Find limit $\lim\limits_{n\rightarrow \infty}P\left(\frac{X_1^3+\ldots+X_n^3}{X_1+\ldots+X_n}<\frac{1}{2}\right)$

Question

I will be very glad for help with following task: We are given $X_n$ which are all i.i.d. $U(0,2)$. We need to find: $$\lim_{n\rightarrow \infty}P\left(\frac{X_1^3+\ldots+X_n^3}{X_1+\ldots+X_n}<\frac{1}{2}\right)$$ It looks like we should apply Law of large numbers, yet I am clueless how to do that.

Answer 1

$X$~$U(0,2)$ and all are i.i.d. So $E(X)=1$. Now to calculate $E(X^{3})$.

$E(X^{3})=\int_{0}^{2}\frac{x^{3}}{2}dx=2$. Hence by weak law of large numbers we have $Y_{n}=\frac{\sum_{1}^{n}X_{i}^{3}}{\sum_{1}^{n}X_{i}}\rightarrow 2$ in probability. Since $\frac{\sum_{1}^{n}X_{i}^{3}}{n}$ goes to $2$ and $\frac{\sum_{1}^{n}X_{i}}{n}$ goes to $1$ in probability. Hence $\lim P(|Y_{n}-2|\leq \epsilon)=1 \Rightarrow \lim P(2-\epsilon\leq Y_{n}\leq2+\epsilon)=1$. Now $P(2-\epsilon\leq Y_{n}\leq2+\epsilon)\leq P(2-\epsilon\leq Y_{n})\Rightarrow 1=\lim P(2-\epsilon\leq Y_{n}\leq2+\epsilon)\leq\lim P(2-\epsilon\leq Y_{n})$

But we know that $P(2-\epsilon\leq Y_{n})\leq1\Rightarrow \lim P(2-\epsilon\leq Y_{n})\leq1\Rightarrow \lim P(2-\epsilon\leq Y_{n})=1$. Now the above is true $\forall \epsilon>0$. Hence taking $\epsilon=\frac{3}{2}$ we have $\lim P(\frac{1}{2}\leq Y_{n})=1 \Rightarrow \lim P(Y_{n}<\frac{1}{2})=0 $