Let $N$ be a random variable taking values $1,2,...,n$, with known probabilities $p_1,p_2,...,p_n$, where $\sum_i p_i = 1$. Furthermore let $X \sim binomial(N,\theta)$.
Consider now the estimator $\frac{X}{N}$ and show that $E(\frac{X}{N}) = \theta$, and $Var(\frac{X}{N}) = \theta(1-\theta)E(\frac{1}{N})$
So far Im struggling to find the expected value. I know that $E(\frac{X}{N}) = E(X) \cdot E(\frac{1}{N}) = n \theta E(\frac{1}{N}).$
The formula for $E(\frac{1}{N})$ is $E(\frac{1}{N}) = \sum_i \frac{1}{i} p_i$ but not sure how to determine this sum.
Also not sure what formula to apply to calculate the variance.
Would appreciate any help.
First apply law of total expectation and get
$$\mathbb{E}\left[\frac{X}{N} \right]=\mathbb{E}\left[\mathbb{E}\left[\frac{X}{N}\left|N=n \right. \right] \right]=\mathbb{E}\left[\frac{1}{n}\mathbb{E}[X] \right]=\mathbb{E}\left[\frac{1}{n}\cdot n\theta \right]=\theta$$
then using the definition
$$\mathbb{V}[X]=\mathbb{E}[X^2]-\mathbb{E}^2[X]$$
you can find your variance
$$\mathbb{E}\left[\left(\frac{X}{N} \right)^2 \right]=\mathbb{E}\left[\mathbb{E}\left(\frac{X}{N} \right)^2 |N=n \right]=$$
$$=\mathbb{E}\left[\frac{1}{n^2}\mathbb{E}[X^2] \right]=\mathbb{E}\left[\frac{\theta(1-\theta)}{n}+\theta^2 \right]=\mathbb{E}\left[\frac{1}{N} \right]\theta(1-\theta)+\theta^2$$
which is
$$\mathbb{V}\left[\frac{X}{N} \right]=\mathbb{E}\left[\frac{1}{N} \right]\theta(1-\theta)+\theta^2-\theta^2=\mathbb{E}\left[\frac{1}{N} \right]\theta(1-\theta)$$
...as requested