Variance

If the generator of random variable ${\displaystyle X}$ is discrete with probability mass function ${\displaystyle x_{1}\mapsto p_{1},x_{2}\mapsto p_{2},\ldots ,x_{n}\mapsto p_{n}}$, then

${\displaystyle \operatorname {Var} (X)=\sum _{i=1}^{n}p_{i}\cdot (x_{i}-\mu )^{2},}$

or equivalently,

${\displaystyle \operatorname {Var} (X)=\left(\sum _{i=1}^{n}p_{i}x_{i}^{2}\right)-\mu ^{2},}$

where ${\displaystyle \mu }$ is the expected value. That is,

${\displaystyle \mu =\sum _{i=1}^{n}p_{i}x_{i}.}$

(When such a discrete weighted variance is specified by weights whose sum is not 1, then one divides by the sum of the weights.)

The variance of a collection of ${\displaystyle n}$ equally likely values can be written as

${\displaystyle \operatorname {Var} (X)={\frac {1}{n}}\sum _{i=1}^{n}(x_{i}-\mu )^{2}=\left({\frac {1}{n}}\sum _{i=1}^{n}x_{i}^{2}\right)-\mu ^{2},}$

where ${\displaystyle \mu }$ is the average value. That is,

${\displaystyle \mu ={\frac {1}{n}}\sum _{i=1}^{n}x_{i}.}$

The variance of a set of ${\displaystyle n}$ equally likely values can be equivalently expressed, without directly referring to the mean, in terms of squared deviations of all points from each other:[1]

${\displaystyle \operatorname {Var} (X)={\frac {1}{n^{2}}}\sum _{i=1}^{n}\sum _{j=1}^{n}{\frac {1}{2}}(x_{i}-x_{j})^{2}={\frac {1}{n^{2}}}\sum _{i}\sum _{j>i}(x_{i}-x_{j})^{2}.}$

If the random variable ${\displaystyle X}$ has a probability density function ${\displaystyle f(x)}$, and ${\displaystyle F(x)}$ is the corresponding cumulative distribution function, then

${\displaystyle {\begin{aligned}\operatorname {Var} (X)=\sigma ^{2}&=\int _{\mathbb {R} }(x-\mu )^{2}f(x)\,dx\\[4pt]&=\int _{\mathbb {R} }x^{2}f(x)\,dx-2\mu \int _{\mathbb {R} }xf(x)\,dx+\mu ^{2}\int _{\mathbb {R} }f(x)\,dx\\[4pt]&=\int _{\mathbb {R} }x^{2}\,dF(x)-2\mu \int _{\mathbb {R} }x\,dF(x)+\mu ^{2}\int _{\mathbb {R} }\,dF(x)\\[4pt]&=\int _{\mathbb {R} }x^{2}\,dF(x)-2\mu \cdot \mu +\mu ^{2}\cdot 1\\[4pt]&=\int _{\mathbb {R} }x^{2}\,dF(x)-\mu ^{2},\end{aligned}}}$

or equivalently,

${\displaystyle \operatorname {Var} (X)=\int _{\mathbb {R} }x^{2}f(x)\,dx-\mu ^{2},}$

where ${\displaystyle \mu }$ is the expected value of ${\displaystyle X}$ given by

${\displaystyle \mu =\int _{\mathbb {R} }xf(x)\,dx=\int _{\mathbb {R} }x\,dF(x).}$

In these formulas, the integrals with respect to ${\displaystyle dx}$ and ${\displaystyle dF(x)}$ are Lebesgue and Lebesgue–Stieltjes integrals, respectively.

If the function ${\displaystyle x^{2}f(x)}$ is Riemann-integrable on every finite interval ${\displaystyle [a,b]\subset \mathbb {R} ,}$ then

${\displaystyle \operatorname {Var} (X)=\int _{-\infty }^{+\infty }x^{2}f(x)\,dx-\mu ^{2},}$

where the integral is an improper Riemann integral.

The exponential distribution with parameter λ is a continuous distribution whose probability density function is given by

${\displaystyle f(x)=\lambda e^{-\lambda x}}$

on the interval [0, ∞). Its mean can be shown to be

${\displaystyle \operatorname {E} [X]=\int _{0}^{\infty }\lambda xe^{-\lambda x}\,dx={\frac {1}{\lambda }}.}$

Using integration by parts and making use of the expected value already calculated, we have:

${\displaystyle {\begin{aligned}\operatorname {E} \left[X^{2}\right]&=\int _{0}^{\infty }\lambda x^{2}e^{-\lambda x}\,dx\\&=\left[-x^{2}e^{-\lambda x}\right]_{0}^{\infty }+\int _{0}^{\infty }2xe^{-\lambda x}\,dx\\&=0+{\frac {2}{\lambda }}\operatorname {E} [X]\\&={\frac {2}{\lambda ^{2}}}.\end{aligned}}}$

