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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 $\Pr(X_a=s,X_b=t)$?

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

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  • $\begingroup$ I didn't know that also the negative binomial was "Memorylessness", thanks. $\endgroup$
    – Paul
    Feb 24 '16 at 16:44
  • $\begingroup$ +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? $\endgroup$
    – drhab
    Jun 12 '16 at 18:01
  • $\begingroup$ @drhab sorry, didn't know that :-) $\endgroup$
    – Paul
    Jun 13 '16 at 13:02
  • $\begingroup$ @Paul Ego te absolvo. You also accepted a good custom :). $\endgroup$
    – drhab
    Jun 13 '16 at 13:33

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