Past Exam question?

hi could anybody help me with this question i found in one the past question papers :


f(x)= e^(x)/2 if x < 0
= e^(-x)/2 if x>0

Find E[|X|]

i got 0 as answer but in the solutions it was given as 1....

 

It can't be 0, because you are asked for the expectancy of the modulus of X, which is always >= 0. You have confused E(X) and E(|X|)

 

E[x]=1/2intregate [limit -infinity to o] [(-x)e^x]
+1/2 intregate [limit 0 to infinity][(x)e^(-x)]
=1/2[{0+(1-0)}+{0-(0-1)}]
=1/2[1+1]
=1

 

I think that's the integral of the p.d.f. over (-inf'ty,inf'ty) - which will always be 1! < Comment no longer relevant.

The expected value of |X| is

Integral over (-inf'ty,inf'ty) of |x| f(x) dx ,​

i.e.

(1/2) * [ Integral over (-inf'ty,0) of (-x)e^x dx + Integral over (0,infty) of xe^(-x) dx ] .​

This turns out to be 1 as well (both integrals are integrable by parts, for example - which is what's been done in the previous post).