# Find the variance of p-hat

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.

• I was able to prove the variance of p-hat but am struggling with where to start with part c. Commented Apr 7, 2019 at 23:22
• Can you update your question showing what you have tried so far? Also you should add the self-study tag while you're at it. Commented Apr 8, 2019 at 1:26
• You need the self study tag Commented Apr 8, 2019 at 2:35

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

• @EBusch does this answer your question? Commented Apr 9, 2019 at 14:18