Question

Suppose that a random variable X has a binomial distribution with n=2, p=0.5. Find the mean and variance of Y = X²

Answer 1

Here X has a binomial distribution with n = 2 and p = 0.5

so here Y = X²

first we will find the distribution of X

p(x) = ²C_x (0.5)^x(0.5)^(2-x)

p(x) = 0.5 * 0.5 = 0.25 ; x = 0

= 2 * 0.5 * 0.5 = 0.5 ; x = 1

= 0.5 * 0.5 = 0.25 ; x = 2

p(x) = 0.25 ; x= 0

= 0.5 ; x = 1

= 0.25 ; x = 2

Now as we know Y = X²

sample space of Y would be {0²,1²,2²} or {0, 1, 4}

so here

p(y) = 0.25 ; y = 0

= 0.5 ; y = 1

= 0.25 ; y = 4

E[Y] = = 0.25 * 0 + 0.5 * 1 + 0.25 * 4 = 1.5

VaR[Y] = E[Y²] - E[Y]²

E[Y²] = 0.25 * 0 * 0 + 0.5 * 1 * 1 + 0.25 * 4 * 4 = 4.5

VaR[Y] = 4.5 - 1.5² = 2.25

SD[Y] = sqrt(2.25) = 1.5

so here

Mean of Y = 1.5

Variance of Y = 2.25