Could anyone help me with this question?
Let $Y$ be a random variable on $\{-1, 0, 1\}$ with $P(Y = -1) = P(Y = 1) = 1/4$ and $P(Y = 0) = 1/2$ and let be a sequence of random variables $(X_n)_{n\ge 1}$ given by $X_n = (-1)^n Y $.
(a) Show that there exists a random variable $X$ such that $X_n$ converges to $X$ in distribution .
b) Show that there is no random variable $Z$ such that $X_n$ almost surely converges to $Z$.
I know that convergence in Distribution means that
$\begin{align}%\label{eq:union-bound}
\lim_{n \rightarrow \infty} F_{X_n}(x)=F_X(x)
\end{align}$
but isn't $ F_{X_n} = (-1)^n F_Y = (-1)^n$ ? and this is not convergent.
for b) I have no idea how to solve it 🙁
Best Answer
For a), notice that by symmetry $X_n$ has the same distribution as $Y$ for all $n$, so $X_n$ converges in distribution to $Y$.
For b) notice that if $X_n$ converges almost surely to some random variable $Z$, then we must have that for all $\epsilon, \delta>0$, there exists and $N_\epsilon$ sufficiently large so that for all $n\ge N_\epsilon$, $P(|X_n - Z|> \epsilon) < \delta$ (in other words, almost sure convergence implies convergence in probability). This works if for example $\delta < 1/4$ and $\epsilon < 2$. Notice that according to the definition of $X_n$, $P(|X_n - X_{n+1}| = 2 ) = 1/2$. But by the triangle inequality $|X_n - X_{n+1}| \le |X_n - Z| + |X_{n+1}-Z|$, and so for $n\ge N_\epsilon$, $P(|X_n - X_{n+1}|> \epsilon ) \le P(|X_n - Z| + |X_{n+1}-Z| > \epsilon) \le P(|X_n - Z| > \epsilon) + |X_{n+1}-Z| > \epsilon) \le 2 \delta$, which is a contradiction. Hence $X_n$ cannot converge almost surely.