In probability theory and statistics, a conditional variance could be the variance of a conditional probability submission. That is, it's the variance of a random variable given the value(s) of one or more other variables. Specially in econometrics, the conditional variance is generally known as the scedastic function or skedastic function
For example, when Y = Height and X
= sex for persons in a very certain population, then Var(height | sex) could be the variable which assigns to everyone in the population this variance of height to the person's sex.

The conditional variance of X given Y = yj is given by

V(X/Y = y$_{j}$) = E[{X - E(X|Y = y$_{j}$)$^{2}$}|Y = y$_{j}]$
In a similar manner conditional variance of Y given X = x$_{i}$ can also be defined.

For a continuous case,

The conditional variance of X may be defined as

V(X|Y = y) = E[{X - E(X/Y = y)$^{2}$}|Y = y]
Similarly, V(Y/X = x) = E[{Y - E(Y/X = x)$^{2}$}|X = x]

The variance of X can be regarded as consisting of two parts, the expectation of the conditional variance and the variance of the conditional expectation.

Symbolically, V(X) = E[V(X|Y)] + V[E(X|Y)]

Proof: E[V(X|Y)] + V[E(X|Y)]

 =E[E(X$^{2}$|Y)-{E(X|Y)}$^{2}$] + E[{E(X|Y)}$^{2}$] - [E{E(X|Y)}]$^{2}$

= E{E(X$^{2}$|Y)}-E{E(X|Y)$^{2}$} + E{E(X|Y)}$^{2}$ - [E{E(X|Y)}]$^{2}$

= E{E(X$^{2}$|Y)} - {E(X)}$^{2}$

= E{$\sum_{i}x_{i}^{2}$P(X = x$_{i}$|Y = y$_{i}$)} - {E(X)}$^{2}$

= E{$\sum_{i}$x$_{i}^{2}$ $\frac{P(X = x_{i} \cap Y = y_{i})}{P(Y= y_{i})}$} - {E(X)}$^{2}$

= $\sum_{j}$[{$\sum_{i}$x$_{i}^{2}$ $\frac{P(X = x_{i}\cap Y = y_{i})}{P(Y=y_{i})}$} P(Y = y$_{i}$)] - {E(X)}$^{2}$

= $\sum_{i}${x$_{i}^{2}$ $\sum_{j}$P(X = x$_{i}$ $\cap$ Y = y$_{j}$)} - [E(X)]$^{2}$

= $\sum_{i}$x$_{i}^{2}$P(X = x$_{i}$) - [E(x)]$^{2}$

= E(X$^{2}$) - {E(X)}$^{2}$

= Var (X)
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 (Y|X = - 1)

Solution: V(Y|X = -1) = E(Y|X = -1)$^{2}$ - {E(Y|X = -1)}$^{2}$

E(Y|X = - 1) = $\sum_{y}$yP(Y = y|X = -1)

= (-1)0 + 0(0.2)+ 1(0)

= 0

E(Y|X  = - 1)$^{2}$ = $\sum_{y}y^{2}P(Y=y|X = -1)$

= 1(0)+ 0(0.2)+ 0

= 0

V(Y|X = -1) = 0

Example 2: Let f(x,y) = 8xy, 0  < x < y < 1;  f(x,y) = 0 elsewhere. Find
E(Y|X=x)
E(XY|X = x)
Var(Y|X = x)

Solution:

f$_{X}(x)$ = $\int_{-\infty}^{\infty}$ f(x,y)dy

= 8x $\int_{x}^{1}$y dy

= 4x (1-x$^{2}$), 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$_{X|Y}$(x|y) = $\frac{f(x,y)}{f_{Y}y}$

= $\frac{2x}{y^{2}}$,

f$_{Y|X}$(y|x)= $\frac{2y}{1-x^{2}}$, 0<x<y<1

E(Y|X = x) = $\int_{x}^{1}$ y($\frac{2y}{1-x^{2}})$dy

= $\frac{2}{3}(\frac{1 - x^{3}}{1 - x^{2}})$

= $\frac{2}{3}(\frac{1+x+x^{2}}{1+x})$

E(XY|X =x) = xE(Y|X = 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

= $\frac{1}{2}(\frac{1-x^{4}}{1 - x^{2}})$

= $\frac{1 + x^{2}}{2}$

Var(Y|X =x) = E(Y$^{2}$|X = x)- {E(Y|X = x)}$^{2}$

= $\frac{1 + x^{2}}{2}$ - $\frac{4}{9}$ $\frac{(1+x+x^{2})^{2}}{(1+x)^{2}})$