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Suppose that $X_n$ is a sequence of random variables where $X_n = n^2$ with probability $\frac{1}{n}$ and $X_n = \frac{1}{n}$ with probability $1- \frac{1}{n}$. I am wondering how to show that the sequence $X_n$ converges in distribution to $0$. I have started with $P(X_n \leq t)$ but when I apply the limit, I have trouble specifying the $t$ ranges. Does anyone have any ideas?

Is it valid that $X_n \to \mathbb{1}_{t \leq 0}$ in distribution?

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    You don't need to have a range for t. You need to fix t at any point of continuity of the CDF F_0(t) for 0. (i.e. any $t\neq 0$ ). Then you need to show the CDFs $F_n(t)$ converge to $F_0(t)$ at those values of $t$ – 2017-01-04
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    I think the best approach would be to consider $t<0$ and $t>0$ separately (since those are the sets on which $F_0(t)$ has different values). ? And no, it's not right that $X_n\rightarrow \ 1_{t\leq0}.$ I think you're confusing $X_n$ for its CDF. As the problem states, $X_n\rightarrow 0$ in distribution. – 2017-01-04

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For any fixed $t<0$ you have that $P(X_n\le t)=0$ $($since $X_n\ge 0$ with probability $1)$ and therefore $$\lim_{n\to \infty} P(X_n\le t)=0\tag1$$ For any fixed $t>0$ you have that \begin{align}P(X_n\le t)&=P\left(X_n=\frac1n\right)\cdot \mathbf 1_{ \left\{\frac1n \le t\right\}}+P\left(X_n=n^2\right)\cdot \mathbf 1_{\left\{n^2 \le t\right\}}\\[0.2cm]&=\left(1-\frac1n\right)\mathbf 1_{\left\{\frac1n \le t\right\}}-\frac1n \mathbf 1_{\{n^2\le t\}}\end{align} Hence $$\lim_{n\to \infty }P(X_n\le t)=\lim_{n\to \infty}\left[\left(1-\frac1n\right)\mathbf1_{\left\{\frac1n0\end{cases}$$ On the other hand, the distribution function $F$ of the degenerate random variable $X\equiv 0$ is $$F_X(t)=\begin{cases}0, & t<0\\1, &t= 0\\1, &t>0\end{cases}$$ However, according to the definition of convergence in distribution we do not need to consider $t=0$ since it is not a continuity point of $F$ and the conclusion follows.

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    Not true that $\lim_{n\rightarrow\infty} P(X_n=0) = 1.$ $P(X_n=0)=0$ for any n, thus the limit is 0. – 2017-01-04
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    @spaceisdarkgreen Yes, of course. I erroneously jumped to this conclusion. Hope that my correction works. Thanks for noticing. – 2017-01-04
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    Yes, looks right. – 2017-01-04
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    @JimmyR. normally indicator functions are defined on random variables but above the indicator is on something like $\mathbf1_{\left\{\frac1n – 2017-01-09
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    @user321627 Yes, above the indicator concerns the sequence $(n)_{n\in \mathbb N}$ and the fixed $t$. It is valid and rigorous to my knowledge. Since $n$ varies, the indicator is now a function of $n$ ($t$ is fixed). But you are right, since you have used the indicator only with random variables, it may be the case that your lecturer will be uncomfortable with it. – 2017-01-10
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Since $P(|X_n|>\epsilon)=1/n$, as $n>1/\epsilon$, so $X_n\to 0$ in probability and $X_n\to 0$ in distribution too.