Question

Suppose \(X\) is a Poisson random variable with mean \(\lambda .\) The parameter \(\lambda\) is itself a random variable whose distribution is exponential with mean \(1 .\) Show that \(P[X=\) \(n\\}=\left(\frac{1}{2}\right)^{n+1}\)

Short answer

In summary, given that \(X\) is a Poisson random variable with mean \(\lambda\), and \(\lambda\) is an exponential random variable with mean 1, the probability \(P[X=n]\) can be computed as: \[P[X=n] = \left(\frac{1}{2}\right)^{n+1}\]

Step-by-step solution

Write down the PMF of Poisson and PDF of exponential distribution

Recall the PMF of a Poisson distribution with mean \(\lambda\) is given by: \[P(X=n|\lambda) = \frac{e^{-\lambda}\lambda^n}{n!}\] And the PDF of an exponential distribution with mean 1 is given by: \[f(\lambda) = e^{-\lambda}\]

Use the law of total probability to express \(P[X=n]\)

We want to find the probability \(P[X=n]\). Since \(X\) depends on \(\lambda\), which is itself a random variable, we should use the law of total probability, integrating over all possible values of \(\lambda\): \[P[X=n] = \int_0^{\infty} P(X=n|\lambda) f(\lambda) d\lambda\] Now, substitute the PMF of Poisson and PDF of exponential distributions into the expression: \[P[X=n] = \int_0^{\infty} \frac{e^{-\lambda}\lambda^n}{n!} \cdot e^{-\lambda} d\lambda\]

Calculate the integral and simplify the expression

Next, we need to calculate the integral. To do this, first simplify the expression inside the integral: \[P[X=n] = \int_0^{\infty} \frac{e^{-2\lambda}\lambda^n}{n!} d\lambda\] Now, use integration by substitution. Let \(u = 2\lambda\), so \(du = 2d\lambda\) and \(d\lambda = \frac{1}{2} du\): \[P[X=n] = \int_0^{\infty} \frac{e^{-u} (u/2)^n}{n!} \cdot \frac{1}{2} du\] \[P[X=n] = \frac{1}{2^{n+1}n!} \int_0^{\infty} e^{-u} u^n du\] We recognize the integral as the Gamma function, which is defined as: \[\Gamma(n+1) = \int_0^{\infty} e^{-u} u^n du\] Since \(\Gamma(n+1) = n!\) for any integer \(n\), we can replace the integral by the factorial. Therefore, the final expression for \(P[X=n]\) is: \[P[X=n] = \frac{1}{2^{n+1}n!} \cdot n!\] Cancel out the factorials, and we are left with: \[P[X=n] = \left(\frac{1}{2}\right)^{n+1}\] This is the desired probability.