Question

6.21 For a Poisson process with parameter λ show that for s <r, the correlation between N, and N, is Corr(N,, N)

Answer 1

A Stochastic Process {$N_{t}\, , t\geq 0$} with state space {0,1,...} is said to be a Poisson process with parameter $\lambda \, (\lambda > 0)$ if:

1) $N_{0}=0$

2) the process has independent increments

3) the number of events in any interval of length 't' is poisson distributed with mean $\lambda t$ i.e.

$N_{t&plus;s}-N_{s} \,\sim Poisson(\lambda t)\, , \; \forall \, s,t \geq 0$.

Proof.

$N_{s} \,\sim Poisson(\lambda s)\, \therefore E\left [ N_{s} \right ]=\lambda s=V\left [ N_{s} \right ]\; \; \; \; \; \; .........(1)$

$N_{t} \,\sim Poisson(\lambda t)\, \therefore E\left [ N_{t} \right ]=\lambda t=V\left [ N_{t} \right ]\; \; \; \; \; \; .........(1)$

$V[N_{s} ] =\lambda s= E[N_{s}^2]- \,E^2[N_{s}]=E[N_{s}^2]-(\lambda s)^2=E[N_{s}^2]-\lambda^2s^2$

$\therefore ,E[N_{s}^2]=\lambda s&plus;\lambda^2s^2 \; \; \; ...\;\; \; \;\; \; \; \; \; \; \; \; \; \; \; \; \; \; \;............(2)$

Here, 0<s<t, by definition of Covariance we have,

$\begin{align*} Cov(N_{s},N_{t}) &=E[(N_{s}-E[N_{s}])(N_{t}-E[N_{t}])] \\ &=E[(N_{s}-\lambda s)(N_{t}-\lambda t)] \\ &=E[N_{s} N_{t}-N_{s}\lambda t-N_{t}\lambda s&plus;\lambda s\,\lambda t] \\ &=E[N_{s} N_{t}]-\lambda t E[N_{s}]-\lambda s E[N_{t}]&plus;\lambda s\,\lambda t\\ &=E[N_{s} N_{t}]-\lambda s\,\lambda t-\lambda s\,\lambda t&plus;\lambda s\,\lambda t \\ &= E[N_{s} N_{t}]-\lambda s\,\lambda t \; \; \; \; \; \; \; \; \; \; \;\; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; .......(3) \end{align*}$

E[N_sN_t] can be rewritten as:

$\begin{align*} E[N_{s}N_{t}] &=E[(N_{t}-N_{s})(N_{s}-N_{0})]&plus;E[N_{s}^2]\; \; \; \; \; \; \; \; \; \; \; ......(4)\; \; \; \; \; \; \; \; \; \; \; (Since, N_{0}=0) \\ &=E[(N_{s}-\lambda s)(N_{t}-\lambda t)] \\ &=E[N_{s} N_{t}-N_{t}N_{0}-N_{s}^2&plus;N_{s}N_{0} ]&plus;E[N_{s}^2]\\ &=E[N_{s} N_{t}]-E[N_{t}N_{0}]-E[N_{s}^2]&plus;E[N_{s}N_{0} ]&plus;E[N_{s}^2]\\ &=E[N_{s} N_{t}]-0&plus;0 =E[N_{s} N_{t}]\\ \end{align*}$

Since, (0,s) and (s,t) are disjoint and independent,

$\begin{align*} E[(N_{t}-N_{s})(N_{s}-N_{0})] &=E[(N_{t}-N_{s})]\: \; E[(N_{s}-N_{0})] \\ &=(\lambda t-\lambda s)\,(\lambda s-\lambda 0) \: \; \; \; \; \; \; \; \; \; \; \; (Since\, property\, of\, difference\, of \,two \,poisson\, process)\\ &=\lambda (t-s)\,\lambda s\; \; \; \; \; \; \; ........(5)\end{align*}$From (4) and (5),

$\begin{align*} E[N_{t}N_{s}]&=\lambda (t-s)\,\lambda s&plus; E[(N_{s}^2] \\ &=(\lambda^2 s t-\lambda^2 s^2)&plus;\,(\lambda s&plus;\lambda^2 s^2)\; \; \; \; \;.... From (2) \\ &=\lambda^2 s t&plus;\,\lambda s\; \; \; \; \; \; \; \; \; \; \; \;\; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; ........(6)\end{align*}$

Thus, from (3) and (6),

$Cov(N_{s},N_{t})=\lambda s \,...\; \; \; \; \; \; \; \; \; \; \; \; ......(7)\\ Corr(N_{s},N_{t})=\frac{Cov(N_{s},N_{t})}{\sqrt{V(N_{s})}\sqrt{V(N_{t})}}\\ Corr(N_{s},N_{t})=\frac{\lambda s}{\sqrt{\lambda s \lambda t}}\\ Corr(N_{s},N_{t})=\sqrt{\frac{s}{t}} \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; From (7) and (1)$

Hence the proof.