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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 :)

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

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