An urn contains white and black balls with $p_w=p$ and $p_b=1−p$. If I made some extractions with replacement, what are the support and the probability function of $X_a$, where $X_a$ is the random variable representing the number of extraction made for get the a-th white ball? How can I get the conditional probability $P(X_a=s|X_b=t)$ (i.e. $X_9=21|X_5=11)$?

I have no idea of how to proceed...

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    $\begingroup$ Assuming $a \gt b$ and $s \gt t$, and that $p$ is known, I would have guessed $P(X_a=s|X_b=t)=P(X_{a-b} = s-t)$, e.g. $X_4=10$ in your example, with a negative binomial distribution $\endgroup$ – Henry Feb 9 '16 at 17:33
  • $\begingroup$ Probability should be between 0 and 1, so I think this is not correct. $\endgroup$ – Paul Feb 9 '16 at 17:45
  • $\begingroup$ @Paul, Henry means $\Pr(X_9=21\mid X_5=11)=\Pr(X_4=10)$ where $X_k\sim\mathcal{NB}(k, p_w)$ $\endgroup$ – Graham Kemp Feb 10 '16 at 1:00

A start: The number of draws until we get the first white is at least $a$, and could be very large. Let us find $\Pr(X_a=k)$ for $k\ge a$.

We have $X_a=k$ if (i) we have exactly $a-1$ white on the first $a-1$ draws and (ii) we get a white on the $a$-th draw.

The probability of exactly $a$ white in the first $k-1$ draws is (binomial distribution) $\binom{k-1}{a-1}p^{a-1}(1-p)^{(k-1)-(a-1)}$. For $\Pr(X_a=k)$, multiply by $p$. We get $$\Pr(X_a=k)=\binom{k-1}{a-1}p^a(1-p)^{k-a}.$$ For more detail about the distribution of $X_a$, please search under negative binomial distribution.

Now perhaps you can tackle the condition prrobability problems on your own. You will not necessarily need this, but recall that $$\Pr(X_a=s\mid X_b=t)=\frac{\Pr(X_a=s\cap X_b=t)}{\Pr(X_b=t)}.$$ It will be worthwhile to tackle first a specific numerical example, like the one in the post.

  • $\begingroup$ Hi, I've looked for a while about conditional negative binomial without any useful result and now I'm stuck on $P(X_a=s\cap X_b=t)$ $\endgroup$ – Paul Feb 14 '16 at 18:07
  • $\begingroup$ I had mentioned you would not necessarily need the probability in your comment above. There are various cases, but the conditional probability is essentially taken care of by the comment of Henry. $\endgroup$ – André Nicolas Feb 14 '16 at 18:20
  • $\begingroup$ I need to think about it, but I was wondering if it is possible to solve it in another way. $\endgroup$ – Paul Feb 14 '16 at 18:27
  • $\begingroup$ Maybe. But the procedure mentioned by Henry is the natural way. $\endgroup$ – André Nicolas Feb 14 '16 at 18:33

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