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The task is to use SLLN to find limits:

a. $\lim_{n \to \infty} \int_{0}^{1} ... \int_{0}^{1} \frac{x_1^3 + ... + x_n^3}{x_1 + ... + x_n} dx_1 ... dx_n$

b. $\lim_{n \to \infty} \int_{0}^{1} ... \int_{0}^{1} f(\sqrt[n]{x_1 ... x_n}) dx_1 ... dx_n$ where $f : [0,1] \to R$ is continuous.

What I've tried: If $a_n$ is a series of sample means size $n$ from population with mean $\mu$ then $a_n \to \mu$, so I tried to write integrands as some functions g(x) of sample mean. Then limits would be integrals of $g(\mu)$. Then I realised that 1. I don't know how to rewrite integrands as functions of sample mean 2. It doesn't make sense because I don't know anything about the distribution the sample comes from and I don't know $\mu$.

Can anyone hint me on how to approach this task?

Brzoskwinia
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1 Answers1

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Hints: Let $X_1,X_2,...$ be an i.i.d sequence with uniform distribution on $(0,1)$. Then the first limit is $\lim_{n \to \infty} E\frac {X_1^{3}+X_2^{3}+\cdots +X_n^{3}} {X_1+X_2+\cdots+X_n}$. Divide numerator and denominator in the ratio by $n$. By SLLN the limit is $\frac {EX_1^{3}} {EX_1}$. [The ratio lies between $0$ and $1$ so DCT can be applied].

For the second part let $g(x)=f(e^{x})$. Then the given limit is $\lim Eg(\frac {\ln X_1+\ln X_2+\cdots+ln X} n)=g(E \ln X_1)$.

Kavi Rama Murthy
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  • $$\lim_{n \to \infty} E \frac{\frac{X_1^3 + ... X_n^3}{n}}{\frac{X_1 + ... X_n}{n}} = $$ $$\lim_{n \to \infty} E \frac{\bar{X^3}}{\bar{X}} = $$ $$E \lim_{n \to \infty} \frac{\bar{X^3}}{\bar{X}} = $$ $$E \frac{\lim_{n \to \infty}\bar{X^3}}{\lim_{n \to \infty} \bar{X}} =$$ $$E \frac{EX^3}{EX} = $$ $$\frac{EX^3}{EX}$$

    Now $EX = \int_{0}^{1}xf(x) dx = \int_{0}^{1}xdx = \frac{1}{2}$ and $EX ^3= \int_{0}^{1}x^3f(x) dx = \int_{0}^{1}x^3dx = \frac{1}{4}$ so the answer is$\frac{\frac{1}{4}}{\frac{1}{2}} = \frac{1}{2}$. Did I anything wrong?

     – Brzoskwinia May 22 '21 at 12:41
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    @Brzoskwinia $\frac 1 2 $ is the correct limit. – Kavi Rama Murthy May 22 '21 at 12:46
  • ok, thanks a lot :) and I think that it should be $EX^3$ instead of $X^3$ in your answer (before "The ratio lies ..."). – Brzoskwinia May 22 '21 at 12:51
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    @Brzoskwinia Sure. That was a typo. – Kavi Rama Murthy May 22 '21 at 12:59