I just started this new section of probability and I'm not so sure on how to resolve these statements:
Let $(X_n)_{n≥1}$ be a sequence of independent random variables such that $\mathbb{P}(X_n=0)=1−\frac1{n^2}$ and $\mathbb{P}(X_n=n^3)=\frac1{n^2}$
a) Show that $X_n\rightarrow 0$ in probability
My attempt: The sequence converges in probability to $0$ if $\forall\epsilon>0 \lim_{n\rightarrow\infty}\mathbb{P}[\left|X_n-0\right|>\epsilon]=0$. Here I have $\mathbb{P}[\left|X_n-0\right|>\epsilon]=\mathbb{P}[\left|X_n\right|>\epsilon]=\mathbb{P}[X_n>\epsilon]=\frac1{n^2}\rightarrow0$ as $n\rightarrow\infty$
b) Show that it is not true that $X_n \rightarrow0$ in $L^p$ for $p≥1$.
Attempt: The sequence converges in $L^p$ to $0$ if $\lim_{n\rightarrow\infty}\mathbb{E}[\left|X_n-0\right|^p]=0$. Well really I don't know how to proceed to show that $\mathbb{E}[\left|X_n\right|^p]\not=0$ since the PDF of $X_n$ is unknown.
c) Show that $X_n\rightarrow 0$ a.s.
Attempt: The sequence converges a.s. to $0$ when $\mathbb{P}[\{\omega\in\Omega :0=\lim_{n\rightarrow\infty}X_n(\omega)\}]=1$. But here I get $\mathbb{P}(X_\infty=0)=1 $ and $\mathbb{P}(X_\infty=\infty)=0$, but since the last one doesn't make sense, we get the result.(??)
Any suggestion/hint/correction is appreciated.
Best Answer
Your solution to (a) looks OK. For (b), notice that you can compute the expectation by: $$E(|X_n|^p) = 0^p P(X_n = 0)+n^{3p}P(X_n=n^3) = n^{3p-2}$$ Now let $n \to \infty$ and conclude.
As a hint for (c): what does the Borel-Cantelli lemma tell us about $P(X_n > 0\, \text{ infinitely often})$?