Massachusetts Institute of Technology
Department of Electrical Engineering & Computer Science
6.041/6.431: Probabilistic Systems Analysis
(Spring 2006)
Recitation 14 Solutions
April 11, 2006
1. We know that:
ρ(X1 , X2 ) =
Cov(X1 , X2 )
σX1 σX2
Therefore we first find the covariance:
Cov(A, B) = E[AB] − E[A]E[B]
= E[W X + W Y + X 2 + XY ]
= E[X 2 ] = 1
and
σA =
�
σB =
�
Var(A) =
√
2
√
Var(B) = 2
and therefore:
1
ρ(A, B) = .
2
We proceed as above to find the correlation of A, C.
Cov(A, C) = E[AC] − E[A]E[C]
= E[W Y + W Z + XY + XZ]
= 0
and therefore
ρ(A, C) = 0.
2. Solution is in the text, pp. 264–265.
3. Solution is in the text, pp. 267–268.
Page 1 of 1