If L(t) and M(t) are independent Poisson Processes, for which a and b is N(t) = aL(t) + bM(t) a Poisson Process. A necessary condition for it to be Poisson is if N(t) has Poisson distribution and a necessary condition for it to be distributed as such is that its mean = variance, so aλ + bμ = a^2λ + b^2μ and we need a and b positive integers for N(t) to be a counting process so we must have a = b = 1. Is this correct ?

  • $\begingroup$ How does this imply $a=b=1$? $\endgroup$
    – raj
    Commented Jan 28 at 16:14


You must log in to answer this question.

Browse other questions tagged .