ii) Consider a pair of random variables X and Y with the following properties: X is a discrete random variable Y is a continuous random variable The unconditional distribution of Y is exponential with parameter λ (> 0) Conditional on Y = y, X follows a Poisson distribution with expected value y. a) Show that the unconditional distribution of X is the distribution described in part (i) for a particular value of p. Here we have P(X|Y=Y)=(e^-y)*(y^x)\x! Then to find marginal distribution of x why are integrating the multiplication of conditional density of x given y with marginal density of y
So we have the following info given \(X|Y \sim \text{Poisson}(Y), \quad Y\sim \text{Expo}(\lambda)\) And we have to find the unconditional distribution of \(X\), we can do this by using law of total probability which says that \(\Pr(X=x) = \sum_{y} \Pr(X=x|Y=y)\Pr(Y=y)\) this only valid if both \(X,Y\) are discrete rvs but here \(Y\) is a continuous rv so we can't use the above definition instead we need to modify this expression i.e. \(\Pr(X=x) = \int_{y} \Pr(X=x|Y=y)f_Y(y)dy\) Where \(f_Y(y)\) is the density of \(Y\) Also \(\Pr(X=x|Y=y)\) really means \(\lim_{h\rightarrow 0} \Pr(X=x|y-h < Y < y+h)\)
