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?

  • $\begingroup$ 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 $0.40$. $\endgroup$
    – whuber
    Commented Sep 8, 2016 at 18:00
  • $\begingroup$ 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. $\endgroup$
    – no nein
    Commented Sep 8, 2016 at 18:45
  • $\begingroup$ @Whuber would you mind sharing your simulation code? $\endgroup$
    – no nein
    Commented Sep 9, 2016 at 13:01

1 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)

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 $X(t)$.

  • $\begingroup$ 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. $\endgroup$
    – no nein
    Commented Sep 8, 2016 at 10:07
  • $\begingroup$ @nonein see my edit $\endgroup$
    – Taylor
    Commented Sep 8, 2016 at 15:29
  • $\begingroup$ 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? $\endgroup$
    – no nein
    Commented Sep 8, 2016 at 15:54
  • $\begingroup$ @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 $\endgroup$
    – Taylor
    Commented Sep 8, 2016 at 16:00
  • $\begingroup$ 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? $\endgroup$
    – no nein
    Commented Sep 8, 2016 at 16:20

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