# Why E is the same for hypergeometric and binomial laws

I'm confused a bit, not sure if it's tiredness or something but... Why do the mathematical expectation of the hypergeometric law is the same as the binomial one?

After all, the hypergeometric one correspond to sampling without replacement, while for binomial it's with replacement.

In my book it's justified quickly this way:

$$E[V] = E[X_{1} + X_{2} + ... + X{n}] = E[X_{1}]+E[X_{2}]+...+E[X_{n}] = \frac{N}{N+M}+\frac{N}{N+M}+...+\frac{N}{N+M} = n \frac{N}{N+M}$$

Say we have N black balls and M white balls, then that would mean the probability to pick a black ball is the same at the ith sampling as it is at the first?! But how can it still be N+M when in fact at the 2nd sampling there is N+M-1 balls in the box?

Shouldn't the probability differ depending on the sampling iteration? I thought it was a conditional case...

I thought the last equality would more resemble:

$$E = \frac{N}{N+M} + \frac{N-1}{N+M-1} + ... + \frac{N-k}{N+M-k}$$

Maybe someone could shed some light on this for me? Why is the expectation $np$ for $n$ samplings without replacement?

Thanks!

• The last formula in your question, the one you thought $E$ would resemble, seems to assume that each of the drawn (and not replaced) balls has the same color. – Andreas Blass Oct 21 '15 at 11:52
• @AndreasBlass don't quote me on that it was a kind of intuition more than anything, I'm trying to demonstrate I am confused because of the conditionnal nature of the probability, and I don't get how it can be treated as a conditional one. – Yannick Oct 21 '15 at 11:55

## 1 Answer

This can be regarded as a consequence of linearity of expectation, even for non-independent random variables. The marginal (i.e. unconditional) probability for each ball remains the same even without replacement.

For example, the first ball has a probability $\dfrac{N}{N+M}$ of being black and $\dfrac{M}{N+M}$ of being white. So the second ball has a probability $\dfrac{N-1}{N+M-1} \times \dfrac{N}{N+M}+\dfrac{N}{N+M-1} \times \dfrac{M}{N+M} = \dfrac{N}{N+M}$ of being black, since the first ball can be black or white, and something similar is true for all the later balls.

• Hi @Henry could you explain where you come from for the probability of the 2nd ball? Sorry it seems I am MASSIVELY confused for some obscure reason. – Yannick Oct 21 '15 at 12:05
• @Yannick: If the first ball is black, the conditional probability the second ball is black is $\frac{N-1}{N+M-1}$. If the first ball is white, the conditional probability the second ball is black is $\frac{N}{N+M-1}$. Multiply, add and simplify to get the answer (as shown). – Henry Oct 21 '15 at 12:11