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
sjc
Dec 3, 2017

#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#

Steve M
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#

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