Conditional random vector distribution

Question

$(X_n,Y_n)$ has a density $f_{X_n,Y_n}(s,t)=n(n-1)(t-s)^{n-2}1_{0<s<t<1}$.
It is independent of variable $N$ with $P(N=n)=\frac{1}{(e-2)n!}, n=2,3,...$

How to show that $(X_N,Y_N)$ has density $f(s,t)=\frac{e^{t-s}}{e-2}1_{0<s<t<1}$?

Answer 1

By Law of total probability, $$F_{X_N,Y_N}(s,t)=\sum_{n=2}^\infty \mathbb P(X_N\leq s,Y_N\leq t ~|~ N=n)\mathbb P(N=n)=\sum_{n=2}^\infty F_{X_n,Y_n}(s,t)\mathbb P(N=n).$$ We also used here that $(X_n,Y_n)$ does not depend on $N$, and therefore $$\mathbb P(X_N\leq s,Y_N\leq t~|~ N=n) = \mathbb P(X_n\leq s,Y_n\leq t~|~ N=n) = \mathbb P(X_n\leq s,Y_n\leq t).$$ Next find $f_{X_N,Y_N}(s,t)=\frac{\partial^2}{\partial s\partial t}F_{X_N,Y_N}(s,t)$ and get $$f_{X_N,Y_N}(s,t)=\sum_{n=2}^\infty f_{X_n,Y_n}(s,t)\mathbb P(N=n).$$ Calculate the last sum and get answer.