Question

Let \((X(t), Y(t))\) describe a stochastic process in two-dimensional space where \(X(t)\) is a Poisson process with parameter \(\lambda_{1}\) and \(Y(t)\) is a Poisson process independent of \(X(t)\) with parameter \(\lambda_{2} .\) Given that the process is in the state \(\left(x_{0}, y_{0}\right)\) at time \(t=0, x_{0}+y_{0}

Short answer

The probability that the process \((X(t), Y(t))\) intersects the line \(x+y=z\) at the point \((x, y)\) is given by the joint probability: \[P(X(t) = x, Y(t) = y) = \frac{(\lambda_1 t)^x e^{-\lambda_1 t}}{x!} \cdot \frac{(\lambda_2 t)^y e^{-\lambda_2 t}}{y!}\]

Step-by-step solution

Define the Poisson Process

A Poisson process can be defined as a stochastic process that models the number of events occurring in a fixed interval of time or space with a constant average rate. A Poisson process with parameter \(\lambda\) has the probability distribution: \[P(N(t) = n) = \frac{(\lambda t)^n e^{-\lambda t}}{n!}\] Where: - \(N(t)\) is the number of events in the process up to time \(t\), - \(n \in \{0,1,2,\dots\}\) is a non-negative integer, - \(\lambda > 0\) is the constant average rate of events, - And \(t \ge 0\) is a non-negative time.

Find the Probabilities for X(t) and Y(t)

Let's find the probability that \(X(t)\) takes the value \(x\): \[P(X(t) = x) = \frac{(\lambda_1 t)^x e^{-\lambda_1 t}}{x!}\] Similarly, let's find the probability that \(Y(t)\) takes the value \(y\): \[P(Y(t) = y) = \frac{(\lambda_2 t)^y e^{-\lambda_2 t}}{y!}\]

Find the Joint Probability using Independence of X(t) and Y(t)

Given that \(X(t)\) and \(Y(t)\) are independent, the joint probability that they take values \(x\) and \(y\) respectively can be calculated as: \[P(X(t) = x, Y(t) = y) = P(X(t) = x) \cdot P(Y(t) = y)\] Plugging in the probabilities from Step 2: \[P(X(t) = x, Y(t) = y) = \frac{(\lambda_1 t)^x e^{-\lambda_1 t}}{x!} \cdot \frac{(\lambda_2 t)^y e^{-\lambda_2 t}}{y!}\]

Find the Probability that the Processes Intersect the Line x + y = z

The process intersects the line \(x+y=z\) at the point \((x, y)\) if and only if the number of events in \(X(t)\) up to time \(t\) is \(x\) and the number of events in \(Y(t)\) up to time \(t\) is \(y\). We need to find the probability that \(X(t)=x\) and \(Y(t)=y\) for an arbitrary point \((x,y)\) on the line \(x+y=z\). Given the conditions \(x_0+y_0<z\) and \(x+y=z\), we must find: \[P(X(t) = x, Y(t) = y | x+y = z)\] We are interested in the joint probability of this event: \[P(X(t) = x, Y(t) = y) = \frac{(\lambda_1 t)^x e^{-\lambda_1 t}}{x!} \cdot \frac{(\lambda_2 t)^y e^{-\lambda_2 t}}{y!}\] Which is the solution to the given problem.