suppose a probability density function pdf of a random varia

suppose a probability density function (pdf) of a random variable x has given by

             0, otherwise }

a) The mean E(X) and the variance Var(X)

b) The Variance Var(Y) with Y = 3X - 1

Solution

a) PDF = ( 4 - 3*x^2 ) / 3...0 < x < 1..
E( X) = Integration from 0 to 1 [ x* ( 4 - 3*x^2) / 3 * dx ] = 5 / 12..

E ( X^2) = Integration from 0 to 1 [ x^2 * ( 4 - 3*x^2) / 3 * dx ] = 11 / 45 = 0.4166666667..
VAR ( X) = E ( X^2) - ( E(X) ) ^2 = ( 11 / 45 ) - ( 5/12)^2 = 0.07083333 ..

B) Var ( Y) = VAR ( 3X - 1 ) = 9 * VAR ( X) = 0.6375....