Assume that two r.v. $W$ and $Y|W=w$ with
(1) $W \sim \text{N}(\mu_w,\sigma_w^2)$ (iid)
(2) $Y|W=w \sim \text{N}(w,\sigma_y^2)$ (iid)
Further we only observe $Y$ if $Y$ is less then $W$, i.e.,
(3) $Y|Y\le W$
Goal: Find the pdf of the censored observations, i.e., of $Y|Y\le W$ and from that deduce the uncensored pdf and the first two moments (so i.m.h.o. we have to find$f_Y(y)$). The first two moments of this uncensored pdf are supposed to depend upon $E(Y|Y\le W)$ and $Var(Y|Y\le W)$.
By definition of conditional pdf we have that:
(4) $f_{Y|W}(y|W = w)= \frac{f_{Y,W}(y,w)}{f_W(w)}$
Next, the definition of a truncated density gives for a abitrary value of $W$:
(5) $ f_{Y|Y\le W}(y|y\le w) = \frac{f_Y(y)}{P(Y\le W)}$
I would simply rewrite (4) to
$f_{Y|W}(y|W = w)f_W(w) = f_{Y,W}(y,w)$
then integration over $f_{Y,W}(y,w)$ w.r.t $w$ should yield $f_Y(y)$, i.e.,
(a) $\int_{-\infty}^{\infty} f_{Y,W}(y,w) dw = \int_{-\infty}^{\infty} f_Y(y|W = w)f_W(w) dw = f_Y(y)$
Plugin in $f_Y(y)$ into (5), ($P(Y\le W)$ will also be given by $f_Y(y)$) I will se how the moments of $f_{Y|Y\le W}(y|y\le w)$ will look and how the moments of $f_Y(y)$ depend upon them.
So (a) will look like
$f_Y(y) = \int_{-\infty}^{\infty}\frac{1}{\sqrt{2\pi\sigma^2_y}}\text{exp}\big(-\frac{(y-w)^2}{2\sigma_y^2}\big)\frac{1}{\sqrt{2\pi\sigma^2_w}}\text{exp}\big(-\frac{(w-\mu_w)^2}{2\sigma_w^2}\big)dw$
Except for the $w$ in the first $\text{exp}$, this looks very easy but since there is a $w$ im a little bit stuck how to solve this...