Exercise on law of total expectation

[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.
c) What about $E(Y)$?

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 $Y\in C(0,1)$?

 

[Orodruin]

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).