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.

  • $\begingroup$ Poissons on dishjoint intervals are independent, so covariance is... $\endgroup$
    – A.S.
    Dec 14, 2015 at 16:04
  • $\begingroup$ If they are independent the covariance is $0$, right? However, are they independent no matter what the density function is? $\endgroup$
    – Eff
    Dec 14, 2015 at 16:13
  • $\begingroup$ What density function? $\endgroup$
    – A.S.
    Dec 14, 2015 at 16:18
  • $\begingroup$ No, it works only for memoryless $f(t)$ - without memorylessness, information will "seep" through". $\endgroup$
    – A.S.
    Dec 14, 2015 at 16:39
  • $\begingroup$ @A.S. Yeah, I thought so too. Do you know how one can approach the problem with the given density? $\endgroup$
    – Eff
    Dec 14, 2015 at 16:41

1 Answer 1


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 $\tau$.

  • $\begingroup$ +1 Thanks a lot for the answer. I have one question, and it might be simple, but how do you get $$E[X(Z-X)] = \frac16 E(Z^2)?$$ $\endgroup$
    – Eff
    Dec 15, 2015 at 12:54
  • $\begingroup$ It is from the assumption that X is evenly distributed in [0, Z], given Z. $\endgroup$
    – Jay.H
    Dec 15, 2015 at 14:20
  • $\begingroup$ @Jay.H Hmm.. Okay. It is just weird, because the result just does not seem to align with my simulations. It is correct to say that $$E(Z) = \int_0^\infty f(t)\,\mathrm{d}t $$ and $$E(Z^2) = \int_0^\infty t^2 f(t)\,\mathrm{d}t $$ right? Can we say that even though we are more likely to hit a long interval? $\endgroup$
    – Eff
    Dec 15, 2015 at 14:42
  • $\begingroup$ You should use $f_Z$ instead of $f$ in the calculation. $\endgroup$
    – Jay.H
    Dec 15, 2015 at 15:14
  • $\begingroup$ @Jay.H Thank you both, now I understand everything! $\endgroup$
    – Eff
    Dec 15, 2015 at 16:43

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