Condition for convergence of random variables

Question

I am working with the various types of convergence of random variables. I came across this exercise:

Consider the following sequence of independent r.vs : $(X_i, i\geq1) : Pr(X_i=i^a)=\frac{1}{i^b}, Pr(X_i=0) = 1-\frac{1}{i^b}$ with $a,b>0$ What are the minimal condition for $a,b$ s.t. $X_i\to 0$ considering quadratic convergence, in probabilty and almost sure convergence. I thought of something among these lines: For the quadratic: I need $E(X_i^2)\to 0, i \to +\infty$ hence, $ \frac{(i^a)^2}{n^b} \to 0, i\to +\infty$ which means $a < b/2$ Is that right?

As for the almost sure convergence, is it ok for every $a,b$? Because I would say that I need $a<0$ in order to be sure to have always $X_i=0$ but since I can't, it will always be $\neq 0$ with prob $1/i^b$ I need to nullify the probability of it being different then $0$ and thus, $b>0$ is ok. What happens if the condition are not verified? It will still converge, right? Even if not almost surely?

For the convergence in probability I have no idea, honestly, on how to estimate $Pr(|X_i|>\epsilon)$ as $i\to \infty$. Is it $Pr(i^a>\epsilon)$? And to have it go to $0$ I need $a>log_i(\epsilon)$ and $b>0$?

Answer 1

We have $$ \mathbb E[X_i^2] = \frac{(i^a)^2}{i^b} = i^{2a-b}, $$ so $\mathbb E[X_i^2]\stackrel{i\to\infty}\longrightarrow 0$ iff $2a-b<0$. It is clear that $X_i$ converges in probability to zero for any $b>0$ since $i^{-b}\stackrel{i\to\infty}\longrightarrow 0$. The $X_i$ converge to zero almost surely iff $$ \mathbb P\left(\liminf_{i\to\infty}X_i<\varepsilon\right)=1,\ \forall\varepsilon>0. $$ For any positive integer $i$, there exists $k>i$ such that $\mathbb P(X_k<1)=0$ - take $k > e^{-a}$. It follows that \begin{align} \mathbb P\left(\liminf_{i\to\infty}X_i<1\right) &= \mathbb P\left(\bigcup_{i=1}^\infty\bigcap_{j=i}^\infty \{X_j<1\}\right)\\ &\leqslant\sum_{i=1}^\infty \mathbb P\left(\bigcap_{j=i}^\infty \{X_j<1\}\right)\\ &\leqslant \sum_{i=1}^\infty \mathbb P(X_k<1) = 0, \end{align} and hence $X_i$ does not converge almost surely to zero.