How to show that 0.5Y is an unbiased estimator of alpha?

Question

How do I show that 0.5Y in part (a) down below is an unbiased estimator of alpha? If n, the number of observations, is very large then Y would equal to alpha+alpha=2*alpha an then 0.5Y=alpha so E(0.5Y)=E(alpha)=alpha but the size of n is not stated so how is this question supposed to be solved?

Answer 1

How do I show that 0.5Y in part (a) down below is an unbiased estimator of alpha?

The rv $Y\cdot n$ is evidently a binomial

$$nY\sim Bin(n;2\alpha)$$

thus

$$\mathbb{E}[nY]=2n\alpha$$

$$\mathbb{E}[Y]=2\alpha$$

and evidently,

$$\mathbb{E}[\frac{Y}{2}]=\alpha$$

Answer 2

The probability that a given observation is nonzero is $\alpha+\alpha$ so

$$\mathbb E\left(\frac{1}{2}Y\right)=\frac{1}{2}\mathbb E(Y)=\frac{1}{2}(\alpha+\alpha)=\alpha$$

Similarly

$$\mathbb E\left(\frac{1}{3}\overline{X}\right)=\frac{1}{3}\mathbb E\left(\overline X\right)=\frac{1}{3}\left(\alpha+2\alpha\right)=\alpha$$

Can you go from here to show that $\frac{1}{2}Y$ is more efficient? Since they're both unbiased estimators, you just need to determine which estimator has a smaller variance.