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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?

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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 $H(t)=\{N(s):0\leq s <t\}$. 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 $$\lambda(t|H(t))= \frac{1+\phi N(t^{-})}{1+\phi \mu(t)} \rho(t).$$

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