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#
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#