Question

For the conditional Poisson process, let \(m_{1}=E[L], m_{2}=E\left[L^{2}\right] .\) In terms of \(m_{1}\) and \(m_{2}\), find \(\operatorname{Cov}(N(s), N(t))\) for \(s \leqslant t .\)

Short answer

The covariance between \(N(s)\) and \(N(t)\) for a conditional Poisson process is given by: \[ \operatorname{Cov}(N(s), N(t)) = \frac{s}{t}(m_2 - m_1^2). \]

Step-by-step solution

Write down the definition of covariance

The covariance between two random variables \(X\) and \(Y\) is given by the following formula: \[ \operatorname{Cov}(X,Y) = E[XY] - E[X]E[Y] \] In our case, the random variables are \(N(s)\) and \(N(t)\).

Rewrite the terms of the covariance in terms of \(m_{1}\) and \(m_{2}\)

We need to find the terms \(E[N(s)N(t)]\), \(E[N(s)]\), and \(E[N(t)]\). We know that these conditional expectations are related to \(m_{1}\) and \(m_{2}\), where: \[ m_{1} = E[L] = E[N(t)], \] \[ m_{2} = E[L^{2}] = E\left[ N(t)^2 \right]. \] For \(s \leq t\), we have \(E[N(s)N(t)] = E[N(s)L]\). Also, since \(E[N(s)]=E[L(s/t)N(t)]=m_{1}(s/t)\) from the properties of the conditional Poisson process. Now let's replace these terms in the expression for the covariance: \[ \begin{aligned} \operatorname{Cov}(N(s), N(t)) &= E[N(s)N(t)] - E[N(s)]E[N(t)] \\ &= E[N(s)L] - m_1(s/t)m_1. \end{aligned} \]

Simplify the expression

The only remaining term to simplify is \(E[N(s)L]\). Using the property of conditional expectations, \[ E[N(s)L] = E\left[ E\left[ N(s)L \mid N(t) \right] N(t) \right] = E\left[ E[N(s) \mid N(t)] N(t) \right] = E\left[ (s/t) N(t)^2 \right]. \] Now, we substitute this term into the covariance expression: \[ \begin{aligned} \operatorname{Cov}(N(s), N(t)) &= E\left[ (s/t) N(t)^2 \right] - m_1(s/t)m_1 \\ &= (s/t) m_2 - m_1(s/t)m_1. \end{aligned} \] Finally, the covariance between \(N(s)\) and \(N(t)\) for a conditional Poisson process is therefore: \[ \operatorname{Cov}(N(s), N(t)) = \frac{s}{t}(m_2 - m_1^2). \]