# How to show the Hansen-Hurwitz estimator is unbiased?

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?

• 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 $i\in S$. Commented Feb 26, 2022 at 20:41

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