Editing Algebra of random variables (section)

== Variance algebra for random variables ==
The variance <math>\operatorname{Var}[Z]</math> of the random variable <math>Z</math> resulting from an algebraic operation between random variables can be calculated using the following set of rules:

* [[Addition]]: <math display="block">\operatorname{Var}[Z] = \operatorname{Var}[X+Y] = \operatorname{Var}[X] + 2 \operatorname{Cov}[X,Y] + \operatorname{Var}[Y].</math>Particularly, if <math>X</math> and <math>Y</math> are [[Independence (probability theory)|independent]] from each other, then: <math display="block">\operatorname{Var}[X+Y] = \operatorname{Var}[X] + \operatorname{Var}[Y].</math>
* [[Subtraction]]: <math display="block">\operatorname{Var}[Z] = \operatorname{Var}[X-Y] = \operatorname{Var}[X] - 2 \operatorname{Cov}[X,Y] + \operatorname{Var}[Y].</math>Particularly, if <math>X</math> and <math>Y</math> are independent from each other, then: <math display="block">\operatorname{Var}[X-Y] = \operatorname{Var}[X] + \operatorname{Var}[Y].</math> That is, for [[independent random variables]] the variance is the same for additions and subtractions: <math display="block">\operatorname{Var}[X+Y] = \operatorname{Var}[X-Y] = \operatorname{Var}[Y-X] = \operatorname{Var}[-X-Y].</math>
* [[Multiplication]]: <math display="block">\operatorname{Var}[Z] = \operatorname{Var}[XY] = \operatorname{Var}[YX].</math> Particularly, if <math>X</math> and <math>Y</math> are independent from each other, then: <math display="block">\begin{align}
\operatorname{Var}[XY] &= \operatorname{E}[X^2] \cdot \operatorname{E}[Y^2] - {\left(\operatorname{E}[X] \cdot \operatorname{E}[Y]\right)}^2 \\[2pt]
&= \operatorname{Var}[X] \cdot \operatorname{Var}[Y] + \operatorname{Var}[X] \cdot {\left(\operatorname{E}[Y]\right)}^2 + \operatorname{Var}[Y] \cdot {\left(\operatorname{E}[X]\right)}^2.
\end{align}</math>
* [[Division (mathematics)|Division]]: <math display="block">\operatorname{Var}[Z] = \operatorname{Var}[X/Y]
= \operatorname{Var}[X \cdot (1/Y)]
= \operatorname{Var}[(1/Y) \cdot X].</math> Particularly, if <math>X</math> and <math>Y</math> are independent from each other, then: <math display="block">\begin{align}
\operatorname{Var}[X/Y]
&= \operatorname{E}[X^2] \cdot \operatorname{E}[1/Y^2] - {\left(\operatorname{E}[X] \cdot \operatorname{E}[1/Y]\right)}^2 \\[2pt]
&= \operatorname{Var}[X] \cdot \operatorname{Var}[1/Y] + \operatorname{Var}[X] \cdot {\left(\operatorname{E}[1/Y]\right)}^2 + \operatorname{Var}[1/Y] \cdot {\left(\operatorname{E}[X]\right)}^2.
\end{align}</math>
* [[Exponentiation]]: <math display="block">\operatorname{Var}[Z] = \operatorname{Var}[X^Y] = \operatorname{Var}[e^{Y\ln(X)}]</math>

where <math>\operatorname{Cov}[X,Y] = \operatorname{Cov}[Y,X]</math> represents the covariance operator between random variables <math>X</math> and <math>Y</math>.

The variance of a random variable can also be expressed directly in terms of the covariance or in terms of the expected value:

<math display="block">\operatorname{Var}[X] = \operatorname{Cov}(X,X) = \operatorname{E}[X^2] - \operatorname{E}[X]^2</math>

If any of the random variables is replaced by a deterministic variable or by a constant value (<math>k</math>), the previous properties remain valid considering that <math>\Pr(X = k) = 1</math> and <math>\operatorname{E}[X] = k</math>, <math>\operatorname{Var}[X] = 0</math> and <math>\operatorname{Cov}[Y,k] = 0</math>. Special cases are the addition and multiplication of a random variable with a deterministic variable or a constant, where:

* <math>\operatorname{Var}[k+Y] = \operatorname{Var}[Y]</math>
* <math>\operatorname{Var}[kY] = k^2 \operatorname{Var}[Y]</math>

If <math>Z</math> is defined as a general non-linear algebraic function <math>f</math> of a random variable <math>X</math>, then:

<math display="block">\operatorname{Var}[Z] = \operatorname{Var}[f(X)] \neq f(\operatorname{Var}[X])</math>

The exact value of the variance of the non-linear function will depend on the particular probability distribution of the random variable <math>X</math>.