Law of total variance is also known as the variance decomposition formula or Eve’s law. According to Eve’s law if we are M and N which are two random variables and lies in the same probability space and also that ‘N’ has a finite variance.

In such case we define the law of total variance as below:

$Var N$ = $E_M$ ($Var [N | M]$) + $Var_M$ ($E [N | M]$)
Where $E [N | M]$ is nothing but the conditional expected value and its value depends on the value of M.

If we have ‘n’ partitioned space with $B_1, …, B_n$ partitions being mutually exclusive as well as exhaustive, then we have,

$Var [Y]$ = $\sum_{i = 1}^n Var (Y | B_i) P (B_i)$ + $\sum_{i = 1}^n E (Y | B_i)^2 (1 – P (B_i)) P (B_i)$ - 2 $\sum_{i = 1}^n \sum_{j = 1}^{i – 1} E (Y | B_i) P (B_i) E (Y | B_j) P (B_j)$ (This is a special case). Here will discuss about law of variance in detail.

Proof

We make use of law of total expectation to prove the law of total variance. 

From the variance definition we have,

 $Var [N] = E [N^2] – [E [N]]^2$
 
When we apply the law of total expectation to every term by applying conditions on the M that is a random variable, we have,

$Var [N]$ = $E_M [E [N^2 | M]] – [E_M [E [N | M]]]^2$

When we redo the conditional Y in second moment that too in terms of its variance and also the first moment we have,

$\rightarrow$ $Var [N]$ = $E_M [Var [N | M]$ + $[E [N | M]]^2]$ – $[E_M [E [N | M]]]^2$ 

We know that the expectation of any sum is also the sum of expectations, so we can regroup the terms as follows:

$Var [N]$ = $E_M [Var [N | M]] + (E_M [[E [N | M]]^2] – [E_M [E [N | M]]]^2)$

We see that the terms that are in parentheses are equal to the variance of E [N |M], the conditional expectations, so we get,

$\rightarrow$ $Var [N]$ = $E_M [Var [N | M]]$ + $Var_M [E [N | M]]$

Hence the law of total variance is proved.

Examples

Example on law of variance is given below:

Example: Suppose the number of visitors to a library in a given day is N. We know Var [N] and E [N]. Let $Y_i$ represent the number of books the ith number of person is reading. Taking assumption that all $Y_i$’s are independent of N and also of each other, also assuming that they have same variance and same mean as given below

$E Y_i$ = $E\ Y$.

And  $Var (Y_i)$ = $Var (Y)$

If S be the total sales given by S = $sum_(i = 1)^N Y_i$, find the values for E S and  Var (S).

Solution:

$E S$ = $E [E [S | N]]$

= $E [E [sum_(i = 1)^N (Y_i | N)] ]$

= $E [sum_(i = 1)^N E [Y_i | N]]$

= $E [sum_(i = 1)^N E [Y_i]]$

= $E [N E [Y]]$

= $E [Y] E [N]$

$Var (S)$ = $E (Var (S | N)) + Var (E [S | N])$

= $E (Var (S | N)) + Var (N E Y)$

= $E (Var (S | N)) + (E Y)^2 Var (N)$

Now $Var (S | N)$ = $\sum_(i = 1)^N\ Var (Y_i | N)$ = $\sum_(i = 1)^N\ Var (Y_i)$ = $N\ Var (Y)$

So we get,

$E (Var (S | N))$ = $E (N\ Var (Y))$ = $E N\ Var (Y)$

So we get

$Var (S)$ = $E N\ Var (Y)$ + $(E Y)^2\ Var (N)$