Page 121 - Fundamentals of Communications Systems

Review of Probability and Random Variables 3.35

Problem 3.28.
(a) If X and Y are independent, we can compute E[XY ]as

∞ ∞

E[XY ] = xyf XY (x, y)dxdy
−∞ −∞

∞ ∞
= xwf X (x) f Y (y)dxdy
−∞ −∞
∞ ∞

= xf X (x)dx yf Y (y)dy
−∞ −∞
= E[X]E[Y ]
In general cov(X, Y ) = E[XY ] − E[X]E[Y ] = 0, so if X and Y are
independent, then X and Y are uncorrelated.
(b) X ∈{−1, 2} Y ∈{−1, 0, 1}
First, let’s compute the marginal PMFs
1 1 2 1

P X (−1) = P XY (−1, y) = + = P X (2) = P XY (2, y) =
3 3 3 3
y∈ y y∈ y
1 1 1
P Y (−1) = P Y (0) = P Y (1) =
3 3 3
X and Y are independent if and only if P X (x) · P Y (y) = P XY (x, y) ∀x, y
If x = 2 and y = 0,

P X (2) · P Y (0) = P XY (x, y)
1 1 1 1
· = = (3.58)
3 3 9 3
X and Y are not independent.

E[XY ] = x · y · P XY (x, y)
x,y
1 1 1
= 2 · 0 · + (−1) · (−1) · + (−1) · 1 ·
3 3 3
1 1
= 0 + − = 0
3 3
1 2

E[X] = x · P X (x) = 2 · − 1 · = 0
3 3
x
1 1 1

E[Y ] = y · P Y (y) = 0 · − 1 · + 1 · = 0
3 3 3
y
cov(X, Y ) = 0 − 0 · 0 = 0