Change in intensity for poisson variable

Question

Assume that we have two independent Poisson random variables, $X_1 \sim Poi(\lambda)$ and $X_2 \sim Poi(\lambda\psi)$. My question regards inference about $\psi$, which could be seen as the relative change in Poisson intensity from one time period to another. I especially want the MLE of $\psi$. What I have been trying to do is to get the joint likelihood and marginalize $\lambda$, but I haven't been successful. Instead I always end up with an expression where I would have to know $\lambda$ to be able to get the MLE of $\psi$. Hope someone here has a good idea of how to approach this.

Answer 1

Let $L(\lambda, \psi)$ be the likelihood function or joint probability density function. (I think you already got it.)

Get $\frac{\partial \log(L(\lambda, \psi))}{\partial \lambda}$ and $\frac{\partial \log(L(\lambda, \psi))}{\partial \psi}$

Set two partial derivatives to zero. Get the solution of $\lambda$ and $ \psi$ in term of $X_1$ and $X_2$. That is your answer.

If the analytical form of the solution is unavailable, you need to use computer software to get the numerical solutions.

Answer 2

A numerical solution to the multiplicative parameter is possible, indeed. Here is a simple solution simulated in R:

    > set.seed(4242)
    > 
    > lambda <- runif(1, min = 0, max = 10)
    > mu <- runif(1, min = 0, max = 10)
    > 
    > lambda
    [1] 9.861143
    > 
    > mu 
    [1] 3.477161
    > 
    > x <- rpois(n = 1000, lambda = lambda)
    > y <- rpois(n = 1000, lambda = mu*lambda)
    > 
    > ll <- function(lambda,mu) {
    +   -sum(dpois(x = x, lambda = lambda)*dpois(x = y, lambda = lambda*mu))
    + }
    > 
    > m <- stats4::mle(ll, start = list(lambda = 5 , mu = 5), method = "L-BFGS-B",
    +                lower = c(0.001, .001))
    > ab <- stats4::coef(m)
    > ab
      lambda       mu 
    9.183705 3.674439 

Answer 3

This is the solution I've reached, based on the answer from a_statistician:

The joint density is given by $$f_{X_1},_{X_2}(x_1, x_2)=e^{-\lambda} e^{-(1+\psi)}\lambda^{x_1+x_2}\psi^{x_2}/x_1!x_2!$$ and the log-likelihood is given by $$l(\lambda, \psi)=-n\lambda-n(1+\psi)+\ln\lambda\sum(x_{i1}+x_{i2})+\ln\psi\sum x_{i2}-\ln(\prod x_{i1}!x_{i2}!).$$ The partial derivative of $l(\lambda, \psi)$ with respect to $\psi$ is $-n+\frac {\sum x_{i2}}\psi$. Setting this equal to zero and solving for $\psi$ yields the MLE of $\psi$: $$\hat\psi=\frac {\sum x_{i2}}n.$$