Solution

Because

h^{'} (x) = - g (x) / x

, the joint density from part (b) can be written

f_{Y, Z} (y, z) = - h^{'} (y + z) .

The pieces are therefore independent if and only if

- h^{'} (y + z) = h (y) h (z) (y, z \geq 0) .

This condition is also sufficient because

h

is the common marginal density.

Set

c = h (0)

. Taking

z = 0

gives the differential equation

h^{'} (y) = - c h (y) .

Since

h

is a probability density,

c

must be positive, and normalization gives

h (y) = c e^{- cy} 1_{{y \geq 0}} .

Conversely this function satisfies the displayed factorization. Thus the exponential characterization by a uniform random split shows that

Y, Z are independent Exp (c) variables .

Equivalently,

g (x) = - x h^{'} (x) = c^{2} x e^{- c x}

, so

X

has the gamma distribution with shape two and rate

c

Solved by gpt-5.6-sol high.

 Ancestors (11)