Find the variance of p-hat

Question

We have not discussed $\hat p$ in my probability and statistics course and a problem involving it is on our hw this week after learning about discrete distributions. The problem states "Let the random variable $Y\sim \text{Binomial}(n,p)$ and let $\hat p = \frac{Y}{n}$.

a. Find the mean of $\hat p$.

b. Find the variance of $\hat p$.

c. Use this and Chebyshevs theorem limit as $n$ goes to infinity of $Pr(\vert\hat p-p\vert < a)$ for any $a>0$.

I was able to find the mean of $\hat p$ as $p$ and I know that the variance of $\hat p$ should be $p(1-p)/n$ but I have been unable to prove that or do part c.

Answer 1

a.

$$ \mathbb{E}(\hat{p}) = \mathbb{E}\left(\dfrac{Y}{n}\right)= \dfrac{np}{n}= p$$

b.

$$ \operatorname{Var}(\hat{p}) = \dfrac{\operatorname{Var}(Y)}{n^2} = \dfrac{p(1-p)}{n} $$

Note here that

$$\lim_{n\rightarrow \infty} \operatorname{Var}(\hat{p}) =\lim_{n\rightarrow \infty} \sigma^2 = 0$$

c. I'll assume you mean Chebyshev's inequality

The inequality says

$$ \operatorname{Pr}(\vert \hat{p} - p| > k\sigma) \leq \dfrac{1}{k^2} $$

In your case $\alpha = k\sigma$ so $1/k = \sigma/\alpha$. Since $\alpha$ is non -zero, we have no problems thus far. Substituting, we have

$$ \operatorname{Pr}(\vert \hat{p} - p| >a) \leq \dfrac{\sigma^2}{\alpha^2} $$

If $\alpha$ does not vary with $n$, what can you conclude about

$$\lim_{n\rightarrow \infty} \operatorname{Pr}(\vert \hat{p} - p| >a)$$

Hint: See the law of large numbers