Questions: Transformations of Random Variables

Question 1

X is uniform on (0, 1) and Y = -ln X. What is the distribution of Y?

• A. Uniform on (0, ∞)
• B. Exponential with rate 1
• C. Normal with mean 0 and variance 1
• D. Uniform on (-∞, 0)

Answer: (B)

For y > 0, F_Y(y) = P(-ln X ≤ y) = P(X ≥ e⁻ʸ) = 1 - e⁻ʸ. That is the CDF of an exponential distribution with rate 1. There is no uniform distribution on an infinite interval.

Question 2

If X has density f_X and Y = 2X, then f_Y(y) = f_X(y/2).

• T. True
• F. False

Answer: False

The correct density is f_Y(y) = (1/2) f_X(y/2). Doubling X spreads its probability over an interval twice as long, so the density must be halved for the total area to stay 1.

Question 3

Let Z be standard normal and Y = Z². Use the CDF method to find f_Y(y) for y > 0.

Model answer: f_Y(y) = (1/√(2πy)) e^(-y/2) for y > 0, the chi-square distribution with 1 degree of freedom.

F_Y(y) = P(-√y ≤ Z ≤ √y) = Φ(√y) - Φ(-√y). Differentiating gives φ(√y)·(1/(2√y)) + φ(-√y)·(1/(2√y)) = φ(√y)/√y, which equals (1/√(2πy)) e^(-y/2). Both preimages ±√y contribute.

Question 4

X is continuous with strictly increasing CDF F. What is the distribution of U = F(X)?

Model answer: Uniform on (0, 1).

For 0 < u < 1, P(F(X) ≤ u) = P(X ≤ F⁻¹(u)) = F(F⁻¹(u)) = u, which is the uniform CDF. This is the probability integral transform, and running it backward (X = F⁻¹(U)) is a standard way to generate samples from a given distribution.