Let (X,Y)(X,Y)(X,Y) be a bivariate random variable then conditional expectation of g(X,Y)g(X,Y)g(X,Y) given Y=yY=yY=y is denoted by E(g(X,Y)/Y=y)E(g(X,Y)/Y=y)E(g(X,Y)/Y=y) and defined as (provided sum/integral converges absolutely)
E(g(X,Y)/Y=y)={∑x∈SXg(x,y)pX/Y(x/y)if (X,Y) is of discrete type and pY(y)>0∫−∞∞g(x,y)fX/Y(x/y)dxif (X,Y) is of continuous type and fY(y)>0E(g(X,Y)/Y=y)=\left\{\begin{matrix} \underset{x\in S_X}{\sum}g(x,y)p_{X/Y}(x/y)& \text{if } (X,Y) \text{ is of discrete type and } p_Y(y)>0 \\ \int_{-\infty}^{\infty}g(x,y)f_{X/Y}(x/y)dx& \text{if } (X,Y) \text{ is of continuous type and } f_Y(y)>0 \end{matrix}\right.E(g(X,Y)/Y=y)={x∈SX∑g(x,y)pX/Y(x/y)∫−∞∞g(x,y)fX/Y(x/y)dxif (X,Y) is of discrete type and pY(y)>0if (X,Y) is of continuous type and fY(y)>0
similarly you can define E(g(X,Y)/X=x)E(g(X,Y)/X=x)E(g(X,Y)/X=x)