# Convergence in distribution example

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1. Apr 25, 2017

### AlexF

1. The problem statement, all variables and given/known data

2. Relevant equations

Definition: A sequence $X_1,X_2,\dots$ of real-valued random variables is said to converge in distribution to a random variable $X$ if $\lim_{n\rightarrow \infty}F_{n}(x)=F(x)$ for all $x\in\mathbb{R}$ at which $F$ is continuous. Here $F_n, F$ are the cumulative distributions functions of the random variables $X_n$ and $X$ respectively.

3. The attempt at a solution

I'm trying to understand/recreate the following solution to the problem.

My working so far is that
$$F_{X}(x)=P(X\leq x)=\begin{cases} 0, &x<-1 \\ 1/2, &x\in[-1,1) \\ 1, &x\geq 1\end{cases}$$ and since $X$ only takes values 1 and -1 then $X_n = (-1)^{n+X}+\frac1n=(-1)^{n+1}+\frac1n$ and so $$F_{X_n}(x)=P(X_n\leq x)=\begin{cases} 0, &x<(-1)^{n+1}+\frac{1}{n} \\ 1, &x\geq (-1)^{n+1}+\frac{1}{n}\end{cases}$$ I can't understand how the limits to this have been achieved in the solution. Why does $F_{X_n}(x)\rightarrow 1/2$ for $t\in(-1,1)$, say?

Last edited: Apr 25, 2017
2. Apr 25, 2017

### andrewkirk

I agree with your analysis. It looks like the problem has been incorrectly stated. The $X_n$ are not even random, since $X_n=(-1)^{n+1}$ for all integer $n$.

The $X_n$ do not converge in distribution to $X$ because $F_X$ is continuous at 0 and equal to $1/2$, but $F_{X_n}(0)$ is alternately $0$ and $1$ as $n$ increments, hence $F_n(0)$ does not converge to $1/2$.

The specific error in the text's attempted proof is the statement that 'for large enough $n$, $F_{X_n}(t)=1/2$' (for $t\in(-1,1)$) .

3. Apr 26, 2017

### AlexF

That makes sense, thanks a lot! I thought I was going crazy xD