Thus, the variance of X is given by

${\displaystyle \operatorname {Var} (X)=\operatorname {E} \left[X^{2}\right]-\operatorname {E} [X]^{2}={\frac {2}{\lambda ^{2}}}-\left({\frac {1}{\lambda }}\right)^{2}={\frac {1}{\lambda ^{2}}}.}$

A fair six-sided die can be modeled as a discrete random variable, X, with outcomes 1 through 6, each with equal probability 1/6. The expected value of X is ${\displaystyle (1+2+3+4+5+6)/6=7/2.}$ Therefore, the variance of X is

${\displaystyle {\begin{aligned}\operatorname {Var} (X)&=\sum _{i=1}^{6}{\frac {1}{6}}\left(i-{\frac {7}{2}}\right)^{2}\\[5pt]&={\frac {1}{6}}\left((-5/2)^{2}+(-3/2)^{2}+(-1/2)^{2}+(1/2)^{2}+(3/2)^{2}+(5/2)^{2}\right)\\[5pt]&={\frac {35}{12}}\approx 2.92.\end{aligned}}}$

The general formula for the variance of the outcome, X, of an n-sided die is

${\displaystyle {\begin{aligned}\operatorname {Var} (X)&=\operatorname {E} \left(X^{2}\right)-(\operatorname {E} (X))^{2}\\[5pt]&={\frac {1}{n}}\sum _{i=1}^{n}i^{2}-\left({\frac {1}{n}}\sum _{i=1}^{n}i\right)^{2}\\[5pt]&={\frac {(n+1)(2n+1)}{6}}-\left({\frac {n+1}{2}}\right)^{2}\\[4pt]&={\frac {n^{2}-1}{12}}.\end{aligned}}}$

The following table lists the variance for some commonly used probability distributions.

Name of the probability distribution	Probability distribution function	Mean	Variance
Binomial distribution	${\displaystyle \Pr \,(X=k)={\binom {n}{k}}p^{k}(1-p)^{n-k}}$	${\displaystyle np}$	${\displaystyle np(1-p)}$
Geometric distribution	${\displaystyle \Pr \,(X=k)=(1-p)^{k-1}p}$	${\displaystyle {\frac {1}{p}}}$	${\displaystyle {\frac {(1-p)}{p^{2}}}}$
Normal distribution	${\displaystyle f\left(x\mid \mu ,\sigma ^{2}\right)={\frac {1}{\sqrt {2\pi \sigma ^{2}}}}e^{-{\frac {(x-\mu )^{2}}{2\sigma ^{2}}}}}$	${\displaystyle \mu }$	${\displaystyle \sigma ^{2}}$
Uniform distribution (continuous)	${\displaystyle f(x\mid a,b)={\begin{cases}{\frac {1}{b-a}}&{\text{for }}a\leq x\leq b,\\[3pt]0&{\text{for }}x<a{\text{ or }}x>b\end{cases}}}$	${\displaystyle {\frac {a+b}{2}}}$	${\displaystyle {\frac {(b-a)^{2}}{12}}}$
Exponential distribution	${\displaystyle f(x\mid \lambda )=\lambda e^{-\lambda x}}$	${\displaystyle {\frac {1}{\lambda }}}$	${\displaystyle {\frac {1}{\lambda ^{2}}}}$
Poisson distribution	${\displaystyle f(x\mid \lambda )={\frac {e^{-\lambda }\lambda ^{x}}{x!}}}$	${\displaystyle \lambda }$	${\displaystyle \lambda }$

Variance is non-negative because the squares are positive or zero:

${\displaystyle \operatorname {Var} (X)\geq 0.}$

The variance of a constant is zero.

${\displaystyle \operatorname {Var} (a)=0.}$

Conversely, if the variance of a random variable is 0, then it is almost surely a constant. That is, it always has the same value:

${\displaystyle \operatorname {Var} (X)=0\iff \exists a:P(X=a)=1.}$

Variance is invariant with respect to changes in a location parameter. That is, if a constant is added to all values of the variable, the variance is unchanged:

${\displaystyle \operatorname {Var} (X+a)=\operatorname {Var} (X).}$

If all values are scaled by a constant, the variance is scaled by the square of that constant:

${\displaystyle \operatorname {Var} (aX)=a^{2}\operatorname {Var} (X).}$

The variance of a sum of two random variables is given by

${\displaystyle \operatorname {Var} (aX+bY)=a^{2}\operatorname {Var} (X)+b^{2}\operatorname {Var} (Y)+2ab\,\operatorname {Cov} (X,Y),}$

${\displaystyle \operatorname {Var} (aX-bY)=a^{2}\operatorname {Var} (X)+b^{2}\operatorname {Var} (Y)-2ab\,\operatorname {Cov} (X,Y),}$

where ${\displaystyle \operatorname {Cov} (X,Y)}$ is the covariance.

