How do I find the cumulative distribution function of the sequence of random variables $X_n=(-1)^{n+X}+1/n$?

Question

The related question is the following.

The solution is as follows.

I know that $$F_{X}(x)=P(X\leq x)=\begin{cases} 0, &x<-1 \\ 1/2, &x\in[-1,1) \\ 1, &x\geq 1\end{cases}.$$

But how do I find $F_{X_n}(x)=P(X_n\leq x)$?

Answer 1

Notice that $X_n = (-1)^{n+X}+\frac1n=(-1)^{n+1}+\frac1n$ since $X$ only takes value $1$ and $-1$.

That is $X_n$ is a deterministic quantity regardless of the value of $X$.

$P(X_n \leq t)=0$ if $t< (-1)^{n+1} + \frac1n$.

$P(X_n \leq t)=1$ if $t \geq (-1)^{n+1} + \frac1n$.