Given $X_n$ iid random variables with $P(X_n = n^2) = 1/n^2, P(X_n = -1) = 1-1/n^2$ show that $\bar{Y_n} \xrightarrow{P} 0$ where $Y_n = X_n/n^2$

Question

Given $X_n$ iid random variables with $P(X_n = n^2) = \frac{1}{n^2}, P(X_n = -1) = 1-\frac{1}{n^2}$. Let $Y_n = \frac{X_n}{n^2}$, I want to show that $\bar{Y_n} \xrightarrow{P}0$, where $\bar{Y_n} = \frac{1}{n}\sum_{i=1}^{n}Y_i$. This looks like a straightforward application of the LLN, the problem is that I got that $E(Y_n) = \frac{1}{n^2}E(X_n) = \frac{1}{n^4}$ rather than $E(Y_n) = 0$ which is the result I was looking for. Any suggestions?

Answer 1

In fact, $\overline {Y_n} \to 0$ a.s.

Since $\sum P(X_n=n^{2}) <\infty$ Borel Cantelli Lemma gives $X_n =-1$ for all $n$ suffcuently large, with probaility $1$. Hence, $Y_n=-\frac 1 {n^{2}}$ for all $n$ suffcuently large. But $\sum\limits_{k=1}^{n} \frac 1 {k^{2}}$ is bounded, so $\overline {Y_n} \to 0$ a.s..