In general, for the sum of ${\displaystyle N}$ random variables ${\displaystyle \{X_{1},\dots ,X_{N}\}}$, the variance becomes:

${\displaystyle \operatorname {Var} \left(\sum _{i=1}^{N}X_{i}\right)=\sum _{i,j=1}^{N}\operatorname {Cov} (X_{i},X_{j})=\sum _{i=1}^{N}\operatorname {Var} (X_{i})+\sum _{i\neq j}\operatorname {Cov} (X_{i},X_{j}).}$

These results lead to the variance of a linear combination as:

${\displaystyle {\begin{aligned}\operatorname {Var} \left(\sum _{i=1}^{N}a_{i}X_{i}\right)&=\sum _{i,j=1}^{N}a_{i}a_{j}\operatorname {Cov} (X_{i},X_{j})\\&=\sum _{i=1}^{N}a_{i}^{2}\operatorname {Var} (X_{i})+\sum _{i\not =j}a_{i}a_{j}\operatorname {Cov} (X_{i},X_{j})\\&=\sum _{i=1}^{N}a_{i}^{2}\operatorname {Var} (X_{i})+2\sum _{1\leq i<j\leq N}a_{i}a_{j}\operatorname {Cov} (X_{i},X_{j}).\end{aligned}}}$

If the random variables ${\displaystyle X_{1},\dots ,X_{N}}$ are such that

${\displaystyle \operatorname {Cov} (X_{i},X_{j})=0\ ,\ \forall \ (i\neq j),}$

then they are said to be uncorrelated. It follows immediately from the expression given earlier that if the random variables ${\displaystyle X_{1},\dots ,X_{N}}$ are uncorrelated, then the variance of their sum is equal to the sum of their variances, or, expressed symbolically:

${\displaystyle \operatorname {Var} \left(\sum _{i=1}^{N}X_{i}\right)=\sum _{i=1}^{N}\operatorname {Var} (X_{i}).}$

Since independent random variables are always uncorrelated (see Covariance § Uncorrelatedness and independence), the equation above holds in particular when the random variables ${\displaystyle X_{1},\dots ,X_{n}}$ are independent. Thus, independence is sufficient but not necessary for the variance of the sum to equal the sum of the variances.

If a distribution does not have a finite expected value, as is the case for the Cauchy distribution, then the variance cannot be finite either. However, some distributions may not have a finite variance, despite their expected value being finite. An example is a Pareto distribution whose index ${\displaystyle k}$ satisfies ${\displaystyle 1<k\leq 2.}$

One reason for the use of the variance in preference to other measures of dispersion is that the variance of the sum (or the difference) of uncorrelated random variables is the sum of their variances:

${\displaystyle \operatorname {Var} \left(\sum _{i=1}^{n}X_{i}\right)=\sum _{i=1}^{n}\operatorname {Var} (X_{i}).}$

This statement is called the Bienaymé formula[2] and was discovered in 1853.[3][4] It is often made with the stronger condition that the variables are independent, but being uncorrelated suffices. So if all the variables have the same variance σ², then, since division by n is a linear transformation, this formula immediately implies that the variance of their mean is

${\displaystyle \operatorname {Var} \left({\overline {X}}\right)=\operatorname {Var} \left({\frac {1}{n}}\sum _{i=1}^{n}X_{i}\right)={\frac {1}{n^{2}}}\sum _{i=1}^{n}\operatorname {Var} \left(X_{i}\right)={\frac {1}{n^{2}}}n\sigma ^{2}={\frac {\sigma ^{2}}{n}}.}$

That is, the variance of the mean decreases when n increases. This formula for the variance of the mean is used in the definition of the standard error of the sample mean, which is used in the central limit theorem.

To prove the initial statement, it suffices to show that

${\displaystyle \operatorname {Var} (X+Y)=\operatorname {Var} (X)+\operatorname {Var} (Y).}$

The general result then follows by induction. Starting with the definition,

${\displaystyle {\begin{aligned}\operatorname {Var} (X+Y)&=\operatorname {E} \left[(X+Y)^{2}\right]-(\operatorname {E} [X+Y])^{2}\\[5pt]&=\operatorname {E} \left[X^{2}+2XY+Y^{2}\right]-(\operatorname {E} [X]+\operatorname {E} [Y])^{2}.\end{aligned}}}$

Using the linearity of the expectation operator and the assumption of independence (or uncorrelatedness) of X and Y, this further simplifies as follows:

${\displaystyle {\begin{aligned}\operatorname {Var} (X+Y)&=\operatorname {E} \left[X^{2}\right]+2\operatorname {E} [XY]+\operatorname {E} \left[Y^{2}\right]-\left(\operatorname {E} [X]^{2}+2\operatorname {E} [X]\operatorname {E} [Y]+\operatorname {E} [Y]^{2}\right)\\[5pt]&=\operatorname {E} \left[X^{2}\right]+\operatorname {E} \left[Y^{2}\right]-\operatorname {E} [X]^{2}-\operatorname {E} [Y]^{2}\\[5pt]&=\operatorname {Var} (X)+\operatorname {Var} (Y).\end{aligned}}}$

