Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I'm stuck with this problem.
Let $X_1, X_2, ...$ be a sequence of independent Bernoulli random variables. Show that if $$\sum_{i=1}^n \frac{p_i}{n} \to l \; \text{ as } \; n \to \infty$$ then $$\sum_{i=1}^n \frac{X_i}{n} \to l \; \text{ stochastically }$$

I have noticed some facts, if we call $X'_n = \sum_{i=1}^n \frac{X_i}{n}$, then $\mu_n = E[X'_n] = \sum_{i=1}^n \frac{p_i}{n}$, so in some sense the $\mu_n$ converges to $l$.
Also using the usual method of proving the Bernoulli law of large numbers I have arrived at the following $$P[|X'_n - \mu_n| < \epsilon] \geq 1 - \frac{\sum_{i=1}^np_iq_i}{\epsilon^2n^2}$$ And the rightmost term I think it goes to $0$ as $n \to \infty$, so this would almost give me the result except for the $\mu_n$ which would have to be replaced by $l$ (which seems reasonable given that the $\mu_n$ converges to it). I'm a little rough in this area so I'm unsure of which manipulations I can do, and also I am unaware of a lot of theorems (maybe one can lead to the result I want).
So anyway, my question is mainly, I am going in the right direction, and if so, how can I conclude the result?
Or are the other standard ways (like using moment generating functions or calculating the cumulative distribution of the $X'_n$ and taking limits) better approaches to this problem?
Any help would be appreciated :)

share|improve this question
this isn't true. it certainly isn't true if the $p_i$ can be zero. –  mike Oct 19 '12 at 11:38

1 Answer 1

up vote 2 down vote accepted

First, looking at your rightmost term (note that $p_iq_i = p_i(1-p_i) \le \frac 14$ for all $i$, we have \[ 1 - \frac{\sum_i p_iq_i}{\epsilon^2 n^2} \ge 1 - \frac 1{4\epsilon^2 n} \] and this indeed goes to 1.

For the $\mu_n \leadsto l$ we observe that by convergence, given $\epsilon > 0$, for $n$ large enough we will have $|\mu_n - l| \le \frac \epsilon 2$. For these $n$ it holds \[ \left\{|X_n' - \mu_n| < \frac \epsilon 2 \right\} \subseteq \left\{|X_n' - l| < \epsilon\right\} \] by the triangle inequality. As the probability of the former set converges to 1, as you proved, the one of the latter also does.

share|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.