# conditional distribution X_1|X_1+X_2=r

suppose $f(x_1,x_2)=p^2q^{x_2},\ x_1=0,1,\ldots,x_2,\ x_2=0,1,2\ldots$ how can find $\mathbb{Pr}(X_2-X_1\leq1)$? also if $(X_1,X_2,X_3)\sim M(n,P_1,P_2,P_3)$ find conditional distribution $X_1|X_1+X_2=r$?

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0 out of 9.  –  Did Mar 18 '12 at 14:44

Hint (for first question): To get a grasp of what the joint distribution is, you might take an explicit value of $p$, like $p=1/3$, and write down various probabilities, such as $f(2,5)$, the probability that $X_1=2$ and $X_2=5$, and $f(5,2)$, and a few others until what the formula really says becomes clear. For what follows, I will assume this preliminary work has been done.
How can we have $X_2-X_1 \le 1$? Note that it is built in that $X_1 \le X_2$. So $X_2-X_1\le 1$ occurs if either (i) $X_1=X_2$ or (ii) $X_1=X_2-1$.
For (i) the possibilities are $X_2=0, X_1=0$; $X_2=1,X_1=1$; $X_2=2,X_1=2$, and so on forever. The probabilities are $p^2q^0$, $p^2q^1$, $p^2q^2$, and so on forever. Add up. We get an infinite geometric series. The sum simplifies nicely, since $q=1-p$. (It is best to define what you mean by various symbols, so that the question you ask is as clear as possible.)
For (ii), the possibilities are $X_2=1, X_1=0$; $X_2=2,X_1=1$; $X_3=3,X_2=2$, and so on forever. You should be able to write down the relevant probabilities. They will look quite a bit like the probabilities in (i). Add up.
Finally, add together your sums (i) and (ii). For completeness, one should check whether the summations make sense. This means separating out the possibilities $p=0$ and $p=1$ for special attention.