Some examples on conditional variance are given below:
Example 1: Let X and Y be two random variables each taking three values 1,0, and 1, and having the joint probability distribution
Y X

1

0

1

Total 
1 
0 
0.1 
0.1 
0.2 
0 
0.2 
0.2 
0.2 
0.6 
1 
0 
0.1 
0.1 
0.2 
Total 
0.2 
.4 
0.4 
1.0

a) Find Var (YX =  1)
Solution: V(YX = 1) = E(YX = 1)$^{2}$  {E(YX = 1)}$^{2}$
E(YX =  1) = $\sum_{y}$yP(Y = yX = 1)
= (1)0 + 0(0.2)+ 1(0)
= 0
E(YX =  1)$^{2}$ = $\sum_{y}y^{2}P(Y=yX = 1)$
= 1(0)+ 0(0.2)+ 0
= 0
V(YX = 1) = 0
Example 2: Let f(x,y) = 8xy, 0 < x < y < 1; f(x,y) = 0 elsewhere. Find
E(YX=x)
E(XYX = x)
Var(YX = x)
Solution: f$_{X}(x)$ = $\int_{\infty}^{\infty}$ f(x,y)dy
= 8x $\int_{x}^{1}$y dy
= 4x (1x$^{2}$), 0 < x < 1
f$_{Y}(y)$ = $\int_{\infty}^{\infty}$ f(x,y)dy
= 8y $\int_{0}^{y}$x dx
= 4y$^{3}$, 0 < y < 1
f$_{XY}$(xy) =
$\frac{f(x,y)}{f_{Y}y}$=
$\frac{2x}{y^{2}}$,
f$_{YX}$(yx)=
$\frac{2y}{1x^{2}}$, 0<x<y<1
E(YX = x) = $\int_{x}^{1}$ y
($\frac{2y}{1x^{2}})$dy
=
$\frac{2}{3}(\frac{1  x^{3}}{1  x^{2}})$=
$\frac{2}{3}(\frac{1+x+x^{2}}{1+x})$E(XYX =x) = xE(YX = x) =
$\frac{2}{3}$ *
$\frac{x(1+x+x^{2})}{1+x}$E(Y$^{2}$X=x)= $\int_{x}^{1}y^{2}$
$\frac{2y}{1  x^{2}}$ dy
=
$\frac{1}{2}(\frac{1x^{4}}{1  x^{2}})$=
$\frac{1 + x^{2}}{2}$Var(YX =x) = E(Y$^{2}$X = x) {E(YX = x)}$^{2}$
=
$\frac{1 + x^{2}}{2}$ 
$\frac{4}{9}$ $\frac{(1+x+x^{2})^{2}}{(1+x)^{2}})$