I have worked out $f_X(x),f_Y(y),f_{X|Y}(x|y) \ \text{and} \ f_{Y|X}(y|x)$
I wish to verify that $E(E(X|Y))=E(X)$
For $E(X)$ I will be calculating:
$$\int_{I_x}xf_X(x) \ dx$$
and for $E(E(X|Y))$ I am not so sure, but currently I think:
$$E(X|Y)=\int_{I_x}xf_{X|Y}(x|y) \ dx $$
but I am not sure what to do from here. Any help would be good.
($I_x$ is the domain on which $x$ is defined for the continuous random variable)
(Beware: case sensitivity.)
$\mathrm I_X$ appears to be the support for the random variable $X$. Likewise you would have $\mathrm I_Y$ as the support for the random variable $Y$.
Similarly you should use $\mathrm I_{X\mid Y=y}$ to be the conditional support for the variable given a particular value, $y$, for the random variable $Y$.
And thus:
$$\begin{align}\mathsf E(\mathsf E(X\mid Y)) ~&=~ \int_{\mathrm I_Y}\;f_Y(y)\color{navy}{\int_{\mathrm I_{X\mid Y=y}} x\,f_{X\mid Y}(x\mid y)\operatorname d x}\operatorname d y \\ &=~ \iint_{\mathrm I_{X\times Y}} x\,f_{X,Y}(x,y)\operatorname d (x,y) & \text{Fubini/Tonelli} \\ &=~ \text{something}\end{align}$$
And then you are done, okay.
