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,−1≤x≤1,−1≤a≤1 then the expectation of XX is? (a) 6/a6a (b) a /3a3 (c) a /2a2 (d) 3/a3a
2 Answers
Explanation:
for a pdf
E(X) =a/3
making (b) the correct answer
Explanation:
By definition if
E(X) =int_(-oo)^(oo) \ xf(x) \ dx
So given that
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