Expected Value and Variance of Random Variable Divided by Another Random Variable

Question

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.

Answer 1

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