Let X and Y be independent random variables. They both have a gamma distribution with mean 3 and variance 3. (a) Find the joint probability density function (pdf) of X, Y . Solution: Since they are independent it is just the product of a gamma density for X and a gamma density for Y . For the gamma distribution, μ = w/λ , σ 2 = w/λ 2 . Since the mean and variance are both 3, λ = 1 and w = 3. So f X,Y ( x, y ) = 1 Γ(3) 2 x 2 y 2 e − x − y if x ≥ 0 , y ≥ 0 0 , otherwise (b) Express P (3 X + Y ≤ 3) as an integral. Do not try to do the integral. Solution: The region where 3 x + y ≤ 3 , x ≥ 0 , y ≥ 0 is the triangle in the upper right quadrant below the line y ≤ 3 − 3 x . So we get Z 1 0 Z 3 − 3 x 0 1 Γ(3) 2 x 2 y 2 e − x − y dy dx 2. Let X have an exponential distribution with E [ X ] = 1. Let Y = X 2 − 2. (a) Find the mean and variance of Y . Solution: First we compute some moments of X for later use. The mgf for X is m ( t ) = 1 / (1 − t ). m ′ ( t ) = 1 (1 − t ) 2 , E [ X ] = m ′ (0) = 1 , m (2) ( t ) = 2 (1 − t ) 3 , E [ X 2 ] = m (2) (0) = 2 , m (3) ( t ) = 6 (1 − t ) 4 , E [ X 3 ] = m (3) (0) = 6 , m (4) ( t ) = 24 (1 − t ) 5 , E [ X 4 ] = m (4) (0) = 24 Now E [ Y ] = E [ X 2 ] − 2 = 2 − 2 = 0 E [ Y 2 ] = E [( X 2 − 2) 2 ] = E [ X 4 − 4 X 2 + 4] = 24 − 4 · 2 + 4 = 20