Covariance between previous and next occurrence. Consider a Poisson process where the time between occurrences is random from a distribution with density function $f(t)$. Assume that we are at a random point in time $T$, so likely between two occurrences, say $O_1$ and $O_2$. Then my question is

What is the covariance between the length of $[O_1,T]$ and the length of $[T,O_2]$?

Thanks for any help.
 A: Let $\tau$ be distribution of inter-arrival times and let $Z=O_2-O_1=O_2-T+(T-O_1)=Y+X$ where $X,Y|Z\sim Unif(0,Z)$. The probability of $T$ landing in an inter-arrival period of length $t$ is proportional to $tf_{\tau}(t)$, hence $$f_Z(t)=\frac {tf_{\tau}(t)}{\mu_\tau}$$
Then conditioning on $Z$ yields $$E(XY)-E(X)E(Y)=E(X(Z-X))-\frac 1 4 E(Z)^2=\frac 1 6E(Z^2)-\frac 1 4E(Z)^2=\frac 1 {\mu_\tau}(\frac {E(\tau^3)
}6-\frac {E(\tau^2)^2}{4\mu_\tau})$$
at the same time $$var(X)=var(Y)=E(\frac 1 {12}Z^2)+var(\frac 1 2 Z)=\frac {E(\tau^3)}{12\mu_\tau}+\frac {E(\tau^3)} {4\mu_\tau}-\frac {E(\tau^2)^2}{4\mu^2_{\tau}}=\frac 1 {\mu_\tau}(\frac {E(\tau^3)}{3}-\frac{E(\tau^2)^2}{4\mu_{\tau}})$$
Hence correlation coefficient is
$$\rho=\frac{2E(\tau^3)E(\tau)-3E(\tau^2)^2}{4E(\tau^3)E(\tau)-3E(\tau^2)^2}=\frac {2\kappa\sigma^3\mu-\mu^4-3\sigma^4}{4\kappa\sigma^3\mu+6\sigma^2\mu^2+\mu^4-3\sigma^4}=\frac{2\kappa c_v-c_v^4-3}{4\kappa c_v+6c_v^2+c_v^4-3}<\frac 1 2$$
To check correctness, plug in $\sigma=0$ to get $\rho=-1$ as expected. For $\tau$ exponential, $\kappa=2$, $c_v=1$ and $\rho=0$ as expected. We also see that by making $\kappa$ large we can approach $\rho = \frac 1 2$, but for $\kappa=0$, $\rho<0$ - as expected.

Since $\rho\ge -1$, we also get probabilistic proof that $$E(\tau^2)^2\le E(\tau^3)E(\tau)$$
for non-negative $\tau$.