In general, the variance of the sum of n variables is the sum of their covariances:

${\displaystyle \operatorname {Var} \left(\sum _{i=1}^{n}X_{i}\right)=\sum _{i=1}^{n}\sum _{j=1}^{n}\operatorname {Cov} \left(X_{i},X_{j}\right)=\sum _{i=1}^{n}\operatorname {Var} \left(X_{i}\right)+2\sum _{1\leq i<j\leq n}\operatorname {Cov} \left(X_{i},X_{j}\right).}$

(Note: The second equality comes from the fact that Cov(X_i,X_i) = Var(X_i).)

Here, Cov(⋅, ⋅) is the covariance, which is zero for independent random variables (if it exists). The formula states that the variance of a sum is equal to the sum of all elements in the covariance matrix of the components. The next expression states equivalently that the variance of the sum is the sum of the diagonal of covariance matrix plus two times the sum of its upper triangular elements (or its lower triangular elements); this emphasizes that the covariance matrix is symmetric. This formula is used in the theory of Cronbach's alpha in classical test theory.

So if the variables have equal variance σ² and the average correlation of distinct variables is ρ, then the variance of their mean is

${\displaystyle \operatorname {Var} \left({\overline {X}}\right)={\frac {\sigma ^{2}}{n}}+{\frac {n-1}{n}}\rho \sigma ^{2}.}$

This implies that the variance of the mean increases with the average of the correlations. In other words, additional correlated observations are not as effective as additional independent observations at reducing the uncertainty of the mean. Moreover, if the variables have unit variance, for example if they are standardized, then this simplifies to

${\displaystyle \operatorname {Var} \left({\overline {X}}\right)={\frac {1}{n}}+{\frac {n-1}{n}}\rho .}$

This formula is used in the Spearman–Brown prediction formula of classical test theory. This converges to ρ if n goes to infinity, provided that the average correlation remains constant or converges too. So for the variance of the mean of standardized variables with equal correlations or converging average correlation we have

${\displaystyle \lim _{n\to \infty }\operatorname {Var} \left({\overline {X}}\right)=\rho .}$

Therefore, the variance of the mean of a large number of standardized variables is approximately equal to their average correlation. This makes clear that the sample mean of correlated variables does not generally converge to the population mean, even though the law of large numbers states that the sample mean will converge for independent variables.

There are cases when a sample is taken without knowing, in advance, how many observations will be acceptable according to some criterion. In such cases, the sample size N is a random variable whose variation adds to the variation of X, such that,

Var(∑X) = E(N)Var(X) + Var(N)E²(X).[5]

If N has a Poisson distribution, then E(N) = Var(N) with estimator N = n. So, the estimator of Var(∑X) becomes nS²_X + nX² giving

standard error(X) = √[(S²_X + X²)/n].

Define ${\displaystyle X}$ as a column vector of ${\displaystyle n}$ random variables ${\displaystyle X_{1},\ldots ,X_{n}}$, and ${\displaystyle c}$ as a column vector of ${\displaystyle n}$ scalars ${\displaystyle c_{1},\ldots ,c_{n}}$. Therefore, ${\displaystyle c^{\mathsf {T}}X}$ is a linear combination of these random variables, where ${\displaystyle c^{\mathsf {T}}}$ denotes the transpose of ${\displaystyle c}$. Also let ${\displaystyle \Sigma }$ be the covariance matrix of ${\displaystyle X}$. The variance of ${\displaystyle c^{\mathsf {T}}X}$ is then given by:[6]

${\displaystyle \operatorname {Var} \left(c^{\mathsf {T}}X\right)=c^{\mathsf {T}}\Sigma c.}$

This implies that the variance of the mean can be written as (with a column vector of ones)

${\displaystyle \operatorname {Var} \left({\bar {x}}\right)=\operatorname {Var} \left({\frac {1}{n}}1'X\right)={\frac {1}{n^{2}}}1'\Sigma 1.}$

The scaling property and the Bienaymé formula, along with the property of the covariance Cov(aX, bY) = ab Cov(X, Y) jointly imply that

${\displaystyle \operatorname {Var} (aX\pm bY)=a^{2}\operatorname {Var} (X)+b^{2}\operatorname {Var} (Y)\pm 2ab\,\operatorname {Cov} (X,Y).}$

This implies that in a weighted sum of variables, the variable with the largest weight will have a disproportionally large weight in the variance of the total. For example, if X and Y are uncorrelated and the weight of X is two times the weight of Y, then the weight of the variance of X will be four times the weight of the variance of Y.

