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Let $X(T)$ be a Poisson process. What is $ \mathbb{P}(X(t) - X(s) = 1 \mid X(t) = 4) \,? $

I split this up into $\mathbb{P}(X(s) = 3, X(t) - X(s) = 1)$ and found the answer. However, it did not work out to be the solution which is $ \frac{4(t-s) s^3}{t^4} \,. $ I am unsure why I am not able to match the solution. Could someone please help me work this out? Thanks so much!

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    @icobes: Sure. Cheers.2011-03-24

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One approach is to follow cardinal's hint, ${\rm P}(A|B) = \frac{{{\rm P}(A \cap B)}}{{{\rm P}(B)}}$. The solution then follows straightforwardly noting that $X(t)-X(s)$ and $X(s)$ are independent.

The quickest approach, however, is to rewrite the solution as $ \frac{{4(t - s)s^3 }}{{t^4 }} = {4 \choose 1}\bigg(\frac{{t - s}}{t}\bigg)^1 \bigg(1 - \frac{{t - s}}{t}\bigg)^3 . $ Does the right-hand side look familiar?

EDIT: More generally, $ {\rm P}(X(t) - X(s) = k|X(t) = n) = {n \choose k}\bigg(\frac{{t - s}}{t}\bigg)^k \bigg(1 - \frac{{t - s}}{t}\bigg)^{n-k} = {n \choose k} \frac{{(t - s)^k s^{n - k} }}{{t^n }}. $ (Put $p=(t-s)/t$, and consider the binomial$(n,p)$ distribution.)

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    This answers the question "What is $\mathbb{P}(X(t) - X(s) = 1 \mid X(t) = 4)$?"2011-03-24