# Exercise on law of total expectation

• psie
psie
Homework Statement
[First, see the relevant equations.] The object of this exercise is to show that if we not assume that ##E|Y|<\infty## in theorem 2.1, then the conclusion does not necessarily hold. Namely, suppose that ##X\in\Gamma(1/2,2)(=\chi^2(1))## and that $$f_{Y\mid X=x}\left(y\right)=\frac{1}{\sqrt{2\pi }}x^{\frac{1}{2}}e^{-\frac{1}{2}xy^2},\quad -\infty<y<\infty.$$

a) Compute ##E(Y\mid X=x), E(Y\mid X)## and, finally ##E(E(Y\mid X))##.
b) Show that ##Y\in C(0,1)##, i.e. is standard Cauchy.
Relevant Equations
Theorem 2.1: Suppose that ##E|Y|<\infty##. Then ##E(E(Y\mid X))=E(Y)##.
I feel like I'm doing something wrong. I have computed $$E(Y\mid X=x)=\int_\mathbb{R}y f_{Y\mid X=x}\, dy,$$with pen and paper, and I get the same that WolframAlpha gets, namely ##0##. Can this be right? If this is indeed true, then is ##E(Y\mid X)=E(E(Y\mid X))=0## too?

How do I go about showing ##Y\in C(0,1)##?

The distribution of ##Y## conditioned on ##X = x## is a normal distribution centered at zero, so yes, it will have zero expectation value.

Regarding (b), you have ##f_{Y|X=x}## and you have ##f_X## given. You need to show that marginalising ##f_{X,Y}## gives the distribution function for C(0,1).

psie

Replies
4
Views
508
Replies
1
Views
286
Replies
1
Views
516
Replies
1
Views
507
Replies
6
Views
575
Replies
16
Views
1K
Replies
3
Views
1K
Replies
20
Views
976
Replies
3
Views
336
Replies
1
Views
964