Alternative Hypergeometric Proof

I'm trying to prove the hypergeometric distribution the alternative way using algebra from this link http://www.math.uah.edu/stat/urn/Hypergeometric.html stated as where n is the sample size, r is the subset of total population m.

$P(Y=y)=\frac{\binom{n}{y}r^{y}(m-r)^{n-y}}{m^n}$

I can prove this using bernoulli trials, however this requires independent with replacement.

The above website gives an explanation regarding drawing from ordered sequences, if you're drawing from ordered sequences then selecting one $P(Y=1)=\binom{n}{1}\frac{r}{m}^{1}(1-\frac{r}{m})^{n-1}$ From here I can derive the hypergeometric in the 'binomial' alternative definition.

but this I think is not right because what about without replacement in the model?

Any help will be great.

thank you

1 Answer

No, the expression given is somewhat analogous to the Binomial case, but not exactly the same thing at all.   Those brackets are not just for associating the terms in the exponents.$$\mathbb P(Y{=}y)=\dfrac{\dbinom{n}{y}r^{(y)}(m-r)^{(n-y)}}{m^{(n)}}$$

Here $r^{(y)}, m^{(n)},$ et cetera, are used to indicate the falling factorial: \begin{align}r^{(y)} & = r(r-1)\cdots(r-y+1) \\[1ex] & =\frac{r!}{(r-y)!} \\[1ex] & = {}^r\mathrm P_y\end{align}

That counts the number of ways to select an ordered list of $y$ items from a set of $r$.

Your favoured event is the combination of two ordered lists: of $y$ from $r$ 'success' items, and $n-y$ from $m-r$ 'fail' items.   $r^{(y)}$ counts the way to select the first ordered list, $(m-r)^{(n-y)}$ counts the ways to select the second, and $\binom ny$ counts the ways to blend the two lists together into one ordered list.

The total space is the selection of an ordered list of $n$ from $m$ items; counted by $m^{(n)}$.

\begin{align}\mathbb P(Y{=}y) ~&=~ \dfrac{{}^n\mathrm C_y\cdot{}^r\mathrm P_y\cdot{}^{m-r}\mathrm P_{n-y}~{}}{{}^m\mathrm P_n} \\[1ex] &=~\dfrac{\dbinom{n}{y}r^{(y)}(m-r)^{(n-y)}}{m^{(n)}} \end{align}