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T is total population, where $t = \sum^N_{i=1}y_i$, and $y_i$ is the value of the unit at index $i$. Let $\hat{t} = \frac{N}{n}\sum^N_{i=1}Q_iy_i$ ($\hat{t}$ is a estimator of total population) where $Q_i$ is the number of times a unit i appears in a sample. Using the properties of the multinomial distribution, find $V[\hat{t}]$

I found that $E[\hat{t}] = t$, but I'm having trouble find the $V[\hat{t}]$.

I know that:

$V[\hat{t}] = E[\hat{t} - E[\hat{t}]^2]$

$ = E[\frac{N}{n}\sum^N_{i=1}Q_iy_i - (\sum^N_{i=1}y_i)^2]$

but I'm not sure how to continue, I'm particularly lost on how to factor the summations

I've also tried $E[\hat{t}^2] - E[\hat{t}]^2$, but still not sure how to get rid of the summations in the expectation

$E[\hat{t}^2] - E[\hat{t}]^2$ =$\frac{N^2}{n^2}E[(\sum^N_{i=1}Q_iy_i)^2] - t^2$

Could I evaluate it directly like this?

$V[\hat{t}] = V[\frac{N}{n}\sum^N_{i=1}Q_iy_i] = \frac{N^2}{n^2}V[\sum^N_{i=1}Q_iy_i] = \frac{N^2}{n^2}\sum^N_{i=1}V[Q_iy_i] = $

$\frac{N^2}{n^2}\sum^N_{i=1}y_iV[Q_i] = \frac{N^2}{n^2}\sum^N_{i=1}y_i (n\frac{1}{N})(1-\frac{1}{N}) = \frac{N}{n}t(1-\frac{1}{N})$

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    What is $t$? (Apart from a refreshing drink.) – 2017-02-04
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    @copper.hat total population – 2017-02-05
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    It may depend on whether you are sampling with or without replacement. In any case, your expression for the variance should perhaps be $V[\hat{t}] = E[(\hat{t} - E[\hat{t}])^2]= E[\hat{t}^2] - (E[\hat{t}])^2$ – 2017-02-05

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