Consider a population of size $N$ and draw i.i.d. a random sample $S=(i_1,\dots,i_n)$ of $\{1,...,N\}$ with replacement. We define the Hansen-Hurwitz estimator as $$ \hat{\tau}= \frac{1}{n}\sum_{j=1}^n \frac{y_{i_j}}{p_{i_j}} $$ where $p_j$ is the probability of selecting unit $j$.

I want to prove this estimator is an unbiased estimator of the population total $\tau = \sum_{j=1}^N y_j$.

When showing the Horwitz-Thompson estimator is unbiased (https://en.wikipedia.org/wiki/Horvitz%E2%80%93Thompson_estimator#Proof_of_Horvitz-Thompson_Unbiased_Estimation_of_the_Mean) you introduce an indicator function to be able to take the sum out of the expectation. Here the index set is not random but I still thinking it might be useful to introduce a similar indicator function. What is random here is the index $i_j$ so we have to get rid of this somehow. Initial steps yield:

$$ E[\hat{\tau}]= \frac{1}{n}\sum_{j=1}^n E\left[\frac{y_{i_j}}{p_{i_j}}\right] $$

Thus it would suffice to prove $E\left[\frac{y_{i_j}}{p_{i_j}}\right]= \tau$. Can anyone help me do this?

  • $\begingroup$ The estimator is simply $\hat\tau=\frac1n\sum_{i\in S}\frac{y_i}{p_i}$. And $E\left(\frac{y_i}{p_i}\right)=\sum_{j=1}^N \frac{y_j}{p_j}\cdot p_j=\sum_{j=1}^N y_j$ for $i\in S$. $\endgroup$ Feb 26, 2022 at 20:41

1 Answer 1


To avoid confusing indexes with values, and to be appropriate for the intended applications, generalize to any finite (or even countable) set $S = \{x_1, x_2, \ldots, x_N\}.$ Let the value of one draw be $X$ and define

$$Y = \frac{X}{p_X}.$$

Using the definition of expectation as the sum of the values times their chances, compute

$$E\left[Y\right] = \sum_{i=1}^N \left(\frac{x_i}{p_i}\right)\,p_i = \sum_{i=1}^N x_i = \tau.$$

Consequently, by linearity of expectation, in an iid sample $X_1, \ldots, X_n$ of $X$ (yielding corresponding values $Y_i = X_i/p_i$) we find

$$E\left[\hat \tau\right] = E\left[\frac{1}{n}\sum_{i=1}^n Y_i\right] = \frac{1}{n}\sum_{i=1}^n E[Y_i] = \frac{1}{n}\sum_{i=1}^n \tau = \tau.$$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.