Find the joint density of X Y and XY where X and Y are inde

Find the joint density of X + Y and X/Y, where X and Y are independent exponential random variables with parameter . Show that X + Y and X/Y are independent.

Solution

suppose we have two variables X and Y with joint density f_xy(x,y) Suppose also we have two variables U and V defined as : U = g1(x, y) and V = g2(x, y) and we want to know what the joint density of U and V is, f_uv(u,v) We define the jacobian, J J = det(dg1/dx dg1/dy) (dg2/dx dg2/dy) Then the equation for f_uv(u,v) is : f_uv(u,v) = f_xy(x,y) / J (we substitute in functions of u and v for x and y) So in your question we have : f_xy(x,y) = L^2 exp{-L(x+y)} u = x + y v = x / y Then du/dx = 1, du/dy = 1, dv/dx = 1/y, dv/dy = -x/(y*y) So J = 1/y - (-) x/(y*y) = (x + y)/(y*y) We can also write : y = u - x vy = x so vu - vx = x or x = uv/(1+v) and y = u/(1+v) Hence f_uv(u,v) = L^2 exp{-L(x+y)} / ((x + y)/(y*y)) So f_uv(u,v) = L^2 exp{-Lu} * (u/((1+v)^2)) ------ Now to prove independence it is enough to show that the distribution function F_uv(u,v) is bilaterally separable. ie. Integral(u=0, u=inf) of f_uv(u,v) gives f_u(u) Integral(v=0, v=inf) of f_uv(u,v) gives f_v(v) f_u(u) = L^2 exp{-Lu} * u * Int((1+v)^-2) = L^2 exp{-Lu} * u * [-(1+inf)^-2] - [-(1+0)^-2)] = L^2 exp{-Lu} * u Likewise L^2 exp{-Lu} * u is the equation for a gamma distribution, so f_v(v) = (1+v)^-2 Hence f_uv(u,v) = f_u(u) * f_v(v), so the two are independen