I have two random variables $x$ and $y$ that follow Poisson distribution with rate $\lambda_0$ and $\lambda_1$. Another random variable $z$ is defined as $(x+y)$ so it also follows a Poisson distribution with rate $\lambda_0+\lambda_1=\lambda_{01}$.

Now I want to write $P(z\geq n)=P(n,\lambda_{01}T)$, which means probability of n or more events happening in time T, in terms of $P(x)$ and $P(y)$. I am not sure if following way is correct or not. $$\sum_{z=n}^\infty p(z,\lambda_{01}T)=\sum_{y=0}^\infty \sum_{x=(n-y)^+}^\infty p(x,\lambda_0 T)p(y,\lambda_1 T)$$ or $$\sum_{z=n}^\infty p(z,\lambda_{01}T)=\sum_{y=0}^\infty \sum_{x=(n-y)^+}^\infty p(y,\lambda_1 T)p(x,\lambda_0 T)+\sum_{x=0}^\infty \sum_{y=(n-x)^+}^\infty p(x,\lambda_0 T)p(y,\lambda_1 T)$$ Thanks!!! If it is correct, do you think if there is any other way of writing it?

  • $\begingroup$ Since you know it is Poisson parameter $\lambda=\lambda_0+\lambda_1$, you can write it as $\sum_{k=n}^\infty e^{-\lambda T} \frac{\(\lambda T)^k}{k!}$. $\endgroup$ – André Nicolas Feb 8 '13 at 7:26
  • $\begingroup$ Thanks Andre, but I need it in terms of x and y. I only know x+y is at least n. $\endgroup$ – Eln Feb 8 '13 at 7:30
  • $\begingroup$ Then I would write it as a sum from $k=n$ to infinity of the usual convolution sum. So the inner sum is $\sum_{i=0}^k \Pr(X_0=i)\Pr(X_1=k-i)$, and then use the usual Poisson probabilities. $\endgroup$ – André Nicolas Feb 8 '13 at 7:35

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.