Suppose we have a joint normal distribution: $X\sim N(\mu_X,\sigma_X^2)$ and $Y^*\sim N(\mu_{Y^*},\sigma_{Y^*}^2)$, and the covariance $cov(X,Y^*)=\rho\sigma_X\sigma_{Y^*}$. Define $Y\equiv Y^*$ if $Y^*>X$ and 0 if otherwise.
How do I compute $E(Y|Y\neq 0)$?
Let $f(x,y^*)$ denote the joint density of $X$ and $Y^*$. Define $I_{(Y^*>X)}$ to be an indicator random variable such that it equals one when $Y^*>X$ and zero otherwise.
First, we have $$ E(Y\mid Y\neq 0) = E(Y^*\mid Y^*>X)=E(Y^*\mid I_{(Y^*>X)}=1). $$ Second, we have $$ P(I_{(Y^*>X)}=1)=\int_{-\infty}^\infty\int_{y^*>x} f(x,y^*)dy^*dx. $$ and $$ P(Y^*=y^*,I_{(Y^*>X)}=1)=\int_{y^*>x} f(x,y^*)dx. $$ Third, the expectation can be calculated by \begin{equation}\label{eq_1} \frac{\int_{-\infty}^\infty y^*\cdot\int_{y^*>x} f(x,y^*)dx dy^*}{\int_{-\infty}^\infty\int_{y^*>x} f(x,y^*)dxdy^*}. \end{equation}
Next, by the property of conditional normal distribution, $$f(x, y^*) = f(x\mid y^*) \cdot f(y^*),$$ where $f(x\mid y^*)$ is the density of $\mathcal{N}\left(\mu_X+\frac{\sigma_{X}}{\sigma_{Y^*}} \rho\left(y^*-\mu_{Y^*}\right),\left(1-\rho^2\right) \sigma_{X}^2\right)$ and $f(y^*)$ is the density of $Y^*$.
Therefore, $$ \int_{y^*>x} f(x,y^*)dx = \int_{y^*>x} f(x\mid y^*) f(y^*)dx = CDF_{x\mid y^*}(y^*)\cdot f(y^*), $$ where $CDF_{x\mid y^*}$ is the CDF of $\mathcal{N}\left(\mu_X+\frac{\sigma_{X}}{\sigma_{Y^*}} \rho\left(y^*-\mu_{Y^*}\right),\left(1-\rho^2\right) \sigma_{X}^2\right)$.
Finally, $$\frac{\int_{-\infty}^\infty y^*\cdot\int_{y^*>x} f(x,y^*)dx dy^*}{\int_{-\infty}^\infty\int_{y^*>x} f(x,y^*)dxdy^*} = \frac{\int_{-\infty}^\infty y^*\cdot CDF_{x\mid y^*}(y^*)\cdot f(y^*) dy^*}{\int_{-\infty}^\infty CDF_{x\mid y^*}(y^*)\cdot f(y^*)dy^*}, $$ where recall that $f(y^*)$ is the density of $\mathcal{N}\left(\mu_{Y^*}, \sigma_{Y^*}^2\right)$.