Assume there is a population of $n$ voters. $n/3$ of those voters voted for Bob.

Given exit polls k'/k voters voted for Bob. What is the expected value of k'/k?. The k people are chosen with replacement.

My intuition is that the expected value from the original population is just 1/3 since it appears to be a Bernoulli distribution. Would the expected value of the exit poll population just be:

$\prod_{i=1}^{k} 1/3 = (\frac{1}{3})^k$ ?

Similarly would the variance just be:

$\prod_{i=1}^{k} 2/9 = (\frac{2}{9})^k$ ?


Edit: The post has been changed, and we are to assume that sampling is with replacement. In that case, the binomial mentioned below as an approximation is now exact.

Let $X$ be the number of people who reported voting for Bob in the exit poll (so this is your $k'$, but I prefer to use $X$ to stress that $X$ is a random variable.) Let $Y=\dfrac{X}{k}$.

The random variable $X$ has hypergeometric distribution, since we are sampling, presumably without replacement, from a population of size $n$.

However, since $n$ is probably quite large in comparison with $k$, the error is not large if we assume that $X$ has binomial distribution.

The mean of $X$, whether we use the actual hypergeometric, or the approximation by the binomial, is $k\dfrac{n/3}{n}$, that is, $\dfrac{k}{3}$.

The mean $E(Y)$ of $Y$, which is $\dfrac{1}{k}E(X)$, is therefore equal to $\dfrac{1}{3}$.

For the variance, it is simplest to approximate by the binomial, and it seems likely that you are intended to do so. The variance of $X$ is approximately $kp(1-p)$, where $p=(n/3)/n=1/3$. So $X$ has variance approximately equal to $k(1/3)(2/3)$.

The variance of $Y$, which is the variance of $\dfrac{1}{k}X$, is equal to $\dfrac{1}{k^2}$ times the variance of $X$. Using our approximation by the binomial, we find that the variance of $Y$ is approximately $\dfrac{(1/3)(2/3)}{k}$.

If you want an exact expression for the variance, go to the Wikipedia link given above. In the formula in the given link, use $k$ instead of their $n$, and $n$ instead of their $N$, and $n/3$ instead of their $m$.

Remark: The proposed expressions for mean and variance in the OP are not correct. Let $X_i=1$ if the $i$-th person polled voted for Bob, and let $X_i=0$ otherwise. You are interested in the expectation of $\frac{X_1+X_2+\cdots+X_k}{k}$. The expectation of a sum is the sum of the expectations. From this we get quickly that our expectation is $\dfrac{1}{3}$. Moreover, this is common-sense correct: about one-third of the people exit-polled will report voting for Bob. The answer cannot be $(1/3)^k$.

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  • $\begingroup$ I apologize for not stating this above, but we can assume for simplicity that the sampled people are chosen with replacement. $\endgroup$ – tengo12 Nov 12 '12 at 17:06
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    $\begingroup$ I'm confused because the variance for Y in your text is $\frac{(1/3)(2/3)}{(k^2)}$ when shouldn't it be $\frac{(1/3)(2/3)}{(k)}$ $\endgroup$ – tengo12 Nov 12 '12 at 18:12
  • $\begingroup$ @tengo12: Oops, typo, now fixed. Thanks. I had meant to write $=\frac{k(1/3)(2/3)}{k^2}=\frac{(1/3)(2/3)}{k}$ and shortened wrongly. $\endgroup$ – André Nicolas Nov 12 '12 at 18:21

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