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I am trying to derive the distribution of $\bar{X}_n$ where $X_1, X_2,...,X_n$ are iid $\sim \mathrm{Bern}(p)$. I used two approaches but I am debating myself and questioning which one would be correct (if any).

Method 1: Using MGF I used the moment generating function and ended up with $\bar{X}_n \sim \mathrm{Bern}(p^n)$

Method 2: I used the CLT and ended up with $\bar{X}_n \sim N(p, \sqrt{pq}/n$) for n being large.

I am not sure which one is correct (if any).

Can someone tell me if I am doing this right or not?

Thank you, I appreciate your help

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Neither.

  • For the first one, you are claiming that the average only take binary value.

  • For the second one, what if $n$ is small.

Guide:

  • Recall that sum of IID Bernoulli follows a binomial distribution $Bin(n, p)$.
  • Average is simply dividing the sum by $n$.
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  • $\begingroup$ Won't I get a Bernoulli again if I simply divided by n? Bin(n/n, p/n) => Bin(1, p/n) (one trial) => Bern(p/n)? $\endgroup$ – S A Dec 4 '18 at 1:10
  • $\begingroup$ Suppose $n=2$, the values that you can get are $0, 0.5, 1$. It is possible to get value $0.5$, something that is not modelled by the Bernoulli distribution. $\endgroup$ – Siong Thye Goh Dec 4 '18 at 1:12

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