Higher conditional intensity implies higher variance of the resulting counting process

Question

Suppose $N=(N_t)_{t\geq0}$ is a simple counting process that is driven by the conditional intensity process $(\lambda_t)_{t\geq0}$. That is, for $(\mathcal F_t)_{t\geq0}$ the natural filtration generated by $N$, and possibly by some random marks attached to the events, $(\lambda_t)$ is the $(\mathcal F_t)$-predictable process such that as $\Delta t\downarrow 0$, \begin{align*}\mathbb P(N(t+\Delta t)-N(t)=0|\mathcal F_t)&=1-\lambda_t+o(\Delta t),\\\mathbb P(N(t+\Delta t)-N(t)=1|\mathcal F_t)&=\lambda_t+o(\Delta t),\\\mathbb P(N(t+\Delta t)-N(t)\geq2|\mathcal F_t)&=o(\Delta t).\end{align*}

When $N$ is a (possibly inhomogeneous) Poisson process, then it is clear that a higher intensity function implies a higher variance of the resulting process. My question is: does this property hold as well for process driven by a conditional intensity, i.e. models where the conditional density is influenced by the history of the process.

More precisely, let $N,\tilde N$ be processes driven by conditional intensities $\lambda,\tilde\lambda$, and such that we can find a coupling such that $\lambda$ is always bounded by $\tilde\lambda$. Does it hold true that $\mathrm{Var}(N[0,1])\leq\mathrm{Var}(\tilde N[0,1])$?

I conjecture that this is the case, by comparison to the Poisson process, and because in some sense `the variance seems to be determined by the expectation of the integral of the conditional intensity'. However, in other contexts stochastic dominance does not (at all!) imply dominance of variances.

Any help or reference is much appreciated. Maybe someone knows a reference to a related result?

Answer 1

I do not think that's the case. Let's consider the case of driving proces constant in time $(\lambda_t)_{t\in[0,1]}=\lambda_0$. To calculate the variance I use conditional variance formula and the fact that $N[0,1]|\lambda_0 \sim Poiss(\int_0^1\lambda_tdt) \sim Poiss(\lambda_0)$.

$$Var(N[0,1]) = \mathbb{E}(Var(N[0,1]|\lambda_0)) + Var(\mathbb{E}(N[0,1])|\lambda_0)) = \mathbb{E}\lambda_0 + Var\lambda_0 $$

Now consider two cases (the coupling)

• $\lambda_0$ always equals $a\in\mathbb{R}_+$ then $Var(N[0,1])=a$
• $\lambda_0$ is either $0$ or $a$ (50/50 chance) then $Var(N[0,1])=a/2 + (a/2)^2$

For $a$ big enough the dominanted case of random $\lambda_0$ yields higher variance. I guess the intuition behind this could be that when the driving process is more random you get more deviation in the counting process