A question in Stochastic Convergence

Question

I was reading a proof in a paper and got stuck at the following. Have been trying by best to figure this out for a long time but had no luck.

It says: When we have

\begin{equation}\tag{1} \frac{1}{{\alpha}^{i-2}}Z_{i-1}\overset{d}\rightarrow\sum_{j=1}^{\infty}\alpha^{1-j}u_j\qquad \text{as}\quad i\rightarrow\infty \end{equation}

where $|\alpha|>1$, $Z_i$ is a sequence of integrable random variables, and $u_j$ is i.i.d. with mean 0 and variance 1, it implies that

\begin{equation}\tag{2} \frac{Z_{i-1}^2}{2+Z_{i-1}^2}\overset{p}\rightarrow 1 \qquad \text{as}\quad i\rightarrow\infty \end{equation}

from which it follows that \begin{equation}\tag{3} \frac{1}{n}\sum_{i=1}^{n}\frac{Z_{i-1}^2}{2+Z_{i-1}^2}\overset{p}\rightarrow 1 \qquad \text{as}\quad n\rightarrow\infty. \end{equation}

$\quad$

If anyone has any clue about this please could you share it with me? Thanks in advance.

Answer 1

Here is part of a proof. I assume that the $Z_i$ and $\alpha$ are real valued.

Following kimchi lover's hint, let $X_i = Z_i/\alpha^i$, so that (1) tells us that $X_i$ converges in distribution to a random variable $X$ which is a.s. finite. (to see this, note the sum on the right side of (1) converges in $L^2$). To be lazy here, let's invoke Skorokhod's representation theorem to produce random variables $Y_i, Y$ which are individually (not jointly) identically distributed to their $X$ counterparts, such that $Y_i \to Y$ almost surely.

Then $$\frac{Z_i^2}{2+Z_i^2} \overset{d}{=} \frac{\alpha^{2i} Y_i}{2 + \alpha^{2i} Y_i} = \frac{Y_i^2}{2 \alpha^{-2i} + Y_i^2}. \newcommand{\Var}{\operatorname{Var}}$$ This converges to 1 a.s. on the event $\{Y \ne 0\}$.

We now need to show that $P(Y = 0) = P(X=0) = 0$, i.e. the the right side of (1) is a.s. nonzero. Here I have a gap.

But, if we can show that, then $\frac{Y_i^2}{2 \alpha^{-2i} + Y_i^2} \to 1$ almost surely. This means that $\frac{Z_i^2}{2+Z_i^2} \to 1$ in distribution, and convergence in distribution to a constant is equivalent to convergence in probability.

Moreover, since $0 \le \frac{Z_i^2}{2 + Z_i^2} \le 1$, by dominated convergence we also have $\frac{Z_i^2}{2+Z_i^2} \to 1$ in $L^1$. By the usual proof for convergence of Cesaro means (which extends to Banach spaces), this implies $\frac{1}{n} \sum_{i=1}^n \frac{Z_i^2}{2+Z_i^2} \to 1$ in $L^1$ as well, and in particular, in probability.

Answer 2

The gap in the accepted answer is filled by observing that the distribution of $Y$ is continuous (that is, with a continuous cdf, which is to say, atomless), so for each $x$, $P(Y=x)=0$. In particular, for $x=0$.

This in turn follows from Theorem XIII in Paul Lévy, "Sur les series dont les termes sont des variables eventuelles independantes", Studia Math. 3 (1931), 119-155.

See the first paragraph of P. Hartman's Infinite convolutions on locally compact abelian groups and additive functions a concise statement, or Lukacs's 1970 Characteristic Functions, p.63. This result is mentioned in Breiman's textbook (p.51) and in Stout's 1974 Almost Sure Convergence, p.100.

Let $V_n$ be a sequence of independent rvs. Let $d_n = \max_x P(V_n=x)$ be the largest "jump" in the distribution function of $X_n$.

Suppose $\sum V_n$ converges almost surely to a random variable $Y$.

Lévy's result is that $Y$ is not a continuous rv if and only if $\prod_1^\infty d_n \ne 0$.

In our case there is an iid sequence of $U_n$ with $EU_n=0$, $EU_n^2=1$, such that $V_n = \alpha^{-n} U_n$ with $|\alpha|>1$. This implies all the $d_n$ are equal and (because the $U_n$ have positive variances) strictly less than $1$. Hence $\prod_{n=1}^\infty d_n = 0$, and hence our $Y$ is a continuous random variable, as desired.

Added, 3 May 2023: This argument only needs part of Lévy's result, namely the implication that if $\sum (1-d_n)$ diverges then $Y$ is atomless, which follows directly from Theorem 2 of B.A. Rogozin, "An estimate for concentration functions" in Theory of Probability and its Applications, 6(1961), pp. 94-97 in the English translation.