Let X and Y be independent continuous random variables that are uniformly distributed on (0,1). Let H=(X+2)Y. Find the probability P(lnH≥z) where z is a given number that satisfies ez<2. Your answer should be a function of z.

Question

Hint: Condition on X.

P(lnH≥z1)= ?

Let X be a standard normal random variable, and let FX(x) be its CDF. Consider the random variable Z=FX(X). Find the PDF fZ(z) of Z. Note that fZ(z) takes values in (0,1) .

fZ(z)= ?

Please, can someone share me the correct answers? I've spent a lot of time trying to solve this problem, but I can't.

Answer 1

if X and Y are uniform random (0,1) and H=(X+2)Y, then the distribution for H looks like the piecewise graph y={0.4: 0<x<2, 1.2-.4x, 2<x<3}, which i estimated using simulation

Answer 2

upon further consideration (using math not simulation), the function for H is
{0<x<2: ln(3)-ln(2), 2<x<3: ln(3)-ln(x)} (piecewise)

the graph of H:
desmos.com/calculator/jf384j0mrt

the probability graph that P(ln(H)>z):
desmos.com/calculator/io40jln2nx

Answer 3

1. P(lnH≥z1)= 1-e^z*ln(3/2)
2. fZ(z)= 1