# Intensity function in Poisson random effect model

I have a somewhat general question about intensity functions in Poisson random effect models.

Consider the Poisson random effects model in which conditional on a random effect $$u$$, an individual experiences events according to a Poisson process with intensity function $$u\rho(t)$$. Furthermore, suppose $$u$$ has a gamma density $$g(u)$$, with mean 1 and variance $$\phi$$. Denote $$N(t)$$ the number of events, $$H(t)$$ the history, and $$\rho(t)= \mu^{\prime}(t)$$.

My goal is to show $$\lambda(t|H(t)) = \left(\frac{1+\phi N(t^{-})}{1+\phi \mu(t)}\right) \rho(t).$$

Here, we have the intensity $$\lambda(t|H(t))$$ is given by

$$\lambda(t|H(t)) = \lim\limits_{\Delta t \to 0}\frac{P(\Delta N(t)=1|H(t))}{\Delta t}.$$

My first thought was the following:

$$\lambda(t|H(t)) = \int \lambda(t|H(t),u) g(u)du = \int u\rho(t) g(u)du = \rho(t).$$ This is obviously wrong, but I am not sure why. I figured that $$\lambda(t|H(t),u)=u\rho(t)$$ because conditional on the random effect $$u$$, we have a Poisson process with intensity $$u\rho(t)$$, but I could be mistaken.

Edit: Upon further thinking about it, I assume the following statement is incorrect: $$\lambda(t|H(t)) = \int \lambda(t|H(t),u) g(u)du.$$

The following should hold instead: $$\lambda(t|H(t)) = \int \lambda(t|H(t),u) g(u|H(t))du.$$

Any thoughts?

What follows is my solution.

For small $$\Delta t$$ with $$\Delta N(t)=N(t, t+\Delta t) = N(t+\Delta t)-N(t)$$ holds:

$$P(N(t,t+\Delta t)=1|H(t)) \\=\frac{P(\Delta N(t)=1, H(t))}{P(H(t))} \\=\frac{\int P(\Delta N(t)=1|H(t),u)P(H(t)|u)g(u)du}{\int P(H(t)|u)g(u) du} \\=\frac{\int u\cdot \rho(t) \cdot \Delta t \cdot P(H(t)|u)g(u) du}{\int P(H(t)|u)g(u) du} .$$

The history $$H(t)$$ is defined by $$H(t)=\{N(s):0\leq s . Conditioned on $$u$$, the history $$H(t)$$ follows a Poisson distribution, i.e. $$P(H(t)|u)=P(N(t^{-})|u) =\exp(-\mu(t) u) \frac{(\mu(t) u)^{N(t^{-})}}{N(t^{-})!}.$$

Since $$g(\cdot)$$ is the density of Gamma distribution, $$\int P(H(t)|u)g(u) du$$ is the probability that a negative binomial distribution has the value $$N(t^{-})$$. Moreover, is is straight forward to show $$\int u P(H(t)|u)g(u) du = \frac{N(t^{-})+1}{\mu(t)}P_{NB}(N(t^{-})+1),$$ where $$P_{NB}(\cdot)$$ is the probability mass function of a negative binomial distribution. Finally, it follows that

$$P(N(t,t+\Delta t)=1|H(t))\\ = \frac{N(t^{-})+1}{\mu(t)} \frac{P_{NB}(N(t^{-})+1)}{P_{NB}(N(t^{-}))}\rho(t)\Delta t\\ = \frac{1+\phi N(t^{-})}{1+\phi \mu(t)} \rho(t)\Delta t$$

and therefore the we obtain the intensity $$\lambda(t|H(t))= \frac{1+\phi N(t^{-})}{1+\phi \mu(t)} \rho(t).$$