Question

Suppose X1 and X2 are iid Poisson(θ) random variables and let T = X1 + 2X2.

(a) Find the conditional distribution of (X1,X2) given T = 7.

(b) For θ = 1 and θ = 2, respectively, calculate all probabilities in the above conditional distribution and present the two conditional distributions numerically.

Answer 1

$(a)~P(X_1=x_1,X_2=x_2|T=7)=0~if~x_1+2x_2\neq 7\\ =\frac{P(X_1=x_1,X_2=(7-x_1)/2)}{P(X_1+2X_2=7)}~if~x_1+2x_2=7\\ =\frac{P(X_1=x_1,X_2=(7-x_1)/2)}{P(X_1=1,X_2=3)+P(X_1=3,X_2=2)+P(X_1=5,X_2=1)+P(X_1=7,X_2=0)}\\ =\frac{P(X_1=x_1)P(X_2=(7-x_1)/2)}{P(X_1=1)P(X_2=3)+P(X_1=3)P(X_2=2)+P(X_1=5)P(X_2=1)+P(X_1=7)P(X_2=0)}\\ since~X_1~and~X_2~are~independent\\ =\frac{\frac{e^{-\theta}\theta^{x_1}}{x_1!}\frac{e^{-\theta}\theta^{(7-x_1)/2}}{((7-x_1)/2)!}}{\frac{e^{-2\theta}\theta^4}{1!3!}+\frac{e^{-2\theta}\theta^5}{3!2!}+\frac{e^{-2\theta}\theta^6}{5!1!}+\frac{e^{-2\theta}\theta^7}{7!0!}}=\frac{\frac{\theta^{\frac{x_1-1}{2}}}{x_1!((7-x_1)/2)!}}{\frac{1}{6}+\frac{\theta}{12}+\frac{\theta^2}{120}+\frac{\theta^3}{5040}}\\$$=\frac{\frac{1}{6}}{\frac{1}{6}+\frac{\theta}{12}+\frac{\theta^2}{120}+\frac{\theta^3}{5040}}~when~x_1=1\\\\ =\frac{\frac{\theta}{12}}{\frac{1}{6}+\frac{\theta}{12}+\frac{\theta^2}{120}+\frac{\theta^3}{5040}}~when~x_1=3\\\\ =\frac{\frac{\theta^2}{120}}{\frac{1}{6}+\frac{\theta}{12}+\frac{\theta^2}{120}+\frac{\theta^3}{5040}}~when~x_1=5\\\\ =\frac{\frac{\theta^3}{5040}}{\frac{1}{6}+\frac{\theta}{12}+\frac{\theta^2}{120}+\frac{\theta^3}{5040}}~when~x_1=7\\\\ =0~otherwise\\ (b)~For~\theta=1\\ P(X_1=x_1|T=7)=0.6447~when~x_1=1\\ =0.3223~when~x_1=3\\ =0.0322~when~x_1=5\\ =0.0008~when~x_1=7\\ =0~otherwise\\ For~\theta=2\\ P(X_1=x_1|T=7)=0.4526~when~x_1=1\\ =0.4526~when~x_1=3\\ =0.0905~when~x_1=5\\ =0.0043~when~x_1=7\\ =0~otherwise\\$