The expression above can be extended to a weighted sum of multiple variables:

${\displaystyle \operatorname {Var} \left(\sum _{i}^{n}a_{i}X_{i}\right)=\sum _{i=1}^{n}a_{i}^{2}\operatorname {Var} (X_{i})+2\sum _{1\leq i}\sum _{<j\leq n}a_{i}a_{j}\operatorname {Cov} (X_{i},X_{j})}$

If two variables X and Y are independent, the variance of their product is given by[7]

${\displaystyle \operatorname {Var} (XY)=[\operatorname {E} (X)]^{2}\operatorname {Var} (Y)+[\operatorname {E} (Y)]^{2}\operatorname {Var} (X)+\operatorname {Var} (X)\operatorname {Var} (Y).}$

Equivalently, using the basic properties of expectation, it is given by

${\displaystyle \operatorname {Var} (XY)=\operatorname {E} \left(X^{2}\right)\operatorname {E} \left(Y^{2}\right)-[\operatorname {E} (X)]^{2}[\operatorname {E} (Y)]^{2}.}$

In general, if two variables are statistically dependent, the variance of their product is given by:

${\displaystyle {\begin{aligned}\operatorname {Var} (XY)={}&\operatorname {E} \left[X^{2}Y^{2}\right]-[\operatorname {E} (XY)]^{2}\\[5pt]={}&\operatorname {Cov} \left(X^{2},Y^{2}\right)+\operatorname {E} (X^{2})\operatorname {E} \left(Y^{2}\right)-[\operatorname {E} (XY)]^{2}\\[5pt]={}&\operatorname {Cov} \left(X^{2},Y^{2}\right)+\left(\operatorname {Var} (X)+[\operatorname {E} (X)]^{2}\right)\left(\operatorname {Var} (Y)+[\operatorname {E} (Y)]^{2}\right)\\[5pt]&-[\operatorname {Cov} (X,Y)+\operatorname {E} (X)\operatorname {E} (Y)]^{2}\end{aligned}}}$

The general formula for variance decomposition or the law of total variance is: If ${\displaystyle X}$ and ${\displaystyle Y}$ are two random variables, and the variance of ${\displaystyle X}$ exists, then

${\displaystyle \operatorname {Var} [X]=\operatorname {E} (\operatorname {Var} [X\mid Y])+\operatorname {Var} (\operatorname {E} [X\mid Y]).}$

The conditional expectation ${\displaystyle \operatorname {E} (X\mid Y)}$ of ${\displaystyle X}$ given ${\displaystyle Y}$, and the conditional variance ${\displaystyle \operatorname {Var} (X\mid Y)}$ may be understood as follows. Given any particular value y of the random variable Y, there is a conditional expectation ${\displaystyle \operatorname {E} (X\mid Y=y)}$ given the event Y = y. This quantity depends on the particular value y; it is a function ${\displaystyle g(y)=\operatorname {E} (X\mid Y=y)}$. That same function evaluated at the random variable Y is the conditional expectation ${\displaystyle \operatorname {E} (X\mid Y)=g(Y).}$

In particular, if ${\displaystyle Y}$ is a discrete random variable assuming possible values ${\displaystyle y_{1},y_{2},y_{3}\ldots }$ with corresponding probabilities ${\displaystyle p_{1},p_{2},p_{3}\ldots ,}$, then in the formula for total variance, the first term on the right-hand side becomes

${\displaystyle \operatorname {E} (\operatorname {Var} [X\mid Y])=\sum _{i}p_{i}\sigma _{i}^{2},}$

where ${\displaystyle \sigma _{i}^{2}=\operatorname {Var} [X\mid Y=y_{i}]}$. Similarly, the second term on the right-hand side becomes

${\displaystyle \operatorname {Var} (\operatorname {E} [X\mid Y])=\sum _{i}p_{i}\mu _{i}^{2}-\left(\sum _{i}p_{i}\mu _{i}\right)^{2}=\sum _{i}p_{i}\mu _{i}^{2}-\mu ^{2},}$

where ${\displaystyle \mu _{i}=\operatorname {E} [X\mid Y=y_{i}]}$ and ${\displaystyle \mu =\sum _{i}p_{i}\mu _{i}}$. Thus the total variance is given by

${\displaystyle \operatorname {Var} [X]=\sum _{i}p_{i}\sigma _{i}^{2}+\left(\sum _{i}p_{i}\mu _{i}^{2}-\mu ^{2}\right).}$

A similar formula is applied in analysis of variance, where the corresponding formula is

${\displaystyle {\mathit {MS}}_{\text{total}}={\mathit {MS}}_{\text{between}}+{\mathit {MS}}_{\text{within}};}$

here ${\displaystyle {\mathit {MS}}}$ refers to the Mean of the Squares. In linear regression analysis the corresponding formula is

${\displaystyle {\mathit {MS}}_{\text{total}}={\mathit {MS}}_{\text{regression}}+{\mathit {MS}}_{\text{residual}}.}$

This can also be derived from the additivity of variances, since the total (observed) score is the sum of the predicted score and the error score, where the latter two are uncorrelated.

