Two independent Poisson process question

Question

If I have two independent Poisson processes, X and Y, with X having a lambda 2 and Y having a lambda 3. Given that the starting value for Y is 3 and X is zero, then how do I calculate the probability of X being larger than Y at any given point, even if only for one instant in time. I realize that I can just simulate my way out of it, but that does not really seem like an elegant solution. Any help/directions would be much appreciated.

Besides the solution below, is it possible to use the CDF and pmf of a Poisson distribution instead of only its PMF, see below.

$1-\prod_{t=1}^{\infty}\bigg[\prod_{k=0}^{\infty} \bigg [1-\bigg(1-e^{-\lambda_1*t}\cdot \sum_{i=0}^{k+D+1}\bigg[\frac{(\lambda_1*t)^{i}}{i!}\bigg]\bigg)\cdot \frac{(\lambda_2*t)^{k}*e^{-\lambda_2*t}}{k!}\bigg]\bigg] $

My idea was that we find the probability that Poisson distribution X with Lambda 2 receives k+3+1 more occurrences conditioned on the fact that Poisson distribution Y gets equal to k occurrences. This is done for all k and then subsequently we do it for all points in time. Is this also a correct way of doing it?

Answer 1

Pick some time point $t > 0$. We know that $X(t) \sim \text{Poisson}(2t)$ and $Y(t) - 3 \sim \text{Poisson}(3t)$. You are asking for $P(X(t) > Y(t))$. Independence comes in handy.

\begin{align} P(X(t) > Y(t)) &= \sum_{j=1}^{\infty}\sum_{k=0}^{\infty} P[X(t) = k+j \cap Y(t) = k ] \\ &= \sum_{j=1}^{\infty}\sum_{k=0}^{\infty} P[X(t) = k+j] P[ Y(t) = k ] \\ &= \sum_{j=1}^{\infty}\sum_{k=3}^{\infty} P[X(t) = k+j] P[ Y(t) = k ] \\ &= \sum_{j=1}^{\infty}\sum_{k'=0}^{\infty} P[X(t) = k'+3+j] P[ Y(t) - 3 = k' ] \\ &= \sum_{j=1}^{\infty}\sum_{k'=0}^{\infty} \frac{e^{-2t}(2t)^{k'+3+j}}{(k'+3+j)!} \frac{e^{-3t}(3t)^{k'}}{k'!} \\ \end{align}

Edit: this might actually be simpler...

\begin{align*} 1-P(X(t)\le Y(t)) &= 1 - \sum_{i=0}^j\sum_{j=3}^{\infty}P(X(t)=j-i)P(Y(t)=j) \\ &= 1-\sum_{j=3}^{\infty} P(X(t)\le j)P(Y(t)=j) \\ &= 1-\sum_{j'=0}^{\infty} P(X(t)\le j'+3)P(Y(t)-3=j') \end{align*}

Which you can approximate in R by typing

calcProb <- function(t)
{
  sumVal <- 0
  for(jprime in 0:1000){
    sumVal <- sumVal + ppois(jprime+3, 2*t)*dpois(jprime,3*t)
  }
  return(1-sumVal)
}

#plot
t <- seq(0,50,.1)
probs <- sapply(t, calcProb)
plot(t,probs, type="l")

And you can see how this probability changes as $t$ changes. This makes sense to me because $Y(t)$ starts higher and grows faster than $X(t)$.