Let $p\in (0,1)$ and $X_1, X_2, ...X_n \sim \text{Bern}(p)$ be $n$ i.i.d. Bernoulli random variables, where the probability that $X_i$ is $1$ equals $p$.

Fix $a,b>0$ different from $1$ that satisfy $a^p b^{1-p} = 1$, and define $C_i = X_i(a-b)+b$. In other words, $C_i$ is $a$ when $X_i$ is $1$, and $b$ when $X_i$ is $0$.

I am interested in the behavior of the random variables $$Z_n = \frac{\sum_{i = 1}^n\left(\prod_{j = 1}^iC_j\right)X_i}{\sum_{i = 1}^n\prod_{j = 1}^i C_j} $$
as $n\to \infty$. Does $Z_n$ converge a.s.? Does $Z_n$ converge in distribution?

Note that the product of the $C_j$'s is sometimes very small and sometimes very large, as the central limit theorem says that $\frac{1}{\sqrt n} \sum \log C_i$ converges in distribution to a normal $\mathcal N (0, \sigma^2)$. This follows from the condition that $a^pb^{1-p} = 1$.

I ran some numerical experiments in Mathematica that suggest that $Z_n$ does converge to a constant, but this constant (presumably the limit of the means of $Z_n$) is a non-trivial function of $a$, $b$, and $p$. Indeed, the mean of $Z_n$ is difficult to compute, and does not seem to simplify nicely.

I am not a probabilist, so any resource that deals with this kind of random variable would be helpful.

  • $\begingroup$ The a.s. convergence could be too hard to handle here. Even for the cumulative sums of iid random variables, the strong law of large numbers is not easy to prove. Also, the relevant theorem here, the CLT, is about weak convergence. So, would you be satisfied by the convergence of $Z_n$ in distribution? If so, please edit your post accordingly. $\endgroup$ Commented Jul 12, 2022 at 13:56
  • $\begingroup$ @losif Pinelis, I think I would be satisfied by the convergence of $Z_n$ in distribution. I have updated the post. $\endgroup$ Commented Jul 12, 2022 at 14:21
  • $\begingroup$ All right. Also, do you understand $\prod_{j = 1}^iC_jX_i$ as $X_i\prod_{j = 1}^iC_j$? $\endgroup$ Commented Jul 12, 2022 at 14:23
  • $\begingroup$ Yes, that's correct. See the edit $\endgroup$ Commented Jul 12, 2022 at 14:31
  • $\begingroup$ and you absolutely want $X_i$ in the numerator and not $X_{i+1}$ ? $\endgroup$
    – mike
    Commented Jul 15, 2022 at 6:04

1 Answer 1


James, using the fact that $X_i$ only takes on 2 values, write $C_iX_i = xC_i + y$ . Then your numerator is $$\sum^n x \prod^j C_i + \sum^{n-1} y \prod^j C_i = \sum^n (x+y) \prod^j C_i - yC_n$$. It suffices to show that $$ ~\frac {\prod^n C_i} {\sum^n \prod^j C_i } \rightarrow 0$$, where I am going to show the convergence in probability. $$$$ Lemma: if $S_n$ is a mean 0 random walk, $\frac {e^{S_n}} {\sum^n e^{S_i}} \rightarrow 0$ in probability. $$$$Proof: $$\frac {e^{S_n}} {\sum^n e^{S_i}} = \frac 1 {1 + e^{-X_n} + e^{-X_n - X_{n-1}} + .... + e^{-X_n - X_{n-1} - ... - X_1}}$$. In the exponents in the denominator is the time reversed random walk, which is also a mean 0 random walk, and it pretty obviously goes to 0 in probability, e.g., break it up into returns to 0.$$$$ So the conclusion is, your expression is converging in probability to x+y. To do this explicitly, $X = (C - b)/(a-b)$, $XC = aX = a(C-b)/(a-b)$ and I seem to be claiming that the limit is a(1-b)/(a-b).


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