Similar decompositions are possible for the sum of squared deviations (sum of squares, ${\displaystyle {\mathit {SS}}}$):

${\displaystyle {\mathit {SS}}_{\text{total}}={\mathit {SS}}_{\text{between}}+{\mathit {SS}}_{\text{within}},}$

${\displaystyle {\mathit {SS}}_{\text{total}}={\mathit {SS}}_{\text{regression}}+{\mathit {SS}}_{\text{residual}}.}$

The population variance for a non-negative random variable can be expressed in terms of the cumulative distribution function F using

${\displaystyle 2\int _{0}^{\infty }u(1-F(u))\,du-\left(\int _{0}^{\infty }(1-F(u))\,du\right)^{2}.}$

This expression can be used to calculate the variance in situations where the CDF, but not the density, can be conveniently expressed.

The second moment of a random variable attains the minimum value when taken around the first moment (i.e., mean) of the random variable, i.e. ${\displaystyle \mathrm {argmin} _{m}\,\mathrm {E} \left(\left(X-m\right)^{2}\right)=\mathrm {E} (X)}$. Conversely, if a continuous function ${\displaystyle \varphi }$ satisfies ${\displaystyle \mathrm {argmin} _{m}\,\mathrm {E} (\varphi (X-m))=\mathrm {E} (X)}$ for all random variables X, then it is necessarily of the form ${\displaystyle \varphi (x)=ax^{2}+b}$, where a > 0. This also holds in the multidimensional case.[8]

Unlike expected absolute deviation, the variance of a variable has units that are the square of the units of the variable itself. For example, a variable measured in meters will have a variance measured in meters squared. For this reason, describing data sets via their standard deviation or root mean square deviation is often preferred over using the variance. In the dice example the standard deviation is √2.9 ≈ 1.7, slightly larger than the expected absolute deviation of 1.5.

The standard deviation and the expected absolute deviation can both be used as an indicator of the "spread" of a distribution. The standard deviation is more amenable to algebraic manipulation than the expected absolute deviation, and, together with variance and its generalization covariance, is used frequently in theoretical statistics; however the expected absolute deviation tends to be more robust as it is less sensitive to outliers arising from measurement anomalies or an unduly heavy-tailed distribution.

In general, the population variance of a finite population of size N with values x_i is given by

$${\displaystyle {\begin{aligned}\sigma ^{2}&={\frac {1}{N}}\sum _{i=1}^{N}\left(x_{i}-\mu \right)^{2}={\frac {1}{N}}\sum _{i=1}^{N}\left(x_{i}^{2}-2\mu x_{i}+\mu ^{2}\right)\\[5pt]&=\left({\frac {1}{N}}\sum _{i=1}^{N}x_{i}^{2}\right)-2\mu \left({\frac {1}{N}}\sum _{i=1}^{N}x_{i}\right)+\mu ^{2}\\[5pt]&=\left({\frac {1}{N}}\sum _{i=1}^{N}x_{i}^{2}\right)-\mu ^{2}\end{aligned}}}$$

where the population mean is

${\displaystyle \mu ={\frac {1}{N}}\sum _{i=1}^{N}x_{i}.}$

The population variance can also be computed using

${\displaystyle \sigma ^{2}={\frac {1}{N^{2}}}\sum _{i<j}\left(x_{i}-x_{j}\right)^{2}={\frac {1}{2N^{2}}}\sum _{i,j=1}^{N}\left(x_{i}-x_{j}\right)^{2}.}$

This is true because

$${\displaystyle {\begin{aligned}&{\frac {1}{2N^{2}}}\sum _{i,j=1}^{N}\left(x_{i}-x_{j}\right)^{2}\\[5pt]={}&{\frac {1}{2N^{2}}}\sum _{i,j=1}^{N}\left(x_{i}^{2}-2x_{i}x_{j}+x_{j}^{2}\right)\\[5pt]={}&{\frac {1}{2N}}\sum _{j=1}^{N}\left({\frac {1}{N}}\sum _{i=1}^{N}x_{i}^{2}\right)-\left({\frac {1}{N}}\sum _{i=1}^{N}x_{i}\right)\left({\frac {1}{N}}\sum _{j=1}^{N}x_{j}\right)+{\frac {1}{2N}}\sum _{i=1}^{N}\left({\frac {1}{N}}\sum _{j=1}^{N}x_{j}^{2}\right)\\[5pt]={}&{\frac {1}{2}}\left(\sigma ^{2}+\mu ^{2}\right)-\mu ^{2}+{\frac {1}{2}}\left(\sigma ^{2}+\mu ^{2}\right)\\[5pt]={}&\sigma ^{2}\end{aligned}}}$$

The population variance matches the variance of the generating probability distribution. In this sense, the concept of population can be extended to continuous random variables with infinite populations.

In many practical situations, the true variance of a population is not known a priori and must be computed somehow. When dealing with extremely large populations, it is not possible to count every object in the population, so the computation must be performed on a sample of the population.[9] Sample variance can also be applied to the estimation of the variance of a continuous distribution from a sample of that distribution.

