# Urn balls without replacement and random vector

An urn contains $w$ white and $b$ black balls. $n$ extractions without replacement are made. $X_i$ and $X_j$ are the random variables representing the number of white balls extracted on $i^{th}$ and $j^{th}$ extraction ($j > i$). I need to find:

1. support and probability function for marginal $X_i$ and $X_j$
2. support and joint probability function for random vector $(X_i,X_j)$
3. support and probability function for $Z=X_i+Y_j$

Question 1

I think this may be the solution:

$p(X_i=s)=\frac{\dbinom{w}{s}\dbinom{b}{i-s}}{\dbinom{w+b}{i}}$, $p(X_j=t)=\frac{\dbinom{w}{t}\dbinom{b}{j-t}}{\dbinom{w+b}{j}}$

Question 2

Not so sure:

$p(X_i=s,X_j=t)=\frac{\dbinom{w}{s}\dbinom{b}{i-s}}{\dbinom{w+b}{i}}\frac{\dbinom{w}{t}\dbinom{b}{j-t}}{\dbinom{w+b}{j}}$

Question 3

Maybe:

I set $v=s+t$ and then $P(Z)=P(X_i+X_j)=\frac{\dbinom{w}{v}\dbinom{b}{i+j-v}}{\dbinom{w+b}{i+j}}$

• What are $X$ and $Y$ in Question 3? Feb 21 '16 at 22:27
• Ops, I meant $X_i$ and $X_j$.
– Paul
Feb 21 '16 at 22:28
• From the wording of the question, I would have supposed that each extraction picked exactly one ball, so that on the $i$th extraction we extract one ball that is either black or white, so the number of white balls extracted on the $i$th extraction is either $0$ or $1$. If $X_i$ were the number of balls extracted on the $i$th extraction _and all previous extractions_ then it might be greater than $1$. But then perhaps I misread the intent of the question. Feb 24 '16 at 18:15
• Hi, I'm sorry I wasn't clear. $X_i$ is the number of balls extracted on the $i$th extraction and all previous extractions.
– Paul
Feb 24 '16 at 18:44

These definitions of $X_i$ and $X_j$ are rather confusing. I find indicator variables much easier to understand, so let me define some:

Let $W_i$ be a random variable such that $$W_i = \begin{cases} 0 & \text{if the ith ball drawn is black,}\\ 1 & \text{if the ith ball drawn is white.} \end{cases}$$

Now if you want $X_i$ to be the total number of white balls extracted during the entire time we are extracting the first $i$ balls, then we can define $$X_i = \sum_{k=1}^i W_k.$$

Then for $j > i$, $$X_j = \sum_{k=1}^j W_k = X_i + \sum_{k=i+1}^j W_k,$$ which causes me to strongly suspect that $X_j$ is not independent of $X_i$.

The proposed results for "Question 1" then look good, but the others do not. I think that \begin{align} P(X_i = s, X_j = t) &= P(X_i = s) P\left( \sum_{k=i+1}^j W_k = t - s \mid X_i = s\right) \\ &= \frac{\dbinom{w}{s}\dbinom{b}{i-s}}{\dbinom{w+b}{i}} \left( \frac{\dbinom{w-s}{t-s}\dbinom{b-i+s}{j-i-t+s}}{\dbinom{w+b-i}{j-i}} \right), \end{align} which may simplify further, but I don't immediately see how.

Also, $$X_i + X_j = 2\sum_{k=1}^i W_k + \sum_{k=i+1}^j W_k,$$ whose probability distribution I fear will be ugly, since it is not enough to know how many white balls were drawn during the $j$ extractions in question; the balls drawn during the first $i$ extractions count twice as much as the balls drawn during the remaining $j - i$ extractions. So for $P(X_i + X_j = v)$, it looks like we have to sum all the probabilities $P(X_i = s, X_j = v - s)$ for $0 \leq s \leq \min\{v,i\}$.

• Hi, thanks for helping, indicator variables are very useful but I'm not very comfortable with them.
– Paul
Feb 24 '16 at 21:19
• I find this whole problem a mass of discomfort after part 1. The interdependence of $X_i$ and $X_j$ is awkward. For example, if $w \geq 2$ and $i<j\leq n$ then $P(X_i=1, X_j=2) > 0$ but $P(X_i=2, X_j=1) = 0$. Feb 24 '16 at 21:25