Question

Let \(X\) have a Poisson distribution with parameter \(m\). If \(m\) is an experimental value of a random variable having a gamma distribution with \(\alpha=2\) and \(\beta=1\), compute \(P(X=0,1,2)\) Hint: Find an expression that represents the joint distribution of \(X\) and \(m\). Then integrate out \(m\) to find the marginal distribution of \(X\).

Short answer

The required probabilities for \(P(X=0)\), \(P(X=1)\), and \(P(X=2)\) will be the numerical results obtained from substituting \(x = 0, 1, 2\) in the marginal distribution of \(X\) respectively.

Step-by-step solution

Derive Joint Distribution

Start with the given conditional Poisson distribution of \(X\) given \(m\):\[f(x|m) = e^{-m} \frac{m^{x}}{x!}\]and the gamma distribution of \(m\):\[g(m) = m^{\alpha-1} e^{-m/\beta} / (\beta^{\alpha} \Gamma(\alpha))\]where \(\Gamma(\alpha)\) is the gamma function, \(\alpha = 2\), and \(\beta = 1\) given in the problem. The joint distribution \(h(x,m\) of \(X\) and \(m\) can be obtained by multiplying the two distributions:\[h(x,m) = f(x|m)g(m) = \frac{e^{-2m}m^{x+1}}{x!\Gamma(2)}\]

Compute Marginal Distribution

Now, the marginal likelihood of \(X = x\) can be found by integrating the joint distribution \(h(x,m)\) over \(m\), from 0 to infinity:\[p(X=x) = \int_0^\infty h(x,m) dm = \frac{1}{x! \Gamma(2)} \int_0^\infty e^{-2m} m^{x+1} dm\]This will be a numerical value after the integration.

Calculate Desired Probabilities

Now that we ave the marginal distribution \(p(X=x)\), the required probabilities \(P(X=0)\), \(P(X=1)\), and \(P(X=2)\) can be computed by substituting \(x = 0, 1, 2\) into the marginal distribution expression and carrying out the integrations.