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I wonder if the Weak Law of Large Numbers is only applicable if the random variable is binomially distributed. (The random variable counts the relative frequency of an event A). So, when you describe the Law, do you have to mention as a prerequisite that you only look at Bernoulli experiments?

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up vote 3 down vote accepted

As long as the $X_i$ are independent identically distributed, and their mean exists, the sample mean converges in probability to the mean. There is no need to have the $X_i$ be Bernoulli.

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Thanks, you two. I only find the Weak LLN phrased for Bernoulli experiments in my books. Why is that the case? Because you can easily proof it for Bernoulli experiments, using Chebyshev's inequality? (there is no proof for other cases than X being Bernoulli) – Sai Jun 26 '12 at 13:41
@Sai: I guess the Bernoulli case is the closest to the intuition. The Chebyshev Inequality argument works in exactly the same way for $X_i$ that have a variance. The result holds more generally, but for $X_i$ that don't have a variance the proof involves more advanced ideas. – André Nicolas Jun 26 '12 at 13:59

Check it out here. Both Weak LLN and Strong LLN are stated over any sequence of iid integrable random variables, so it applies not only to Bernoulli random variables, but also to Poisson, Gaussian, exponential etc.

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