_i)du_1\cdots du_n=\lim_{n\rightarrow\infty}\frac{1}{n} \sum_{i=1}^n\int _0^1 \cdots\int _0^1 f(X_i)du_1\cdots du_n\stackrel{a.s.}\rightarrow\int_0^1 f(X) du$

$\lim_{n\rightarrow\infty}\frac{1}{n} \int _0^1 \cdots\int _0^1 \sum_{i=1}^n g(X_i)du_1\cdots du_n=\lim_{n\rightarrow\infty}\frac{1}{n} \sum_{i=1}^n\int _0^1 \cdots\int _0^1 g(X_i)du_1\cdots du_n\stackrel{a.s.}\rightarrow\int_0^1 g(X) du$

with $\int_0^1 g(X) du\neq 0$

According to $X_n\stackrel{a.s.}\rightarrow X, Y_n \stackrel{a.s.}{\rightarrow} Y \implies X_n Y_n \stackrel{a.s.}\rightarrow XY$

We can prove

$$\lim_{n\rightarrow\infty}\int _0^1\cdots\int _0^1 \frac{\sum_{i=1}^n f(X_i)}{\sum_{i=1}^n g(X_i)}du_1\cdots du_n=\frac{\int_0^1 f(X)du}{\int_0^1g(X)du}$$

Thus, $$E\frac{\sum_{i=1}^n f(X_i)}{\sum_{i=1}^n g(X_i)}=\frac{E(f(X))}{E(g(X))}$$ where $Eg(X)\neq 0$.

My question:

Can not make sure my proof, especially:

(1) The relationship between:$\int _0^1 \cdots\int _0^1 f(x_i)dx_1\cdots dx_n$ and $\int_0^1 f(x) dx$ (2) The relationship between: $E(\sum_{i=1}^n f(X_i))$ and $ \int _0^1 \cdots\int _0^1 \sum_{i=1}^n f(X_i)du_1\cdots du_n$

Sebastiano
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Olivia
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1 Answers1

2

You actually have to prove that $$ \lim_{n\to +\infty}\mathbb E\frac{\sum_{i=1}^n f(X_i)}{\sum_{i=1}^n g(X_i)}=\frac{\int_0^1f(x)dx}{\int_0^1g(x)dx}. $$ You noticed that $\frac{\sum_{i=1}^n f(X_i)}{\sum_{i=1}^n g(X_i)}$ converges almost surely to $\frac{\int_0^1f(x)dx}{\int_0^1g(x)dx}$, so it just remains to check that we can switch the expectation and the limit symbol in $\lim_{n\to +\infty}\mathbb E\frac{\sum_{i=1}^n f(X_i)}{\sum_{i=1}^n g(X_i)}$. This can be done thanks to the dominated convergence theorem because due to the assumptions, $$ 0\leqslant\frac{\sum_{i=1}^n f(X_i)}{\sum_{i=1}^n g(X_i)}\leqslant C. $$

Davide Giraudo
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