We take a sample with replacement of n values Y₁, ..., Y_n from the population, where n < N, and estimate the variance on the basis of this sample.[10] Directly taking the variance of the sample data gives the average of the squared deviations:

${\displaystyle \sigma _{Y}^{2}={\frac {1}{n}}\sum _{i=1}^{n}\left(Y_{i}-{\overline {Y}}\right)^{2}=\left({\frac {1}{n}}\sum _{i=1}^{n}Y_{i}^{2}\right)-{\overline {Y}}^{2}={\frac {1}{n^{2}}}\sum _{i,j\,:\,i<j}\left(Y_{i}-Y_{j}\right)^{2}.}$

Here, ${\displaystyle {\overline {Y}}}$ denotes the sample mean:

${\displaystyle {\overline {Y}}={\frac {1}{n}}\sum _{i=1}^{n}Y_{i}.}$

Since the Y_i are selected randomly, both ${\displaystyle {\overline {Y}}}$ and ${\displaystyle \sigma _{Y}^{2}}$ are random variables. Their expected values can be evaluated by averaging over the ensemble of all possible samples {Y_i} of size n from the population. For ${\displaystyle \sigma _{Y}^{2}}$ this gives:

${\displaystyle {\begin{aligned}\operatorname {E} [\sigma _{Y}^{2}]&=\operatorname {E} \left[{\frac {1}{n}}\sum _{i=1}^{n}\left(Y_{i}-{\frac {1}{n}}\sum _{j=1}^{n}Y_{j}\right)^{2}\right]\\[5pt]&={\frac {1}{n}}\sum _{i=1}^{n}\operatorname {E} \left[Y_{i}^{2}-{\frac {2}{n}}Y_{i}\sum _{j=1}^{n}Y_{j}+{\frac {1}{n^{2}}}\sum _{j=1}^{n}Y_{j}\sum _{k=1}^{n}Y_{k}\right]\\[5pt]&={\frac {1}{n}}\sum _{i=1}^{n}\left[{\frac {n-2}{n}}\operatorname {E} \left[Y_{i}^{2}\right]-{\frac {2}{n}}\sum _{j\neq i}\operatorname {E} \left[Y_{i}Y_{j}\right]+{\frac {1}{n^{2}}}\sum _{j=1}^{n}\sum _{k\neq j}^{n}\operatorname {E} \left[Y_{j}Y_{k}\right]+{\frac {1}{n^{2}}}\sum _{j=1}^{n}\operatorname {E} \left[Y_{j}^{2}\right]\right]\\[5pt]&={\frac {1}{n}}\sum _{i=1}^{n}\left[{\frac {n-2}{n}}\left(\sigma ^{2}+\mu ^{2}\right)-{\frac {2}{n}}(n-1)\mu ^{2}+{\frac {1}{n^{2}}}n(n-1)\mu ^{2}+{\frac {1}{n}}\left(\sigma ^{2}+\mu ^{2}\right)\right]\\[5pt]&={\frac {n-1}{n}}\sigma ^{2}.\end{aligned}}}$

Hence ${\displaystyle \sigma _{Y}^{2}}$ gives an estimate of the population variance that is biased by a factor of ${\displaystyle {\frac {n-1}{n}}}$. For this reason, ${\displaystyle \sigma _{Y}^{2}}$ is referred to as the biased sample variance. Correcting for this bias yields the unbiased sample variance, denoted ${\displaystyle s^{2}}$:

${\displaystyle s^{2}={\frac {n}{n-1}}\sigma _{Y}^{2}={\frac {n}{n-1}}\left({\frac {1}{n}}\sum _{i=1}^{n}\left(Y_{i}-{\overline {Y}}\right)^{2}\right)={\frac {1}{n-1}}\sum _{i=1}^{n}\left(Y_{i}-{\overline {Y}}\right)^{2}}$

Either estimator may be simply referred to as the sample variance when the version can be determined by context. The same proof is also applicable for samples taken from a continuous probability distribution.

The use of the term n − 1 is called Bessel's correction, and it is also used in sample covariance and the sample standard deviation (the square root of variance). The square root is a concave function and thus introduces negative bias (by Jensen's inequality), which depends on the distribution, and thus the corrected sample standard deviation (using Bessel's correction) is biased. The unbiased estimation of standard deviation is a technically involved problem, though for the normal distribution using the term n − 1.5 yields an almost unbiased estimator.

The unbiased sample variance is a U-statistic for the function ƒ(y₁, y₂) = (y₁ − y₂)²/2, meaning that it is obtained by averaging a 2-sample statistic over 2-element subsets of the population.

Distribution and cumulative distribution of S²/σ², for various values of ν = n − 1, when the y_i are independent normally distributed.

