$\frac{1}{u_n}W_n-v_n\Rightarrow W$ implies $u_n \to +\infty$

Question

Suppose $(X_n)_n$ are i.i.d. random variables and let $W_n = \sum_{k=1}^n X_k$. Assume that there exist $u_n>0 , v_n \in \mathbb{R}$ such that

$$\frac{1}{u_n}W_n-v_n\Rightarrow W$$

where $W$ is not degenerate. Show that

$$u_n\to \infty , \frac{u_n}{u_{n+1}}\to 1$$

What happens to $u_n$ if $W$ is degenerate?

Hint: You may need to consider ${u_{2n}}/{u_n}$.

Is the following attempt, for the first part, true?

In order to remove $v_n,$ we can consider $\frac{1}{u_n}\sum_{k=1}^n(X_{2k+1}-X_{2k})$ which converges in distribution to a non-degenerate random variable $Y.$ So we can suppose that $\frac{1}{u_n}W_n$ converges in distribution to $W$.

If $W$ is non-degenerate then there exist $x \in \mathbb{R};|\phi_W(x)|<1,$ since $\frac{1}{u_n}W_n-v_n\Rightarrow W$ then there exist $k \in \mathbb{N};|\phi_{X_1}(\frac{x}{u_k})|<1$ which means that $X_1$ is not degenerate, if $(u_n)_n$ is bounded from above then there exist a subsequence $(u_{k_n})$ such that $W_{k_n}$ converges in distribution. Let $(u_{q_n})_n$ be an arbitrary subsequence, since $X_1$ is not-degenerate then $\sum_{l=1}^{q_n}X_l=W_{q_n}$ doesn't converges in distribution, so $u_{q_n}$ is not bounded from above and we can extract a subsequence from $u_{q_n}$ diverging to $+\infty.$

In case $W$ is degenerate, is it possible to know the behavior of $u_n$?

Answer 1

Suppose $\frac {W_n} {c_n} $ converges in distribution to $W$ and $(c_n)$ does not tend to $\infty$. Then there is a subsequence $c_{n_k}$ converging to some real number $c$. This implies that $W_{n_k}$ converges in distribution to $cW$. Hence $(\phi(t))^{n_k} \to Ee^{itcW}$ where $\phi$ is the characteristic function of $X_i$'s. Since $Ee^{itcW}$ does not vanish for $t$ near $0$ it follows that $|\phi (t)|=1$ for al $t$ near $0$. This implies that $X_i$'s are a.s constants. In this case $c_n \sim nc$.

If $W$ is allowed to be degenerate, take $X_n=0$ for all $n$. Obviously nothing can be said about $c_n$ in this case.

Answer 2

In order to finish the solution we should show that $\frac{u_{n+1}}{u_n} \to 1$.

As $\frac{X_{n+1}}{u_n} =\frac{X_{1}}{u_n} $ and $u_n \to \infty$ we have $\frac{X_{n+1}}{u_n} \to 0$ in distribution. We have

$\frac{\sum_{k=1}^{n+1} X_k - X_{n+1}}{u_n} = \frac{\sum_{k=1}^n X_k}{u_n} \to W$, $\frac{X_{n+1}}{u_n} \to 0$, thus $\frac{\sum_{k=1}^{n+1} X_k}{u_n} \to W$. But we know that $\xi_n = \frac{\sum_{k=1}^{n+1} X_k}{u_{n+1}} \to W$. Put $c_n = \frac{u_{n+1}}{u_n}$. Thus $\xi_n \to W$ and $c_n \xi_n = \frac{u_{n+1}}{u_n} \frac{\sum_{k=1}^{n+1} X_k}{u_{n+1}} = \frac{\sum_{k=1}^{n+1} X_k}{u_{n}} \to W$.

So we know that $\xi_n \to W$, $c_n \xi_n \to W$ in distribution, $c_n >0$, $W$ is nondegenerate and we want to show that $c_n \to 1$.

Let us prove if by contradiction. Suppose that there is $c_{n_k} \to c \in [0,1) \cup (1, \infty]$. Instead of $c_{n_k}$ we will write $c_n$.

According to Skorokhod's representation theorem we may assume W.L.O.G. that $\xi_n \to W$ a.s. (and still $c_n \xi_n \to W$ in distribution).

Case 1. $c = \infty$. As $\xi_n \to W$ a.s., $|W| < \infty$, $c_n \to \infty$, we have $|c_n \xi_n(\omega)| \to (+\infty)$ for $\omega: W(\omega) \ne 0$. But $|c_n \xi_n| \to |W| < \infty$ in distribution. Thus $W = 0$ a.s. It's contradiction.

Case 2. $c=0$. As $\xi_n \to W$ and $c_n \to 0$, we have $c_n W_n \to 0$, but $c_n W_n \to W$. Thus $W = 0$ a.s. It's contradiction.

Case 3. $c \in (0,1) \cup (1,\infty)$. As $\xi_n \to W, c_n \xi_n \to W, c_n \to c$, we have $W = cW$ in distribution and hence $\frac{1}{c}W = W$ in distribution.

We see that it's sufficient to show that $W = dW$ in distribution with $d \in (0,1)$ (and $d=c$ or $d = \frac1{c}$) is impossible.

As $W =dW$, we have $$W = dW = d(dW) = d^2 W = \ldots = d^n W $$ in distribution.

Put $\eta_n = d^n W$. Thus $\eta_n \to 0$, because $d^n \to 0$. But $\eta_n = W$ in distribution. Thus $W=0$. It's contradiction.

Hence we got a contradiction in every case, q.e.d.