Joint probability density The question is taken from one of the past year papers. Please tell me if my answer needs any correction/improvement, thanks!

The joint density function of $X$ and $Y$ is given by $f(x,y) =
 \left\{ {\begin{array}{*{20}{c}}    {\frac{K}{y}{e^{ -
 \frac{{{y^2}}}{2} - \frac{x}{y}}}} & {0 < y < x < \infty }  \\    0 &
 {o.w}  \\ \end{array}} \right.$ (i) Determine the value of $K$
   (ii) Find the marginal probability density function of $Y$ .
  (iii) Evaluate $E[Y]$ and $Var(Y)$. (iv) Find the conditional
  density function ${f_{X|Y}}(x|y)$ for $0 < y < x$, and then evaluate
  $E[X|Y]$. (v) Evaluate $E[X]$. (vi) Evaluate $Cov(X,Y)$.

(i)
$\int\limits_0^\infty  {\int\limits_0^x {\frac{K}{y}{e^{ - \frac{{{y^2}}}{2} - \frac{x}{y}}}dydx = } } K\int\limits_0^\infty  {\int\limits_y^\infty  {\frac{1}{y}{e^{ - \frac{{{y^2}}}{2} - \frac{x}{y}}}dxdy = } } K\int\limits_0^\infty  {\frac{1}{y}{e^{ - \frac{{{y^2}}}{2}}}} \int\limits_y^\infty  {{e^{ - \frac{x}{y}}}} dxdy = K\int\limits_0^\infty  {\frac{1}{y}{e^{ - \frac{{{y^2}}}{2}}}y{e^{ - 1}}} dy = K{e^{ - 1}}\sqrt {2\pi } \frac{1}{{\sqrt {2\pi } }}\int\limits_0^\infty  {{e^{ - \frac{{{y^2}}}{2}}}} dy = K{e^{ - 1}}\sqrt {2\pi } \frac{1}{2}$.
The above integral must equal $1$, hence we solve $K = e\sqrt {\frac{2}{\pi }}$
(ii) It follows directly from (i) that ${f_Y}(y) = K\int\limits_y^\infty  {\frac{1}{y}{e^{ - \frac{{{y^2}}}{2} - \frac{x}{y}}}dx = K} {e^{ - \frac{{{y^2}}}{2}}}{e^{ - 1}} = {e^{ - \frac{{{y^2}}}{2}}}\sqrt {\frac{2}{\pi }} $
(iii)$E[Y] = \int\limits_0^\infty  y {e^{ - \frac{{{y^2}}}{2}}}\sqrt {\frac{2}{\pi }} dy = \sqrt {\frac{2}{\pi }} $
Let $Z$ ~ $N(0,1)$ then $E[{Z^2}] = Var(Z) = 1$. Thus $E[{Z^2}] = \frac{1}{{\sqrt {2\pi } }}\int\limits_{ - \infty }^\infty  {{z^2}{e^{ - \frac{{{z^2}}}{2}}}dz = 1}  \Rightarrow \frac{1}{{\sqrt {2\pi } }}\int\limits_0^\infty  {{z^2}{e^{ - \frac{{{z^2}}}{2}}}dz = \frac{1}{2}} $, and then
$E[{Y^2}] = \sqrt {\frac{2}{\pi }} \int\limits_0^\infty  {{y^2}} {e^{ - \frac{{{y^2}}}{2}}}dy = \sqrt {\frac{2}{\pi }} \int\limits_0^\infty  {{z^2}{e^{ - \frac{{{z^2}}}{2}}}dz = } \sqrt {\frac{2}{\pi }}  \times \frac{1}{2} \times \sqrt {2\pi }  = 1$
$Var(Y) = E[{Y^2}] - E{[Y]^2} = 1 - \frac{2}{\pi } = \frac{{\pi  - 2}}{\pi }$
(iv)
${f_{X|Y}}(x|y) = \frac{{f(x,y)}}{{{f_Y}(y)}} = \frac{1}{y}{e^{1 - \frac{x}{y}}}$
$E[X|Y = y] = \frac{e}{y}\int\limits_0^\infty  {x{e^{ - \frac{x}{y}}}} dx = \frac{e}{y}\left( {\frac{{{y^2} + y}}{e}} \right) = y + 1$
(v)
$E[X] = E[E[X|Y]] = E[Y + 1] = 1 + E[Y] = 1 + \sqrt {\frac{2}{\pi }} $
(vi)
$E[XY] = E[E[XY|Y]] = E[YE[X|Y]] = E[Y(Y + 1)] = E[{Y^2}] + E[Y] = 1 + \sqrt {\frac{2}{\pi }} $
$Cov(X,Y) = E[XY] - E[X]E[Y] = 1 + \sqrt {\frac{2}{\pi }}  - \left( {1 + \sqrt {\frac{2}{\pi }} } \right)\sqrt {\frac{2}{\pi }}  = 1 - \frac{2}{\pi }$
 A: These seem mostly correct, but there is a computational mistake in (iv), with consequences in (v) and (vi). Note that the joint density of $(X,Y)$ corresponds to $Y=|Z|$, with $Z$ standard normal, and $X=Y+YT$, with $T$ standard exponential and $(T,Z)$ independent.
Thus, in (iv), $E[X\mid Y]=Y+YE[T]=2Y$ (the integral in the question should start at $x=y$, not at $x=0$). 
The mistake propagates to (v), where the correct answer is $E[X]=E[E[X\mid Y]]=2E[Y]$, that is, $E[X]=2\sqrt{2/\pi}$. 
Likewise, in (vi), $E[XY]=E[E[X\mid Y]Y]=2E[Y^2]$ hence $\mathrm{cov}(X,Y)=2\mathrm{var}(Y)$, and $\mathrm{var}(Y)$ was (correctly) computed in (iii).
Edit: Assume that $(Y,T)$ is as above and consider any bounded measurable function $u$. Then,
$$
E[u(X,Y)]=E[u(Y(1+T),Y)]=\iint u(z(1+s),z)f_Y(z)f_T(s)\mathrm dz\mathrm ds.
$$
Use the change of variable $x=z(1+s)$, $y=z$, with Jacobian $\mathrm dx\mathrm dy=y\mathrm dz\mathrm ds$ to deduce that
$$
E[u(X,Y)]=\iint u(x,y)\color{red}{f_Y(y)f_T((x/y)-1)y^{-1}}\mathrm dx\mathrm dy.
$$
This holds for every bounded measurable function $u$ hence, the density $f_{X,Y}$ of $(X,Y)$ is the part $\color{red}{\text{in red}}$, that is,
$$
f_{X,Y}(x,y)=f_Y(y)f_T((x/y)-1)y^{-1}=\sqrt{2/\pi}\mathrm e^{-y^2/2}\mathbf 1_{y\gt0}\cdot\mathrm e^{-x/y+1}\mathbf 1_{x\gt y}/y,
$$
and the marginal $f_X$, the density of $X$, is
$$
f_X(x)=\int f_{X,Y}(x,y)\mathrm dy=\sqrt{2/\pi}\int_0^x\mathrm e^{-y^2/2-x/y+1}\mathrm dy/y.
$$
Note that this was already (almost fully) given at the very beginning of the question. Anyway, for every $x\geqslant0$,
$$
f_X(x)=\sqrt{2/\pi}\int_0^1\mathrm e^{-x^2u^2/2-1/u+1}\mathrm du/u.
$$
