Let's break down and solve the problem step by step. --- ## **Step 1: Understanding the Problem** - \( X \) and \( Y \) are independent, given \(\lambda\). - \( X | \lambda \sim \text{Poisson}(\lambda) \) - \( Y | \lambda \sim \text{Poisson}(\lambda) \) - \(\lambda \sim \text{Exp}(\theta)\), i.e., \( f_\lambda(\lambda) = \theta e^{-\theta \lambda} \) for \( \lambda \geq \). - We seek the conditional distribution of \( X \mid (X+Y = n) \). --- ## **Step 2: Joint Distribution** The joint pmf of \( (X, Y, \lambda) \) is: \[ P(X = x, Y = y, \lambda) = P(X = x | \lambda) P(Y = y | \lambda) f_\lambda(\lambda) \] - \( P(X = x | \lambda) = \frac{\lambda^x e^{-\lambda}}{x!} \) - \( P(Y = y | \lambda) = \frac{\lambda^y e^{-\lambda}}{y!} \) - \( f_\lambda(\lambda) = \theta e^{-\theta\lambda} \) So, \[ P(X = x, Y = y, \lambda) = \frac{\lambda^x}{x!} e^{-\lambda} \cdot \frac{\lambda^y}{y!} e^{-\lambda} \cdot \theta e^{-\theta\lambda} = \frac{\lambda^{x+y}}{x!y!} e^{-2\lambda} \theta e^{-\theta\lambda} \] \[ = \frac{\lambda^{x+y}}{x!y!} \theta e^{-(\theta+2)\lambda} \] --- ## **Step 3: Marginal for \(X, Y\)** Integrate out \(\lambda\): \[ P(X = x, Y = y) = \int_^\infty \frac{\lambda^{x+y}}{x!y!} \theta e^{-(\theta+2)\lambda} d\lambda \] This integral is of the form: \[ \int_^\infty \lambda^k e^{-a\lambda} d\lambda = \frac{\Gamma(k+1)}{a^{k+1}} \] Here, \(k = x+y\), \(a = \theta+2\). \[ P(X = x, Y = y) = \frac{\theta}{x!y!} \frac{\Gamma(x+y+1)}{(\theta+2)^{x+y+1}} \] But \(\Gamma(x+y+1) = (x+y)!\): \[ P(X = x, Y = y) = \frac{\theta (x+y)!}{x!y! (\theta+2)^{x+y+1}} \] --- ## **Step 4: Conditional Distribution \(P(X = k | X+Y = n)\)** Let \(X = k\), \(Y = n-k\): \[ P(X = k | X+Y = n) = \frac{P(X = k, Y = n-k)}{P(X+Y = n)} \] - Numerator: \(P(X = k, Y = n-k) = \frac{\theta n!}{k! (n-k)! (\theta+2)^{n+1}}\) - Denominator: \(P(X+Y = n) = \sum_{j=}^n P(X = j, Y = n-j)\) \[ = \sum_{j=}^n \frac{\theta n!}{j! (n-j)! (\theta+2)^{n+1}} = \frac{\theta n!}{(\theta+2)^{n+1}} \sum_{j=}^n \frac{1}{j!(n-j)!} = \frac{\theta n!}{(\theta+2)^{n+1}} \cdot \frac{2^n}{n!} = \frac{\theta 2^n}{(\theta+2)^{n+1}} \] --- ## **Step 5: Simplify the Conditional Probability** \[ P(X = k | X+Y = n) = \frac{\frac{\theta n!}{k! (n-k)! (\theta+2)^{n+1}}}{\frac{\theta 2^n}{(\theta+2)^{n+1}}} = \frac{n!}{k! (n-k)!} \cdot \frac{1}{2^n} \] \[ = \binom{n}{k} \frac{1}{2^n} \] --- ## **Step 6: Final Answer** The conditional distribution is: \[ \boxed{X \mid (X+Y = n) \sim \text{Binomial}(n, 1/2)} \] That is, \[ P(X = k \mid X+Y = n) = \binom{n}{k} \frac{1}{2^n} \] --- ### **Explanation** Even though \(X\) and \(Y\) are not independent marginally (because they share the random \(\lambda\)), conditioning on their sum removes the effect of \(\lambda\), and the distribution is symmetric in \(X\) and \(Y\), just as in the fixed-\(\lambda\) case. Thus, the answer is a binomial distribution with probability \(1/2\).