Tom K. answered 07/06/19
Knowledgeable and Friendly Math and Statistics Tutor
U = X/X+Y
V = X+Y
Thus, UV = X/X+Y(X+Y) = X; V - UV = Y
dX/dU = V
dX/dV = U
dY/dU = -V
dY/dV = 1 - U
Thus, as f(x,y) = λ2e-λ(x+y) 0 <= x,y <= ∞
f(u,v) = λ2e-λ(V) |V U |
|-V 1 - U|
= λ2Ve-λ(V) 0 < U < 1 0 < V < ∞ (Note that, for any value of V, U is able to range from 0 to 1, so the domain is "rectangular", also, necessary for independence)
U and V are independent.
Tom K.
I did not provide f(u) and f(v), but they are pretty obvious. f(u) = 1 0<= u <= 1; f(v) = lambda^2 v e^-lambda v07/06/19