# Negative binomial and joint probability distribution

An urn contains $w$ white and $b$ black balls. I made some extractions with replacement and $X_a$ is the random variable representing the number of extraction made to get the $a^\mathrm{th}$ white ball. I need to find joint probability distribution for $\Pr(X_a=s,X_b=t)$, where $b>a$ and $t>s$.

I know this is a negative binomial distribution with probability function:

$$\Pr(X_a=i)=\binom{i-1}{a-1}{p_w}^a(1-p_w)^{i-a}$$

How do I combine this to get $\Pr(X_a=s,X_b=t)$?

• What does "$th$" mean, in the definition of $X_a$? Did you mean "$a^\mathrm{th}$" ?
– user228113
Feb 24 '16 at 16:13
• I meant the first or second or third or $a-th$.
– Paul
Feb 24 '16 at 16:15
• Are you using $b$ to mean two different things? You probably want $\Pr(X_a=s,X_b=t) = \Pr(X_a=s)\Pr(X_{b-a}=t-s)$ Feb 24 '16 at 16:36
• I didn't know that also the negative binomial was "Memorylessness", thanks
– Paul
Feb 24 '16 at 16:45

Notice that the problem has no memory, so $\mathrm{Pr}(X_{a+r}=s+q\mid X_a=s)=\mathrm{Pr}(X_r=q)$.

Then, since $b>a$, $$\mathrm{Pr}(X_a=s, X_b=t)=\mathrm{Pr}(X_a=s)\mathrm{Pr}(X_b=t\mid X_a=s)=\\=\mathrm{Pr}(X_a=s)\mathrm{Pr}(X_{b-a}=t-s)$$

Which you can easily calculate.

• I didn't know that also the negative binomial was "Memorylessness", thanks.
– Paul
Feb 24 '16 at 16:44
• +1) @Paul Make it a custom to accept answers if they are acceptable. You received about $10$ answers on questions, but accepted none of them. Why not? Jun 12 '16 at 18:01
• @drhab sorry, didn't know that :-)
– Paul
Jun 13 '16 at 13:02
• @Paul Ego te absolvo. You also accepted a good custom :). Jun 13 '16 at 13:33