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 Z(t) is E(Z(t)) = 0, and the second moment is E(Z(t)^2) = (\lambda_{1}+\lambda_{2})t.

Step-by-step solution

Express Pn(t) as a Product of Different Probabilities

Since X(t) and Y(t) are independent Poisson processes, we can express the probability Pn(t) as the product of probabilities from X(t) and Y(t): $$ P_{n}(t) = \sum_{k=-\infty}^{\infty} \operatorname{Pr}\\{X(t)=k\\} \operatorname{Pr}\\{Y(t)=k-n\\} $$

Rewrite the Probabilities Using Poisson Distribution Formula

Recall that the probability mass function for a Poisson process is given by $$ \operatorname{Pr}\\{N(t) = k\\} = \frac{(\lambda t)^{k} e^{-\lambda t}}{k!} $$ where N(t) is the Poisson process with parameter λ. Thus, we can rewrite the probabilities in the expression for Pn(t): $$ P_{n}(t) = \sum_{k=-\infty}^{\infty} \frac{(\lambda_{1} t)^{k} e^{-\lambda_{1} t}}{k!} \frac{(\lambda_{2} t)^{k-n} e^{-\lambda_{2} t}}{(k-n)!} $$

Compute the Generating Function of Z(t)

We define the generating function G(z) as $$ G(z) = \sum_{n=-\infty}^{\infty} P_{n}(t) z^{n} $$ By substituting the expression of Pn(t) from step 2: $$ G(z) = \sum_{n=-\infty}^{\infty} \left( \sum_{k=-\infty}^{\infty} \frac{(\lambda_{1} t)^{k} e^{-\lambda_{1} t}}{k!} \frac{(\lambda_{2} t)^{k-n} e^{-\lambda_{2} t}}{(k-n)!} \right) z^{n} $$

Simplify the Generating Function

We switch the order of summation and simplify the terms: $$ G(z) = e^{-\left(\lambda_{1}+\lambda_{2}\right) t} \sum_{k=-\infty}^{\infty} \frac{(\lambda_{1} t)^{k}}{k!} \sum_{n=-\infty}^{\infty} \frac{(\lambda_{2} t)^{k-n}}{(k-n)!} z^{n} $$ Then, we make a change of variables, let m = n - k: $$ G(z) = e^{-\left(\lambda_{1}+\lambda_{2}\right) t} \sum_{k=-\infty}^{\infty} \frac{(\lambda_{1} t)^{k}}{k!} \sum_{m=-\infty}^{\infty} \frac{(\lambda_{2} t)^{m}}{m!} z^{m+k} $$ Now, recall that the sum of exponents is the product of the generating functions; thus, we get the final expression for G(z): $$ G(z) = \exp{-\left(\lambda_{1}+\lambda_{2}\right) t} \exp{\left(\lambda_{1} z t+\left(\lambda_{2} / z\right) t\right)}, \quad|z| \neq 0 $$

Compute E(Z(t)) and E(Z^2(t))

Recall that the expected value can be obtained by differentiating the generating function and evaluating at z = 1: $$ E(Z(t)) = G'(1) = 0 $$ Similarly, we can evaluate the second moment: $$ E\left(Z(t)^{2}\right) = G''(1) + G'(1) = \lambda_{1}t + \lambda_{2}t = (\lambda_{1}+\lambda_{2})t $$ Thus, E(Z(t)) = 0, and E(Z^2(t)) = (\lambda_{1}+\lambda_{2})t.