Question

If the probability density function of `X` is `f(x) = (1+ax) /2, -1 le x le 1, -1 le a le 1` then the expectation of `X` is? (a) `6/a` (b) `a /3` (c) `a /2` (d) `3/a`

Answer 1

`E(X)=alpha/3`

Explanation:

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

`E(X)=int_(all" "x)xf(x)dx`

`E(X)=int_(-1)^1x((1+alphax)/2)dx`

`E(X)=1/2int_(-1)^1(x+alphax^2)dx`

`E(X)=1/2[x^2/2+alphax^3/3]_(-1)^1`

`E(X)=1/2{[x^2/2+alphax^3/3]^1-[x^2/2+alphax^3/3]_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`

Answer 2

`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`