A discrete random variable X is characterized by a pmf (probability mass functio

Question

A discrete random variable X is characterized by a pmf (probability mass function) given by, pZ (x) = {A/2, for x = 2 A/6, for x = 1 2/3/A, for x = - 1. a) Find A element R_+ such that the normalization condition is fulfilled, sigam_x pX (x) = 1; b) Compute the expected value mu_X = F_i [X] of X; c) Compute the variance sigma^2_X = F_i [(X - mu_X)^2] of X; d) Compute the cdf (cumulative distribution function) F (1) of X defined as, F (1) = P (X lessthanorequalto 1) = sigma_x pX (x), with x lessthanorequalto 1.

Explanation / Answer

a) We know that the sum of all probabilities should be equal to 1. Therefore we get:

A/2 + A/6 + 2A/3 = 1

Taking LCM of 6 we get:

(3A + A + 4A) / 6 = 1

8A = 6

A = 3/4 = 0.75

Therefore the required value of A here is 0.75

b) The expected value of X is computed as:

E(X) = 2*P(X=2) + 1*P(X=1) + (-1)*P(X=-1)

E(X) = 2*(A/2) + (A/6) - (2A/3) = A + (A/6) - (2A/3) = (6A + A - 4A)/6 = A/2 = 0.75/2 = 0.375

Therefore the expected value of X is 0.375

c) The variance of X is computed as:

Var(X) = E(X2) - (E(X))2

E(X2) = 22*P(X=2) + 12*P(X=1) + (-1)2*P(X=-1)

E(X2) = 4*(A/2) + (A/6) + (2A/3) = A + (A/6) + (2A/3) = (6A + A + 4A)/6 = 11A/6 = 1.375

Therefore the variance now is computed as:

Var(X) = E(X2) - (E(X))2 = 1.375 - 0.3752 = 1.234375

Therefore the required variance here is 1.234375

d) The CDF of the given distribution is computed as:

F(X) = P(X < = x)

F(X< -1) = 0

F(x= - 1) = 2A/3 = 0.5

F( -1 < = X < 1) = 0.5

F( X = 1) = 0.5 + (A/6) = 0.625

F(1 <= X< 2) = 0.625

F( X > = 2) = 1

Therefore the final CDF is:

F(X< -1) = 0

F( -1 < = X < 1) = 0.5

F(1 <= X< 2) = 0.625

F( X > = 2) = 1

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