Let $\mathbb P(X=1) = \mathbb P(X=-1) = 1/2$. Define
$$X_n = \begin {cases} X & \text{with probability } 1- \frac{1}{n}\\ e^n & \text{with probability } \frac{1}{n} \end {cases}$$
Does $X_n \leadsto X$ (i.e., converge in distribution)? Does $X_n \stackrel{P}{\rightarrow} X$? Does $\mathbb{E}[(X-X_n)^2] \to 0$?
I think $X_n \leadsto X$ because just by picturing the CDF as n goes to infinity, their CDFs converge. Can I get some confirmation on this?
But for convergence in probability, I'm having trouble with the limits.
To do this, what I'm trying to do is find $\mathbb P(|X_n-X| \leq \epsilon)$ and then just take 1 minus that as shown on this site: http://www.statlect.com/prbcon1.htm
If $X_n = X$, then $|X_n - X|$ is just $0$ so for all $n$, $|X_n -X| < \epsilon$ But if $X_n = e^n$ then I have the two cases for $X$. For $X=1$, $|X_n-X| = |e^n-1| \leq \epsilon$. and For $X=-1$, $|X_n-X| = |e^n +1| = e^n + 1 \leq \epsilon$. This is where I am stuck. I'm trying to find conditions relating $n$ and $\epsilon$ similiar to here the example on the site. But I'm not sure how to get that from $|e^n - 1| \leq \epsilon$ and $e^n + 1 \leq \epsilon$ .
Am I even on the right track?