# Divergence rate of geometric sum of random variables

Let $(X_n)_{n\in\mathbb{N}}$ be a sequence of strictly positive and identically distributed random variables and let $\beta\le 1$. I am trying to prove that

$$0<\lim_{\beta\rightarrow 1}(1-\beta)\sum_{n=0}^{\infty}\beta^nX_n <\infty.$$

I have already shown that the limit is finite if $\mathbb{E}\log_+(X_n))<\infty$, but I am having trouble showing the nonzero part. Does anybody have a reference or idea on how to prove this result?


Note that $\mu:=\E X_1\in(0,\infty]$. Suppose first that $\mu<\infty$. Then, for $\be\uparrow1$, \begin{align*} (1-\be)\sum_{n=0}^{\infty}\be^n X_n &=(1-\be)^2\sum_{n=0}^\infty X_n\sum_{j=n}^\infty\be^j \\ & =(1-\be)^2\sum_{j=0}^\infty \be^j\sum_{n=0}^j X_n \\ & \sim(1-\be)^2\sum_{j=0}^\infty \be^j (j+1)\mu =\mu; \end{align*} here we used the strong law of large numbers and the fact that $(1-\be)^2\sum_{j=0}^N \be^j\sum_{n=0}^j X_n\to0$ for each natural $N$.
So, $$\lim_{\be\uparrow1}(1-\be)\sum_{n=0}^{\infty}\be^n X_n=\mu \tag{1}$$ if $\mu<\infty$. By using the truncation $X_n1_{X_n\le C}$, it is now easy to see that (1) holds for $\mu=\infty$ as well.

(In particular, it follows that the limit does not have to be finite if you only have $\E\log_+(X_n)<\infty$.)

Write \begin{align*} S_\beta &=(1-\beta)\big[X_0+X_1+X_2+\dots\big] \\ &=(1-\beta)\big[(1-\beta)X_0 + (\beta-\beta^2)(X_0+X_1) + (\beta^2-\beta^3)(X_0+X_1+X_2) + \dots\big] \\ &=(1-\beta)^2\big[X_0 + \beta(X_0+X_1) + \beta^2 (X_0+X_1+X_2)+ \dots\big]. \end{align*}

Suppose $X_n$ has finite mean $\mu$. Let $\epsilon>0$. By the Strong Law of Large Numbers, with probability $1$, $X_0+X_2+\dots+X_{n-1}>n(\mu-\epsilon)$ for all large enough $n$. In that case, \begin{align*} \liminf_{\beta\uparrow 1} S_\beta &\geq \liminf_{\beta\uparrow 1} (\mu-\epsilon)(1-\beta)^2[1+2\beta+3\beta^2+\dots]\\ &=\mu-\epsilon. \end{align*} In a similar way, with probability $1$, $\liminf_{\beta\uparrow1} S_\beta\leq \mu+\epsilon$. So you have that with probability $1$, the limit $\lim_{\beta\uparrow1} S_\beta$ exists and equals $\mu$.

In the case where the mean is infinite, the same argument gives that with probability $1$, $S_{\beta} \to \infty$ as $\beta\uparrow1$. This seems to contradict your statement that the limit is finite whenever $\mathbb{E} \log_+ X_n<\infty$.

• Ah, I see Iosif posted an almost identical argument a few moments earlier :) – James Martin Sep 4 '18 at 17:29
• Right, practically the same argument at about the same time. :-) – Iosif Pinelis Sep 4 '18 at 17:39