Question

1. (i) Find the probability P(0<X₁<1/3 , 0<X₂<1/3) where X₁, X₂ have the joint pdf

                   f(x₁, x₂) = 4x₁(1-x₂) ,     0<x₁<1  0<x₂<1

                                      0,                  otherwise

(ii) For the same joint pdf, calculate E(X₁X₂) and E(X₁+ X₂)

(iii) Calculate Var(X₁X₂)

Answer 1

(i) The probability here is computed as:

Therefore 5/81 = 0.0617 is the required probability here.

b) The expected value here is computed as:

Therefore E(X₁X₂) = 2/9

The expected value here is computed as:

Therefore E(X₁ + X₂) = E(X₁) + E(X₂) = (2/3) + (1/3) = 1

c) The variance here is computed as:

Var(X₁X₂) = E(X₁²X₂²) - [E(X₁X₂)]²

Therefore, the variance now is computed here as:

Var(X₁X₂) = E(X₁²X₂²) - [E(X₁X₂)]² = (1/12) - (2/9)² = 0.0340

Therefore 0.0340 is the required variance here.