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I'm trying to solve the following task and I'm struggling very much. I don't know if it is correct what I did so far.

Let $(X_n,n\geq1)$ be a sequence of independent random variables such that $\alpha>0$ and

$$P(X_n=0)=1-\frac{1}{n^{\alpha}}\quad \text{and} \quad P(X_n=n)=\frac{1}{n^{\alpha}},\quad n\geq 1$$

Prove that

  1. $X_n\to 0$ as $n\to \infty$ in probability
  2. $X_n\to 0$ as $n\to \infty$ almost surely if $\alpha>1$
  3. For $r>0,X_n\to0$ as $n\to \infty$ in $L^r$ if $\alpha>r$

My attempt:

For 1 we must show $$\lim\limits_{n\to \infty}P(|X_n-0|>\epsilon)\to 0$$

$$\sum\limits_{n=1}^\infty P(X_n=0) = \sum\limits_{n=1}^\infty 1- \frac{1}{n^\alpha}=\infty$$ So Borel-Cantelli says $$P(\limsup X_n=0) = 1\implies \lim\limits_{n\to \infty}P(|X_n-0|>\epsilon)\to 0 $$

For the second one: If $\alpha >1$ $$\sum\limits_{n=1}^\infty P(X_n=n) = \sum\limits_{n=1}^\infty \frac{1}{n^\alpha}<\infty$$ Then Borell Cantelli tells us that $$P(\limsup X_n=n) = 0$$

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Part $1$ is much simpler than what you have. Note that $$P(|X_n| > \epsilon) \leq P(|X_n| > 0) = P(X_n = n) = n^{-\alpha} \to 0.$$

Part $2$ is correct.

For Part $3$, consider the following: $$\|X_n\|_{L^r}^r = n^r \cdot n^{-\alpha} = n^{r - \alpha}.$$ Under what conditions does this go to $0$?

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  • $\begingroup$ Thanks for the answert, I do not understand two things: 1) Why is $P(|X_n|>0)=P(X_n=n)$? I thought $P(|X_n|>0)=\sum\limits_{n=1}^\infty P(|X_n|=n)$ 2) Does $P(\limsup X_n=0)=0$ automatically mean $X_n\to 0$ in probability? $\endgroup$ – Matriz Apr 21 '16 at 15:44
  • $\begingroup$ (1), each $X_n$ only takes values of $0$ and $n$. If $|X_n| > 0$, then we know that $X_n \neq 0$, which means it must be equal to $n$. $\endgroup$ – Marcus M Apr 21 '16 at 15:46
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    $\begingroup$ (2) I don't understand this part. The term: $\limsup X_n$ is a random variable, so it doesn't make sense to write $P(\limsup X_n)$. $\endgroup$ – Marcus M Apr 21 '16 at 15:46
  • $\begingroup$ I'm sorry, I meant Does $P(\limsup X_n=n)=0$ automatically mean $X_n\to 0$ in probability? $\endgroup$ – Matriz Apr 21 '16 at 15:48
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    $\begingroup$ I think you mean $P(\limsup X_n = 0) = 1$. Yes, this in fact means that $X_n \to 0$ almost surely. Since each $X_n$ is non-negative, $\limsup X_n = 0$ implies that $\lim X_n = 0$. Thus, $P(\limsup X_n = 0) = P(\lim X_n = 0).$ $\endgroup$ – Marcus M Apr 21 '16 at 15:50

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