Assume that you have two independent Poisson processes, N1( t ) with rate λ1 and N2( t ) with rate λ2 . What is the probability that n events occur in the first process before m events occur in the second process?
I know this question has been asked in the forum before. In fact the answer available in text book is
$$ \sum_{k=n}^{n+m-1} \binom{n+m-1}{k} (\frac{\lambda _{1}}{\lambda _{1}+\lambda _{2}})(\frac{\lambda _{2}}{\lambda _{1}+\lambda _{2}})^{n+m-1-k} $$
Now my doubt is the following:
Suppose I want to find out the probability that 1st event of type 1 occurred before 2nd event of type 2.
There can be two possibilities:
One event of type A occurs ------ This means no matter what event occurs next (either of type A or of type B), the condition will still be satisfied.
One event of type B occurs, followed by a event of type A. ---- No matter what event occurs next, the condition will still be satisfied.
Now, probability of 1st possibility occurring is $\frac{\lambda _{1}}{\lambda _{1}+\lambda _{2}}$ and probability of 2nd possibility occuring is $(\frac{\lambda _{2}}{\lambda _{1}+\lambda _{2}}) (\frac{\lambda _{1}}{\lambda _{1}+\lambda _{2}})$
So, the required probability I get is
$\frac{\lambda _{1}}{\lambda _{1}+\lambda _{2}} + (\frac{\lambda _{2}}{\lambda _{1}+\lambda _{2}}) (\frac{\lambda _{1}}{\lambda _{1}+\lambda _{2}})$
Whereas, using the equation given in the book, I get the answer
$(\frac{\lambda _{1}}{\lambda _{1}+\lambda _{2}})^{2} + 2(\frac{\lambda _{2}}{\lambda _{1}+\lambda _{2}}) (\frac{\lambda _{1}}{\lambda _{1}+\lambda _{2}})$
Can anyone please tell me where am I doing it wrong?
You are doing nothing wrong.
For every $x$, $x+x(1-x)=x^2+2x(1-x)$. For $x=\lambda_1/(\lambda_1+\lambda_2)$, this shows that the formula $x+x(1-x)$ in the book equals your formula $x^2+2x(1-x)$.