I am posting the complete answer for reference: \begin{eqnarray*} E(X|a<Y<b) &= &\int_a^bE(X|Y=y)f_{Y|a<Y<b}(y)dy \\ & = &\rho_{xy}\frac{\sigma_x}{\sigma_y}\int_a^bf_{Y|a<Y<b}(y)dy \\ & = &\rho_{xy}\frac{\sigma_x}{\sigma_y}\sigma_y\frac{\phi(a-\phi(b)}{\Phi(b)-\Phi(a)} \end{eqnarray*}

With $\rho_{xy}=\frac{E(xy)}{\sigma_x\sigma_y}$. The last step is derived from the properties of the univariate truncated normal distribution, easily found on its wikipedia page. A reference for the first step is here

With $a=0,b=\infty$ we have $$ E(X,Y>0) = \rho_{xy}\sigma_x\frac{\phi(0)}{\Phi(0)}= \rho_{xy}\sigma_x\cdot \frac{2}{\sqrt{2\pi}}$$ Thanks for pointing in the right direction.

I am posting the complete answer for reference: \begin{eqnarray*} E(X|a<Y<b) &= &\int_a^bE(X|Y=y)f_{Y|a<Y<b}(y)dy \\ & = &\rho_{xy}\frac{\sigma_x}{\sigma_y}\int_a^bf_{Y|a<Y<b}(y)dy \\ & = &\rho_{xy}\frac{\sigma_x}{\sigma_y}\sigma_y\frac{\phi(a)-\phi(b)}{\Phi(b)-\Phi(a)} \\ & = &\frac{\sigma_{xy}}{\sigma_y}\frac{\phi(a)-\phi(b)}{\Phi(b)-\Phi(a)} ,\end{eqnarray*} where $\rho_{xy} = E(xy) / (\sigma_x\sigma_y) =\sigma_{xy} / (\sigma_x \sigma_y)$.

The second-to-last step is derived from the properties of the univariate truncated normal distribution, easily found on its wikipedia page. A reference for the first step is here

With $a=0,b=\infty$ we have $$ E(X,Y>0) = \rho_{xy}\sigma_x\frac{\phi(0)}{\Phi(0)}= \rho_{xy}\sigma_x\cdot \frac{2}{\sqrt{2\pi}}= \frac{\sigma_{xy}}{\sigma_y}\cdot \frac{2}{\sqrt{2\pi}} $$ Thanks for pointing in the right direction.

I am posting the complete answer for reference: \begin{eqnarray*} E(X|a<Y<b) &= &\int_a^bE(X|Y=y)f_{Y|a<Y<b}(y)dy \\ & = &\rho_{xy}\frac{\sigma_x}{\sigma_y}\int_a^bf_{Y|a<Y<b}(y)dy \\ & = &\rho_{xy}\frac{\sigma_x}{\sigma_y}\sigma_y\frac{\phi(a-\phi(b)}{\Phi(b)-\Phi(a)} \end{eqnarray*}

With $\rho_{xy}=\frac{E(xy)}{\sigma_x\sigma_y}$. The last step is derived from the properties of the univariate truncated normal distribution, easily found on its wikipedia page. A reference for the first step is here

Thanks for pointing in the right direction.

I am posting the complete answer for reference: \begin{eqnarray*} E(X|a<Y<b) &= &\int_a^bE(X|Y=y)f_{Y|a<Y<b}(y)dy \\ & = &\rho_{xy}\frac{\sigma_x}{\sigma_y}\int_a^bf_{Y|a<Y<b}(y)dy \\ & = &\rho_{xy}\frac{\sigma_x}{\sigma_y}\sigma_y\frac{\phi(a-\phi(b)}{\Phi(b)-\Phi(a)} \end{eqnarray*}

With $\rho_{xy}=\frac{E(xy)}{\sigma_x\sigma_y}$. The last step is derived from the properties of the univariate truncated normal distribution, easily found on its wikipedia page. A reference for the first step is here

With $a=0,b=\infty$ we have $$ E(X,Y>0) = \rho_{xy}\sigma_x\frac{\phi(0)}{\Phi(0)}= \rho_{xy}\sigma_x\cdot \frac{2}{\sqrt{2\pi}}$$ Thanks for pointing in the right direction.