If the probability density function of XX is f(x) = (1+ax) /2, -1 le x le 1, -1 le a le 1f(x)=1+ax2,1x1,1a1 then the expectation of XX is? (a) 6/a6a (b) a /3a3 (c) a /2a2 (d) 3/a3a

2 Answers
Dec 3, 2017

E(X)=alpha/3E(X)=α3

Explanation:

for a pdf f(x)f(x) the expectation of X

E(X)=int_(all" "x)xf(x)dxE(X)=all xxf(x)dx

E(X)=int_(-1)^1x((1+alphax)/2)dxE(X)=11x(1+αx2)dx

E(X)=1/2int_(-1)^1(x+alphax^2)dxE(X)=1211(x+αx2)dx

E(X)=1/2[x^2/2+alphax^3/3]_(-1)^1E(X)=12[x22+αx33]11

E(X)=1/2{[x^2/2+alphax^3/3]^1-[x^2/2+alphax^3/3]_0}E(X)=12{[x22+αx33]1[x22+αx33]0}

E(X)=1/2{(cancel(1/2)+alpha/3)-(cancel(1/2)-alpha/3)}

E(X)=1/2(alpha/3+alpha/3)=1/2xx(2alpha)/3

E(X)=alpha/3

Dec 3, 2017

E(X) =a/3

making (b) the correct answer

Explanation:

By definition if f(x) is a continuous probability density function then:

E(X) =int_(-oo)^(oo) \ xf(x) \ dx

So given that f(x) = (1+ax)/2 for -1 le x le 1 then we have:

E(X) = int_(-1)^(1) \ x \ ((1+ax)/2) \ dx

\ \ \ \ \ \ \ \ = 1/2 \ int_(-1)^(1) \ x (1+ax) \ dx

\ \ \ \ \ \ \ \ = 1/2 \ int_(-1)^(1) \ x+ax^2 \ dx

\ \ \ \ \ \ \ \ = 1/2 [ x^2/2+(ax^3)/3 ]_(-1)^(1)

\ \ \ \ \ \ \ \ = 1/2 {(1/2+a/3)-(1/2-a/3)}

\ \ \ \ \ \ \ \ = 1/2 (1/2+a/3-1/2+a/3)

\ \ \ \ \ \ \ \ = 1/2 (2a)/3

\ \ \ \ \ \ \ \ = a/3