Question

Let \(\\{X(t), t \geq 0\\}\) and \(\\{Y(t), t \geq 0\\}\) be two independent Poisson processes with parameters \(\lambda_{1}\) and \(\lambda_{2}\), respectively. Define $$ Z(t)=X(t)-Y(t), \quad t \geq 0 $$ This is a stochastic process whose state space consists of all the integers. (positive, negative, and zero). Let $$ P_{n}(t)=\operatorname{Pr}\\{Z(t)=n\\}, \quad n=0, \pm 1, \pm 2, \ldots $$ Establish the formula $$ \sum_{n=-\infty}^{\infty} P_{n}(t) z^{n}=\exp \left(-\left(\lambda_{1}+\lambda_{2}\right) t\right) \exp \left(\lambda_{1} z t+\left(\lambda_{2} / z\right) t\right), \quad|z| \neq 0 $$ Compute \(E(Z(t))\) and \(E\left(Z(t)^{2}\right)\)

Short answer

The expected value of the process Z(t) is E(Z(t)) = \((\lambda_1-\lambda_2)t\), and the expected square value of Z(t) is E(Z(t)^2) = \(\lambda_1^2t^2+\lambda_2^2t^2\).

Step-by-step solution

Review Poisson process probability generating functions (PGFs)

Recall that the probability generating function for a Poisson distribution with parameter \(\lambda\) is given by: G(z) = \(\sum_{k=0}^{\infty} e^{-\lambda}\frac{\lambda^k}{k!}z^k = e^{\lambda(z-1)}\) Here, G(z) is the PGF, z is a dummy variable, and k is the number of events. For the given problem, we have two Poisson processes with parameters \(\lambda_1\) and \(\lambda_2\). We will find the related PGFs and combine them to establish the formula for Z(t).

Find the probability generating functions for X(t) and Y(t)

Since X(t) is a Poisson process with parameter \(\lambda_1\), the PGF for X(t) is given by: \(G_X(z) = e^{\lambda_1(z-1)t}\) Similarly, Y(t) is a Poisson process with parameter \(\lambda_2\), so the PGF for Y(t) is given by: \(G_Y(z) = e^{\lambda_2(z-1)t}\)

Find the probability generating function for Z(t)

Z(t) is defined as X(t) - Y(t). We can compute the PGF for Z(t) by taking the product of the PGFs for X(t) and Y(t with reversed sign) since X(t) and Y(t) are independent: \(G_Z(z) = G_X(z).G_Y(\frac{1}{z}) = e^{\lambda_1(z-1)t}.e^{\lambda_2\left(\frac{1}{z}-1\right)t}\) Simplify the expression: \(G_Z(z) = e^{-\left(\lambda_1+\lambda_2\right)t}.e^{\left(\lambda_1zt +\lambda_2\frac{t}{z}\right)}\) This is the probability generating function for Z(t). Our next task is to compute the expected value and the expected square value of the process Z(t).

Compute E(Z(t))

To compute the expected value of Z(t), we need to find the first derivative of the probability generating function G(z) with respect to z, and then evaluate the derivative at z=1: \(E(Z(t)) = \frac{dG_Z(z)}{dz}\Big|_{z=1} = \left(-\left(\lambda_1+\lambda_2\right)t.e^{-\left(\lambda_1+\lambda_2\right)t} + (\lambda_1t.e^{\lambda_1zt}+\lambda_2t.e^{\frac{\lambda_2t}{z}})\right)\Big|_{z=1}\) Evaluating at z=1: \(E(Z(t)) = \left(-\left(\lambda_1+\lambda_2\right)t + (\lambda_1t+\lambda_2t)\right)\Big|(e^0)\) \(E(Z(t)) = (\lambda_1-\lambda_2)t\) Thus, the expected value of Z(t) is \((\lambda_1-\lambda_2)t\)

Compute E(Z(t)^2)

To compute the expected square value of Z(t), we need to find the second derivative of the probability generating function G(z) with respect to z, and then evaluate the derivative at z=1: \(E\left(Z(t)^{2}\right) = \frac{d^2G_Z(z)}{dz^2}\Big|_{z=1}\) Compute the second derivative of G(z) \(E\left(Z(t)^{2}\right) = \frac{d^2}{dz^2}\left(e^{-\left(\lambda_1+\lambda_2\right)t}.e^{\left(\lambda_1zt +\lambda_2\frac{t}{z}\right)}\right)\Big|_{z=1}\) \(=\left(\lambda_1^2t^2e^{\left(\lambda_1zt\right)}+\lambda_2^2t^2e^{\left(\frac{\lambda_2t}{z}\right)}\right)\Big|_{z=1}\) Evaluate the second derivative at z=1: \(E\left(Z(t)^{2}\right) = \left(\lambda_1^2t^2+\lambda_2^2t^2\right)\Big|(e^0)\) \(E\left(Z(t)^{2}\right) = \lambda_1^2t^2+\lambda_2^2t^2\) Thus, the expected square value of Z(t) is \(\lambda_1^2t^2+\lambda_2^2t^2\) In conclusion, we have established the formula for the sum of probabilities of the states of the process Z(t), and computed the expected value and the expected square value of the process as \((\lambda_1-\lambda_2)t\) and \(\lambda_1^2t^2+\lambda_2^2t^2\), respectively.