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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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Why don't you edit your post and show your own work? Many times when you write things out and think about each step carefully, you'll quickly see where you've gone wrong. –  cardinal Mar 24 '11 at 2:02
    
@cardinal I don't know how to use the math notation, so it is going to be one huge mess! But I just applied the Poisson distribution to each of those independent incremements and I got a bunch of e^ terms and lambdas that didn't go away and thus didn't match the solution... –  icobes Mar 24 '11 at 2:04
    
Hint: My guess is you are confusing joint probability and conditional probability. –  cardinal Mar 24 '11 at 2:07
    
@cardinal So is this P(X(s) = 3, X(t) - X(s) = 1) an incorrect restatement of the original problem? If so, then how do I proceed? I would use the binomial theorem but I am only familiar with that when we have P (X(t) = n | X(T) = m) but here we have an an extra s term... –  icobes Mar 24 '11 at 2:10
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@icobes: Bigger hint: For events $A$ and $B$ such that $\mathbb{P}(B) > 0$, $\mathbb{P}(A \mid B) = \frac{\mathbb{P}(A \cap B)}{\mathbb{P}(B)}$. –  cardinal Mar 24 '11 at 2:15
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1 Answer

up vote 3 down vote accepted

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 $\mathbb{P}(X(t) - X(s) = 1 \mid X(t) = 4)$?" –  Shai Covo Mar 24 '11 at 19:13
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