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Let $\hat{p}$ be the proportion of successes in $n$ independent Bernoulli trials each having probability $p$ of success.

(a) Compute the expectation of $\hat{p} (1-\hat{p})$.

Isn't $P(\hat{p} = k) = \binom{n}{k} p^k(1-p)^{n-k}$? So is $\mathbb{E}[\hat{p}(1-\hat{p})] = \sum_{k=0}^n \left[\binom{n}{k} p^k(1-p)^{n-k}\right]\left[1-\binom{n}{k} p^k(1-p)^{n-k}\right]$? Or am I just completely going about this the wrong way?

How do you calculate the approximate mean and variance of $\hat{p}(1-\hat{p})$ using the delta method.

Any advice would be appreciated... Thanks!

Edit: So I got $\mathbb{E} \left[\hat{p}(1-\hat{p})\right] = \frac{n-1}{n} p(1-p)$. How do I use the delta method?...

symmetricuser
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1 Answers1

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Hints:

  • Your first line is $P(\hat p=k/n)$, not $P(\hat p=k)$

  • In full generality, $E(\hat p(1-\hat p))=\sum\limits_k(k/n)(1-(k/n))P(\hat p=k/n)$

Did
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