for some discrete IID random variables $W_{1}, W_{2}, \dots$ with finite and positive mean and variance, define $ X_{n} = \sum_{i=1}^{n} W_{i}^{2}$. How would we be able to show that $X_{n}$ does not converge in probability?
I see that $\mathbb{E}[X_{n}]$ increases as n goes up and $X_{n}$ should diverge, but I was wondering how I could show this under the definition of convergence in probability:
If $X_{n}$ converges in probability to X, then for any $\epsilon > 0$, we have $$lim_{n \to \infty} \mathbb{P}(|X_{n} - X| > \epsilon) = 0$$
Thanks a lot.
Convergence in probability implies convergence in distribution. Since $\mathbb E[W_1^2]>0$ we have $$ \mathbb E[X_n] = \mathbb E\left[\sum_{i=1}^n W_i^2\right] = \sum_{i=1}^n \mathbb E[W_i^2] =\sum_{i=1}^n \mathbb E[W_1^2] = n\mathbb E[W_1^2]\stackrel{n\to\infty}\longrightarrow \infty, $$ and so $X_n$ cannot converge in distribution. It follows that $X_n$ cannot converge in probability.