My question is on an exercise from Probability Essentials by Jacod and Protter:
18.6 Let $(X_n)_{n \geq 1}$ be a sequence of real valued random variables with $\mathcal{L}(X_n)$
uniform on $[−n, n]$. In what sense(s) do $X_n$ converge to a random variable
$X$? [Answer: None.]
I don't understand why the answer is "None". This must mean that $X_n$ does not converge in distribution to any random variable (since convergence in distribution implies almost-sure convergence as well as convergence in probability). But I came up with the following argument that seems to show that the characteristic functions of $X_n$ converges to the characteristic function of the zero function, which would imply that $X_n \to 0$ in distribution by Levy's continuity theorem.
My question: Where is my mistake in the following "proof"?
"Proof": First, let us a fix a probability space $(\Omega, \mathcal{H}, \mathbb{P})$. Now let $X_1,X_2,\ldots: \Omega \to \mathbb{R}$ be a sequence of $\text{Uniform}([-n,n])$ random variables. Let $\mu_n: \mathcal{B}(\mathbb{R}) \to [0,1]$ denote the distribution of $X_n$, and let $\widehat{\mu}_n: \mathbb{R} \to \mathbb{R}$ denote the characteristic function of $X_n$. For $r = 0$, we clearly have $\widehat{\mu}_n(0) = 1$ for all $n$, and for $r \neq 0$ we have
\begin{align*}
\widehat{\mu}_n(r) &= \mathbb{E}[e^{irX_n}] \\[4pt]
&= \int_{\mathbb{R}} \mu_n(dx) \, e^{irx} \\[4pt]
&= \int_{\mathbb{R}} \text{Leb}(dx) \frac{1}{2n} \chi_{[-n,n]}(x) \, e^{irx} \\[4pt]
&= \frac{1}{2n} \int_{-n}^{n} \text{Leb}(dx) e^{irx} \\[4pt]
&= \frac{1}{2n} \left[\frac{1}{ir}e^{irx} \right]_{x=-n}^{x=n} \\[4pt]
&= \frac{1}{2irn} \left(e^{irn} – e^{-irn} \right) \\[4pt]
&= \frac{1}{2irn} \left[\left(\cos(rn) + i \sin(rn)\right) – \left(\cos(-rn) + i \sin(-rn) \right) \right] \\[4pt]
&= \frac{1}{2irn} \left[\left(\cos(rn) + i \sin(rn)\right) – \left(\cos(rn) – i \sin(rn) \right) \right] && (\cos \text{ is even},\, \sin \text{is odd}) \\[4pt]
&= \frac{2i \sin(rn)}{2irn} \\[4pt]
&= \frac{\sin(rn)}{rn}.
\end{align*}
Thus, we have
\begin{align*}
\widehat{\mu}_n(r) =
\begin{cases}
1 & \text{if } r = 0 \\[2pt]
\dfrac{\sin(rn)}{rn} & \text{if } r \neq 0
\end{cases}
\end{align*}
for each $n \in \mathbb{N}$. And since $\lim\limits_{r \to 0} \frac{\sin(rn)}{rn} = 1$ for all $r \neq 0$, this implies that $\widehat{\mu}_n$ is continuous at $0$. Also, $\lim\limits_{n \to \infty} \frac{\sin(rn)}{rn} = 0$ for all $r \neq 0$. Thus,
\begin{align*}
\lim_{n \to \infty} \widehat{\mu}_n(r) &=
\begin{cases}
1 & \text{if } r = 0 \\[2pt]
0 & \text{if } r \neq 0,
\end{cases} \\[4pt]
&= \chi_{\{0\}}(r).
\end{align*}
But $\chi_{\{0\}}$ is the characteristic function of the zero function, i.e. $X(\omega):= 0$ for all $\omega \in \Omega$. Then by Levy's continuity theorem, $X_n \to 0$ in distribution.
…Where did I go wrong?
Best Answer
The characteristic function for the constant $0$ is $1$ for all $r \in \mathbb{R}^1$. In other words, $\chi_{\{0\}}(r)$ in your post is not the characteristic function of $0$, hence the condition of Levy's theorem does not hold.
There are many ways to verify that $X_n$ does not converge weakly (i.e., in distribution) to any random variable. One approach is to check the definition: for any $x \in \mathbb{R}^1$, there exists a sufficiently large positive integer $N$ such that $|x| < N$, whence for all $n > N$, it follows that \begin{align} P[X_n \leq x] = \int_{-n}^x \frac{1}{2n}dt = \frac{x + n}{2n} \to \frac{1}{2} \tag{1} \end{align} as $n \to \infty$. On the other hand, if $X_n \to_d X$ for some random variable $X$, then $P[X_n \leq x] \to P[X \leq x]$ for $x \in C(F)$, where $C(F)$ denotes the set of continuities of the distribution function $F$ of $X$. Since $\lim_{y \to \infty}F(y) = 1$, there exists $x_0 \in \mathbb{R}^1$ such that $P[X \leq x_0] > 7/8$, say, which implies $P[X_n \leq x_0] > 3/4$ for all sufficiently large $n$. This is in contradiction with $(1)$.