Question

Let X be a discrete random variable with the pmf

p(x): 0.8 for x=-4,

0.1 for x=-2,

0.07 for x=0,

0.03 for x=2

a) Find E(2/X)

b) Find E(lXl)

c) Find Var(lXl)

Answer 1

p(x):

0.8 for x=-4,

0.1 for x=-2,

0.07 for x=0,

0.03 for x=2

a) E(2/X) = sum of [ (2/x)*p(x) ]

= (2/(-4))*(0.8) + (2/(-2))*(0.1) + (2/0)*(0.07) + (2/2)*(0.03)

= ∞ (undefined)

since (2/0) is present the value will be tending to infinity {or we can also say undefined}

b) E(lXl) = sum of [ (|x|)*p(x) ]

= (4)*(0.8) + (2)*(0.1) + (0)*(0.07) + (2)*(0.03)

E(lXl) = 3.46

c) Var(lXl)

Var(lXl) = E(lXl^2) - E(lXl)^2

E(lXl^2) = sum of (lXl^2 * p(x))

= 16*0.8 + 4*0.1 + 0*0.07 + 4*0.03

= 13.32

E(lXl)^2 = (3.46)^2 {E(|x|) calculated in previous part}

= 11.9716

Var(lXl) = E(lXl^2) - E(lXl)^2

= 13.32 - 11.9716

Var(lXl) = 1.3484

(please UPVOTE)