Question

If X and Y are independent and identically distributed uniform random variables on (0,1) compute the joint density of

U = X+Y, V = X/(X+Y)

Part A,

The state space of (U,V) i.e. the domain D over which f_U,Y (u,v) is non-zero can be expressed as

(D = {(u,v) $\epsilon$ R x R] 0 < h₁(u,v) < 1, 0 < h₂(u,v) < 1} where x = h₁ (u,v) and y = h₂ (u,v)

Find h₁(u,v) = (write a function in terms of u and v)

Find h₂(1,0.25)

Part B,

For (u,v) $\epsilon$ D, f_U,V(u,v) = (answer)

Answer 1

We are applying transformation g: (0,1)2 + R2 such that g(x,y) = (u, v) = (2+y, +y Using the theorem about the transformation of variables, we have that the density function of random vector (UV) = g(X,Y) can be written as fu.v(u, v) = fx.y (1,7). det(g(x,y)-1 Calculate that volt» – [Fara acelet which yields that det(g(x,y) = 7 (2 + y)2 (2 + y) 2 +y Thus fuv(u, v) = fxy(2,y). (2+ y) = 1+y=u

uv = x
h1(u,v) = uv
y = U - x = U - UV = U(1-V)

hence
h2(u,v) = U(1-V)