So I am in the beginning of taking probability class. I got this problem from the lecture notes of last year, and I want to try all the problems as much as I can before the new semester starts. I really need you guys to check my answer since noone can do that now.
Problem: By using the strong law of large numbers, with $0<p<q$, compute \begin{eqnarray*} \lim_{n\rightarrow\infty}\int_0^1\int_0^1\dotsi\int_0^1\frac{x_1^q+x_2^q+\dotsb+x_n^q}{x_1^p+x_2^p+\dotsb+x_n^p}\mathrm{d}x_1\mathrm{d}x_2\dotsb\mathrm{d}x_n \end{eqnarray*} This is my answer: Let $(X_i)$ be a sequence of i.i.d. uniform random variables on unit interval, then the joint probability density function of $X_1,X_2,\dotsc,X_n$ is $f_{X_1,X_2,\dotsc,X_n}(x_1,x_2,\dotsc,x_n)=1$ on the $n$-dimensional unit cube, and equals to $0$ elsewhere. Hence, we have \begin{eqnarray*} \int_0^1\int_0^1\dotsi\int_0^1\frac{x_1^q+x_2^q+\dotsb+x_n^q}{x_1^p+x_2^p+\dotsb+x_n^p}\mathrm{d}x_1\mathrm{d}x_2\dotsb\mathrm{d}x_n&=&E\left(\frac{X_1^q+X_2^q+\dotsb+X_n^q}{X_1^p+X_2^p+\dotsb+X_n^p}\right)\\ &=&E\left(\frac{\frac{1}{n}\sum_{i=1}^n X_i^q}{\frac{1}{n}\sum_{i=1}^n X_i^p}\right). \end{eqnarray*} By strong law of large numbers, we get for $a>0$, \begin{eqnarray*} \frac{1}{n}\sum_{i=1}^n X_i^a\xrightarrow{a.s}E(X_1^a)=\frac{1}{a+1}. \end{eqnarray*} Then, the answer of the limit of the integral equals to $\frac{p+1}{q+1}$
Is my answer right? If it is wrong, can you point out where I did wrong? And could you please give me some hints to correct it. Thank you so much.
The idea is right, but you need a supplementary argument to justify how the almost sure convergence of $\frac{\frac{1}{n}\sum_{i=1}^n X_i^q}{\frac{1}{n}\sum_{i=1}^n X_i^p}$ implies the convergence of expectations. A dominated convergence argument works here. Indeed, since $ 0\leqslant X_i \leqslant 1$ and $0<p<q$, we have $ X_i ^q\leqslant X_i ^p$ hence $\frac{ \sum_{i=1}^n X_i^q}{ \sum_{i=1}^n X_i^p}$ is smaller than one and almost surely non-negative (unless $X_i=0$ for all $i$, but the probability of this event is zero).