Question

Random variable X has the following cumulative distribution function: 0 x〈1 0.12 1Sx <2 F(x) 0.40 2 x<5 0.79 5 x<9 1x29 a. Find the probability mass function of X. b. Find E[X] c. Find E[1/(2X+3)] d. Find Var[X]

Answer 1

a)

This is discrete probability mass function

p(x) = 0.12   x = 1

      = 0.4 - 0.12 = 0.28    x = 2

0.79 - 0.40 = 0.39   x = 5

1-0.79 =0.21    x= 9

b)

E(X) = 1 * 0.12 + 2 * 0.28 + 5 * 0.39 + 9* 0.21

= 4.52

c)

E(1/(2x+3))

= 1/(2*1 + 3) * 0.12 + 1/(2*2 + 3) * 0.28 + 1/(2*5+3) * 0.39 + 1/(2*9 + 3)* 0.21

=0.104

d)

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

E(X^2) = 1 * 0.12 + 2^2 * 0.28 + 5^2 * 0.39 + 9^2 * 0.21 = 28

Var(X) = 28 - 4.52^2

= 7.5696