Simple setup:
$G\sim N(0,1)$, $x\sim N(0,1)$, $m\sim N(x+G,1)$
I want to find $E(G|m>n)$, that is, the expectation of G conditional on the truncated variable m, where m is above some value n. (The next stage is to look at the finite sample characteristics, but that's for later. Alternatively, it's fairly easy to estimate via simulation.)
My current setup is:
$\int{\int{\int_{x=n}^{ \infty}{P(G|P(m|P(x|\mu_x,\sigma^2_x),\sigma^2_m),\sigma^2_G))}dx} dm} dG$
From there, I find (perhaps correctly, but (hopefully) unnecessarily complexly): $\int{\int{\int_{x=n}^{ \infty}{{\frac{1}{\sqrt{d\pi\sigma^2}}\cdot e^{-\frac{({\frac{1}{\sqrt{d\pi\sigma^2}}\cdot e^{-\frac{({\frac{1}{\sqrt{d\pi\sigma^2}}\cdot e^{-\frac{(x-\mu_x)^2}{2\sigma_x^2}}}-\mu_m)^2}{2\sigma_m^2}}}-\mu_G)^2}{2\sigma_G^2}}dx}}dm}dG}$
I'm trying to figure out either what I did wrong in expressing it this way, or how to express / simplify this more neatly. My first intuition was: can/should I simplify or solve the distribution of $P(x|m>n))$ first?
You can assume, that $m$ comes from $N(0,3)$ as it is a sum of three random variables, each from $N(0,1)$ $(m=G+x+y)$.
Also, consider that $\pmb{X}=(G,x,y)$. Obviously $\pmb{X}$ comes from three dimensional normal distribution with $I_3$ (identity matrix) as covariance matrix.
\begin{equation} \begin{split} E(G|m>n) &= \frac{1}{P(m>n)}\int_{m>n} GdP_{\pmb{X}} = \\ =\frac{1}{1-\Phi(n/\sqrt3)}&\int_{-\infty}^{\infty}\int_{n-G}^{\infty}\int_{n-G-x}^{\infty} G f_{\pmb{X}}(G,x,y)dydxdG= \\ =\frac{1}{1-\Phi(n/\sqrt3)}&\int_{-\infty}^{\infty}\int_{n-G}^{\infty}\int_{n-G-x}^{\infty} G \frac{1}{(2\pi)^{3/2}}e^{\frac{-(G^2+x^2+y^2)}{2}}dydxdG. \end{split} \end{equation}
Computing this integral should not be a problem for you.