The formula for the expected value in a binomial distribution is:

$$E(X) = nP(s)$$ where $n$ is the number of trials and $P(s)$ is the probability of success.

The formula for the expected value in a hypergeometric distribution is:

$$E(X) = \frac{ns}{N}$$ where $N$ is the population size, $s$ is the number of successes available in the population and $n$ is the number of trials.

$$E(x) = \left( \frac{s}{N} \right)n $$ $$P(s) = \frac{s}{N}$$ $$\implies E(x) = nP(s)$$

Why do both the distributions have the same expected value? Why doesn't the independence of the events have any effect on expected value?


For either one, let $X_i=1$ if there is a success on the $i$-th trial, and $X_i=0$ otherwise. Then $$X=X_1+X_2+\cdots+X_n,$$ and therefore by the linearity of expectation $$E(X)=E(X_1)+E(X_2)+\cdots +E(X_n)=nE(X_1). \tag{1}$$ Note that linearity of expectation does not require independence.

In the hypergeometric case, $\Pr(X_i=1)=\frac{s}{N}$, where $s$ is the number of "good" objects among the $N$. This is because any object is just as likely to be the $i$-th one chosen as any other. So $E(X_i)=1\cdot\frac{s}{N}+0\cdot \frac{N-s}{N}=\frac{s}{N}$. It follows that $E(X)=n\frac{s}{N}$.

Essentially the same proof works for the binomial distribution: both expectations follow from Formula $(1)$.

  • $\begingroup$ Sorry, but I am completely lost. Perhaps my question is vague, and an example could help. Consider two scenarios: In scenario A, you are given 13 cards from a deck. In scenario B, you pick one card, check its value and return it to the deck. You repeat this 13 times. Is the expected number of aces same for both cases? $\endgroup$
    – asdf
    Jan 5 '13 at 4:12
  • $\begingroup$ Yes, the expectation is the same. In each case it happens to be $1$. $\endgroup$ Jan 5 '13 at 4:14
  • $\begingroup$ Ah, that makes sense. Thank you for the help. $\endgroup$
    – asdf
    Jan 5 '13 at 4:16
  • $\begingroup$ @asdf: The method of indicator random variables that was used above is very handy. It often makes it possible to calculate $E(X)$ even when we do not have a neat formula for the probability distribution function of $X$. In the case of the hypergeometric, $p_k=\Pr(X=k)=\binom{s}{k}\binom{N-s}{n-k}/\binom{n}{k}$, and $E(X)=\sum_{k=0}^n kp_k$. By a slightly tricky, but not really hard, manipulation, we can simplify this to $\frac{ns}{N}$. But the method of indicator random variables is quite a bit easier to use in this case. $\endgroup$ Jan 5 '13 at 4:24

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