Question

Q7. (2096) Let W be a random variable giving the number of heads minus the number of tails in three tosses of a coin. Assume that the coin is biased so that a tail is twice as likely to occur as a head.List the elements of the sample space for the three tosses of a coin and to each sample point assign a value w of a) Find the probability distribution (p.m.f) of the random variable w. b) Find the cumulative distribution function (c.d.f.) of W. c) Find the expected value of W d) Find the variance of W

Answer 1

a)

The sample space for the three tosses of a coin are

(T, T, T) w = 0 - 3 = -3

(T, T, H) w = 1 - 2 = -1

(T, H, T) w = 1 - 2 = -1

(T, H, H) w = 2 - 1 = 1

(H, T, T) w = 1 - 2 = -1

(H, T, H) w = 2 - 1 = 1

(H, H, T) w = 2 - 1 = 1

(H, H, H) w = 3 - 0 = 3

a)

P(T) = 2 P(H)

Also, P(T) = 1 - P(H)

1 - P(H) = 2P(H)

=> P(H) = 1/3

P(T) = 2 P(H) = 2/3

PMF of W is,

P(W = -3) = (2/3) * (2/3) * (2/3) = 8/27

P(W = -1) = P(T, T, H) + P(T, H, T) + P(H, T, T) = (2/3) * (2/3) + (1/3) + (2/3) * (1/3) * (2/3) + (1/3) * (2/3) * (2/3) = 12/27 = 4/9

P(W = 1) = P(T, H, H) + P(H, T, H) + (H, H, T) = (2/3) * (1/3) * (1/3) + (1/3) * (2/3) * (1/3) + (1/3) * (1/3) * (2/3) = 6/27 = 2/9

P(W = 3) = (1/3) * (1/3) * (1/3) = 1/27

b)

CDF of W is,

P(W < -3) = 0

P(-3 $\le$ W < -1) = 8/27

P(-1 $\le$ W < 1) = 8/27 + 4/9 = 20/27

P(1 $\le$ W < 3) = 20/27 + 2/9 = 26/27

P(W $\le$ 3) = 1

c)

Expected value of W is,

E(W) = (8/27) * (-3) + ( 4/9) * (-1) + (2/9) * 1 + (1/27) * 3 = -1

(d)

E(W²) = (8/27) * (-3)² + ( 4/9) * (-1)² + (2/9) * 1² + (1/27) * 3² = 11/3

Variance of W is,

Var(W) = E(W²) - [E(W)]² = (11/3) - (-1)² = 8/3