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