# Theoretical variance of an unbiased estimator.

Let $n_1, n_2, n_3$ denote sample sizes of $3$ samples of people, where $n_1 < n_2 < n_2$ and let $X_1, X_2, X_3$ denote the number of left handed people in each sample. Let $\hat{p_{1}} = \frac{(X_1)}{n_1}, \hat{p_{2}} = \frac{(X_2)}{n_2}, \hat{p_{3}} = \frac{(X_3)}{n_3}$

Show that $\hat{p_{12}} = \frac{(n_1)(\hat{p_1}) + (n_2)(\hat{p_2})}{n_1 +n_2}$ is an unbiased estimator, and find the variance. In order to show unbiasedness you prove that $E(\hat{p_{12}}) = \mu$, by allowing $\frac{X_1}{n_1}$ etc to = $\mu$ as they follow a normal distribution. In order to find theoretical variance, do i need to prove that $\frac{\sigma^2}{n} = 0$, or find it in terms of $X_1, n_1$ etc by showing that $\frac{\sigma}{n} = \frac{\sum(X_i - X_{bar})}{n}$.

• I would assume that the probability of an individual being left-handed is $\mu$ and then you find $\text{var}(\hat{p}_{12}) = E\left[ (\hat{p}_{12}-\mu)^2\right]$ in terms of $\mu$. This is binomially rather than normally distributed – Henry Oct 18 '16 at 10:47
• Thanks for the reply. It is not specified whether it is normally or binomially distributed. Is there any significant difference in the variance or mean calculation if we assume the sample sizes are large enough? – Jonathan Carver Oct 18 '16 at 11:00
• You are going to have to use the binomial distribution to find the variances of $X_1$ and $X_2$ so you may as well go on using it – Henry Oct 18 '16 at 11:44
• An unbiased estimator of what? Also, your (X) notation is peculiar and should be defined. – wolfies Oct 18 '16 at 14:27