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.

  • $\begingroup$ I assume you mean $X \sim binomial (N,\theta)$? Otherwise it is possible to find counterexamples for the statements to be proven. $\endgroup$ Commented Sep 6, 2021 at 14:38
  • $\begingroup$ @Andreas You are correct, fixed now. $\endgroup$
    – Pame
    Commented Sep 6, 2021 at 14:46

1 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


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

  • $\begingroup$ Im curious how you got $$\mathbb{E}\left[\frac{1}{n}\mathbb{E}[X] \right]=\frac{1}{N}\cdot N\theta?$$ Or did you mean to put a small $n$ in the bracket? $\endgroup$
    – Pame
    Commented Sep 6, 2021 at 14:50
  • $\begingroup$ @Pame : yes in this case is better to do that. Amended $\endgroup$
    – tommik
    Commented Sep 6, 2021 at 15:03
  • $\begingroup$ Also not sure if u noticed but I made a correction in my post so it now says that $X \sim binomial(N,p)$ so that $E(X) = N \theta$. $\endgroup$
    – Pame
    Commented Sep 6, 2021 at 15:05
  • 1
    $\begingroup$ @Pame : your conditional random variable $(X|N=n )\sim Bin(n;\theta)$ thus this is the correct notation. $n$ is lowercase when in the bracket and Capital when considered a random variable... $\endgroup$
    – tommik
    Commented Sep 6, 2021 at 15:09
  • $\begingroup$ @Pame The notation in this answer is incorrect. One should write $E\left[\frac{X}{N}\right]=E\left[E\left[\frac{X}{N}\mid N\right]\right]=E\left[\frac1NE[X\mid N]\right]$. And you are given that $E[X\mid N]=N\theta$. $\endgroup$ Commented Sep 6, 2021 at 16:45

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