Suppose $x=y+\epsilon$, where $y$ and $\epsilon$ are independent and normally distributed and $E[\epsilon]=0$. I know how to derive $E[y|x]$: it is simply $\frac{\frac{1}{Var(y)}}{\frac{1}{Var(y)}+\frac{1}{Var(x)}} \cdot E[y] + \frac{\frac{1}{Var(x)}}{\frac{1}{Var(y)}+\frac{1}{Var(x)}} \cdot x$.
But what is $E[y|x, y<z]$ (where $z$ is some value)? I am not sure how to approach it and was thinking of calculating the conditional distribution of $y|x$ and then truncate it at $z$ and find the mean of this truncated distribution. But I am not sure if this is conceptually correct.
Thank you!