Being a function of random variables, the sample variance is itself a random variable, and it is natural to study its distribution. In the case that Y_i are independent observations from a normal distribution, Cochran's theorem shows that s² follows a scaled chi-squared distribution:[11]

${\displaystyle (n-1){\frac {s^{2}}{\sigma ^{2}}}\sim \chi _{n-1}^{2}.}$

As a direct consequence, it follows that

${\displaystyle \operatorname {E} \left(s^{2}\right)=\operatorname {E} \left({\frac {\sigma ^{2}}{n-1}}\chi _{n-1}^{2}\right)=\sigma ^{2},}$

and[12]

${\displaystyle \operatorname {Var} \left[s^{2}\right]=\operatorname {Var} \left({\frac {\sigma ^{2}}{n-1}}\chi _{n-1}^{2}\right)={\frac {\sigma ^{4}}{(n-1)^{2}}}\operatorname {Var} \left(\chi _{n-1}^{2}\right)={\frac {2\sigma ^{4}}{n-1}}.}$

If the Y_i are independent and identically distributed, but not necessarily normally distributed, then[13]

${\displaystyle \operatorname {E} \left[s^{2}\right]=\sigma ^{2},\quad \operatorname {Var} \left[s^{2}\right]={\frac {\sigma ^{4}}{n}}\left(\kappa -1+{\frac {2}{n-1}}\right)={\frac {1}{n}}\left(\mu _{4}-{\frac {n-3}{n-1}}\sigma ^{4}\right),}$

where κ is the kurtosis of the distribution and μ₄ is the fourth central moment.

If the conditions of the law of large numbers hold for the squared observations, s² is a consistent estimator of σ². One can see indeed that the variance of the estimator tends asymptotically to zero. An asymptotically equivalent formula was given in Kenney and Keeping (1951:164), Rose and Smith (2002:264), and Weisstein (n.d.).[14][15][16]

Samuelson's inequality is a result that states bounds on the values that individual observations in a sample can take, given that the sample mean and (biased) variance have been calculated.[17] Values must lie within the limits ${\displaystyle {\bar {y}}\pm \sigma _{Y}(n-1)^{1/2}.}$

It has been shown[18] that for a sample {y_i} of positive real numbers,

${\displaystyle \sigma _{y}^{2}\leq 2y_{\max }(A-H),}$

where y_max is the maximum of the sample, A is the arithmetic mean, H is the harmonic mean of the sample and ${\displaystyle \sigma _{y}^{2}}$ is the (biased) variance of the sample.

This bound has been improved, and it is known that variance is bounded by

${\displaystyle \sigma _{y}^{2}\leq {\frac {y_{\max }(A-H)(y_{\max }-A)}{y_{\max }-H}},}$

${\displaystyle \sigma _{y}^{2}\geq {\frac {y_{\min }(A-H)(A-y_{\min })}{H-y_{\min }}},}$

where y_min is the minimum of the sample.[19]

If ${\displaystyle x}$ is a scalar complex-valued random variable, with values in ${\displaystyle \mathbb {C} ,}$ then its variance is ${\displaystyle \operatorname {E} \left[(x-\mu )(x-\mu )^{*}\right],}$ where ${\displaystyle x^{*}}$ is the complex conjugate of ${\displaystyle x.}$ This variance is a real scalar.

If ${\displaystyle X}$ is a vector-valued random variable, with values in ${\displaystyle \mathbb {R} ^{n},}$ and thought of as a column vector, then a natural generalization of variance is ${\displaystyle \operatorname {E} \left[(X-\mu )(X-\mu )^{\operatorname {T} }\right],}$ where ${\displaystyle \mu =\operatorname {E} (X)}$ and ${\displaystyle X^{\operatorname {T} }}$ is the transpose of ${\displaystyle X,}$ and so is a row vector. The result is a positive semi-definite square matrix, commonly referred to as the variance-covariance matrix (or simply as the covariance matrix).

If ${\displaystyle X}$ is a vector- and complex-valued random variable, with values in ${\displaystyle \mathbb {C} ^{n},}$ then the covariance matrix is ${\displaystyle \operatorname {E} \left[(X-\mu )(X-\mu )^{\dagger }\right],}$ where ${\displaystyle X^{\dagger }}$ is the conjugate transpose of ${\displaystyle X.}$ This matrix is also positive semi-definite and square.

Another generalization of variance for vector-valued random variables ${\displaystyle X}$, which results in a scalar value rather than in a matrix, is the generalized variance ${\displaystyle \det(C)}$, the determinant of the covariance matrix. The generalized variance can be shown to be related to the multidimensional scatter of points around their mean.[22]

A different generalization is obtained by considering the Euclidean distance between the random variable and its mean. This results in ${\displaystyle \operatorname {E} \left[(X-\mu )^{\operatorname {T} }(X-\mu )\right]=\operatorname {tr} (C),}$ which is the trace of the covariance matrix.

• Correlation
• Distance variance
• Explained variance
• Pooled variance