# Derivation of the Negative Hypergeometric distribution's expected value using indicator variables

I'm trying to understand how to derive the Negative Hypergeometric's expected value using indicator variables. Note, in the problem below, we are only interested in the expected value before the first success:

An urn contains $w$ white balls and $b$ black balls, which are randomly drawn one by one without replacement. The number of black balls drawn before drawing any white balls has a Negative Hypergeometric distribution. Find this expected value.

This is the solution from the textbook (Introduction to Probability by Blitzstein and Hwang):

Let us prove this using indicator r.v.s. Label the black balls as $1, 2, . . . , b$, and let $I_j$ be the indicator of black ball $j$ being drawn before any white balls have been drawn.

Then $\mathbb{P}(I_j = 1) = \displaystyle\frac{1}{w + 1}$ since, listing out the order in which black ball $j$ and the white balls are drawn (ignoring the other balls), all orders are equally likely by symmetry, and $I_j = 1$ is equivalent to black ball $j$ being first in this list.

So by linearity,

$$\mathbb{E}\left(\sum_{j=1}^b I_j\right)=\sum_{j=1}^b \mathbb{E}(I_j)=\frac{b}{w+1}$$

What I don't get is why the probability of the indicator variable is $\mathbb{P}(I_j = 1) = \displaystyle\frac{1}{w + 1}$.

Shouldn't it be $\displaystyle\frac{b}{w+b}$?

Since there are $b$ black balls and $b+w$ total balls.

But then if I use this, I think the expected value I get is Hypergeometric and not the Negative. Also, in the solution it says to ignore the other balls and I am not quite sure why we can do that.

• The probability that $I_j=1$ is the probability that ONE PARTICULAR ball is drawn before any white balls are drawn. That doesn't depend on the colors of the other non-white balls. $\qquad$ – Michael Hardy May 24 '16 at 17:40
• Call the one particular black ball we're concerned with $B$. Then the orders in question are: $$\begin{array}{cccccccccl} W & W & W & \cdots & W & W & W & B & & w!\text{ different orders} \\ W & W & W & \cdots & W & W & B & W & & w!\text{ different orders} \\ W & W & W & \cdots & W & B & W & W & & w!\text{ different orders} \\ & \vdots & \vdots & & \vdots & \vdots \end{array}$$ The question is: Why are the events identified by the rows of this array equally probable? $\qquad$ – Michael Hardy May 24 '16 at 17:49

Call the one particular black ball we're concerned with $B$. Then the orders in question are: $$\begin{array}{cccccccccl} W & W & W & \cdots & W & W & W & B & & w!\text{ different orders} \\ W & W & W & \cdots & W & W & B & W & & w!\text{ different orders} \\ W & W & W & \cdots & W & B & W & W & & w!\text{ different orders} \\ \vdots & \vdots & \vdots & & \vdots & \vdots & \vdots & \vdots \end{array}$$

The question is: Why are the events identified by the rows of this array equally probable?

And why can we ignore the other black balls?

And are these really two different questions?

In each row, there are $w!$ different orders in which the white balls can appear, and there are $(b-1)!$ different orders in which the black balls can appear, thus $w!(b-1)!$ different outcomes corresponding to each one row of this array. The fact that the number of outcomes corresponding to every row is the same for every row is the reason why all the rows are equally probable events. Since there are $w+1$ rows, each row has probability $1/(w+1)$.

Perhaps one should say $w!(b-1)!$ different orders in each row rather than saying "ignore the other black balls". Saying "ignore the other black balls" really just means there are equallly many orders in which the other black balls can appear in each row of the array. In the same way, the orders in which the white balls appear in each row of the array was in effect "ignored" because the number of such orders is the same in every row.

Let's try a concrete example with two balls of each color, called $W_1,W_2,B_1,B_2$, and the particular black ball we're concerned with is $B_1$. The array above then looks like this: $$\begin{array}{ccc} W & W & B \\ W & B & W \\ B & W & W \end{array}$$ So there is a $1/3$ chance that that $B_1$ appears before any white ball. Now look at the list of all $4!=24$ orders $$\begin{array}{c} \left.\begin{array}{cccc} W_1 & W_2 & B_1 & B_2 \\ W_1 & W_2 & B_2 & B_1 \\ W_1 & B_2 & W_1 & B_1 \\ B_2 & W_1 & W_2 & B_1 \\ W_2 & W_1 & B_1 & B_2 \\ W_2 & W_1 & B_2 & B_1 \\ W_2 & B_2 & W_1 & B_1 \\ B_2 & W_2 & W_1 & B_1 \end{array} \right\} \text{ This is } W\ W\ B \\[6pt] \left.\begin{array}{cccc} W_1 & B_1 & W_2 & B_2 \\ W_1 & B_1 & B_2 & W_2 \\ W_1 & B_2 & B_1 & W_2 \\ B_2 & W_1 & B_1 & W_2 \\ W_2 & B_1 & W_1 & B_2 \\ W_2 & B_1 & B_2 & W_1 \\ W_2 & B_2 & B_1 & W_1 \\ B_2 & W_2 & B_1 & W_1 \end{array} \right\} \text{ This is } W\ B\ W \\[6pt] \left.\begin{array}{cccc} B_1 & W_1 & W_2 & B_2 \\ B_1 & W_1 & B_2 & W_2 \\ B_1 & B_2 & W_1 & W_2 \\ B_2 & B_1 & W_2 & W_1 \\ B_1 & W_2 & W_1 & B_2 \\ B_1 & W_2 & B_2 & W_1 \\ B_1 & B_2 & W_2 & W_1 \\ B_2 & B_1 & W_2 & W_1 \end{array} \right\} \text{ This is } B\ W\ W \end{array}$$

• Where is the $(b-1)!$ from? Isn't there just one $B$ in each row so there are only $1!$ different orders in which the black balls can appear? – Justin Liang May 24 '16 at 19:02
• @JustinLiang : $(b-1)!$ is the number of orders in which the other black balls can be arranged. $\qquad$ – Michael Hardy May 24 '16 at 19:50
• @JustinLiang : I've added a concrete example to my answer. $\qquad$ – Michael Hardy May 24 '16 at 20:14
• Wow thanks Michael, this makes a lot of sense now! – Justin Liang May 24 '16 at 22:01
• @MichaelHardy TY for your great contribution. I still do not understand one part of your answer. On the top of your answer you said that all the row has w! different order. for example we have $W_1 W_2 W_3 B_1$ then there is w! = 3! order that W of that specific row can be arranged with $B_1$ hold its same position which is = 3! = 6. then you said on the bottom that it is w!(b-1)! and from your example with $W_1, W_2, B_1, B_2$ from the row WWB there is 8 orders if I use your formula which is w!(b-1)! I get 2!(2-1)! = 2!1!=2 because we have 2 W and 2 B but actually we have 8 arrangement. Why? – user3270418 Dec 26 '16 at 8:42