# Two independent Poisson process 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?

• You really should run a simulation, if only to check any analytical answers. I might be misunderstanding your question, but my simulations produce values around $0.40$.
– whuber
Commented Sep 8, 2016 at 18:00
• Hmm probably true, currently I get values around 0.30 with analytical, this is the probability that process X exceeds process Y at any given point. Commented Sep 8, 2016 at 18:45
• @Whuber would you mind sharing your simulation code? Commented Sep 9, 2016 at 13:01

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 $X(t)$.

• Thanks a bunch for the help! I have a code snippet from R that I created before I saw your solution, we get almost the same result but not quite: for (j in 1:100){ for (i in 0:100){ val=val*(1-(1-ppois(i+3,2*j))*ppois(0+i,3*j)) print(1-val) } } As far as I can see we are doing pretty much the same thing right? But the results differ a bit. Commented Sep 8, 2016 at 10:07
• @nonein see my edit Commented Sep 8, 2016 at 15:29
• Awesome! This is somewhat what I ended up doing, I do have one question though when using ppois wouldn't it be 1-ppois? As we are interested in calculating the probability of exceeding? Lastly, how come we are adding the probabilities and not multiplying them? Commented Sep 8, 2016 at 15:54
• @nonein last question: because $P(A \cup B) = P(A) + P(B)$ if $A \cap B = \emptyset$. First question, no, the code inside the for loop estimates the value of the sum of the last line, which has cdfs of X and pmfs of Y-3 Commented Sep 8, 2016 at 16:00
• Sorry for not understanding completely. Normally when wanting to the probability of event A happening conditional on event B isn't that then P(A)*P(B)? Just like we do with the pmf and CDF. Regarind the R-code I am unsure what you mean, in the code above we multiply the probability of scoring jprime +3 or below multiplied by the probability of scoring jprime. Aren't we interested in the probability of scoring above jprime multiplied by the probability of scoring exactly jprime in poisson process Y? Commented Sep 8, 2016 at 16:20