Variance

 ( Moments ) 

In probability theory and statistics, variance is the expected value of the squared deviation from the mean of a random variable. The standard deviation (SD) is obtained as the square root of the variance. Variance is a measure of dispersion, meaning it is a measure of how far a set of numbers is spread out from their average value. It is the second central moment of a distribution, and the covariance of the random variable with itself, and it is often represented by $\backslash sigma^2$, $s^2$, $\backslash operatorname\{Var\}(X)$, $V(X)$, or $\backslash mathbb\{V\}(X)$.

An advantage of variance as a measure of dispersion is that it is more amenable to algebraic manipulation than other measures of dispersion such as the expected absolute deviation; for example, the variance of a sum of uncorrelated random variables is equal to the sum of their variances. A disadvantage of the variance for practical applications is that, unlike the standard deviation, its units differ from the random variable, which is why the standard deviation is more commonly reported as a measure of dispersion once the calculation is finished. Another disadvantage is that the variance is not finite for many distributions.

There are two distinct concepts that are both called "variance". One, as discussed above, is part of a theoretical probability distribution and is defined by an equation. The other variance is a characteristic of a set of observations. When variance is calculated from observations, those observations are typically measured from a real-world system. If all possible observations of the system are present, then the calculated variance is called the population variance. Normally, however, only a subset is available, and the variance calculated from this is called the sample variance. The variance calculated from a sample is considered an estimate of the full population variance. There are multiple ways to calculate an estimate of the population variance, as discussed in the section below.

The two kinds of variance are closely related. To see how, consider that a theoretical probability distribution can be used as a generator of hypothetical observations. If an infinite number of observations are generated using a distribution, then the sample variance calculated from that infinite set will match the value calculated using the distribution's equation for variance. Variance has a central role in statistics, where some ideas that use it include descriptive statistics, statistical inference, hypothesis testing, goodness of fit, and Monte Carlo sampling.

$$\backslash operatorname\{Var\}(X)\; =\; \backslash operatorname\{Cov\}(X,\; X).$$ The variance is also equivalent to the second cumulant of a probability distribution that generates $X$. The variance is typically designated as $\backslash operatorname\{Var\}(X)$, or sometimes as $V(X)$ or $\backslash mathbb\{V\}(X)$, or symbolically as $\backslash sigma^2\_X$ or simply $\backslash sigma^2$ (pronounced "sigma squared"). The expression for the variance can be expanded as follows:

In other words, the variance of is equal to the mean of the square of minus the square of the mean of . This equation should not be used for computations using floating-point arithmetic, because it suffers from catastrophic cancellation if the two components of the equation are similar in magnitude. For other numerically stable alternatives, see algorithms for calculating variance.

$$\backslash operatorname\{Var\}(X)\; =\; \backslash sum\_\{i=1\}^n\; p\_i\; \backslash cdot\; \{\backslash left(x\_i\; -\; \backslash mu\backslash right)\}^2,$$

where $\backslash mu$ is the expected value. That is,

$$\backslash mu\; =\; \backslash sum\_\{i=1\}^n\; p\_i\; x\_i\; .$$

(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 $n$ equally likely values can be written as

$$\backslash operatorname\{Var\}(X)\; =\; \backslash frac\{1\}\{n\}\; \backslash sum\_\{i=1\}^n\; (x\_i\; -\; \backslash mu)^2$$

where $\backslash mu$ is the average value. That is,

$$\backslash mu\; =\; \backslash frac\{1\}\{n\}\backslash sum\_\{i=1\}^n\; x\_i\; .$$

The variance of a set of $n$ equally likely values can be equivalently expressed, without directly referring to the mean, in terms of squared deviations of all pairwise squared distances of points from each other:

$$\backslash operatorname\{Var\}(X)\; =\; \backslash frac\{1\}\{n^2\}\; \backslash sum\_\{i=1\}^n\; \backslash sum\_\{j=1\}^n\; \backslash frac\{1\}\{2\}\; \{\backslash left(x\_i\; -\; x\_j\backslash right)\}^2\; =\; \backslash frac\{1\}\{n^2\}\; \backslash sum\_i\; \backslash sum\_\{j>i\}\; \{\backslash left(x\_i\; -\; x\_j\backslash right)\}^2.$$

$$\backslash begin\{align\}$$

  \operatorname{Var}(X) = \sigma^2 &= \int_{\R} {\left(x - \mu\right)}^2 f(x) \, dx \\[4pt]
    &= \int_{\R} x^2 f(x)\,dx -2\mu\int_{\R} xf(x)\,dx + \mu^2\int_{\R} f(x)\,dx \\[4pt]
    &= \int_{\R} x^2 \,dF(x) - 2 \mu \int_{\R} x \,dF(x) + \mu^2 \int_{\R} \,dF(x) \\[4pt]
    &= \int_{\R} x^2 \,dF(x) - 2 \mu \cdot \mu + \mu^2 \cdot 1 \\[4pt]
    &= \int_{\R} x^2 \,dF(x) - \mu^2,
\end{align}
     

or equivalently,

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

where $\backslash mu$ is the expected value of $X$ given by

$$\backslash mu\; =\; \backslash int\_\{\backslash R\}\; x\; f(x)\; \backslash ,\; dx\; =\; \backslash int\_\{\backslash R\}\; x\; \backslash ,\; dF(x).$$

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

If the function $x^2f(x)$ is Riemann-integrable on every finite interval $a,b\backslash subset\backslash R,$ then

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

where the integral is an improper Riemann integral.

Using integration by parts and making use of the expected value already calculated, we have: $$\backslash begin\{align\}$$

 \operatorname{E}\left[X^2\right] &= \int_0^\infty x^2 \lambda e^{-\lambda x} \, dx \\
   &= {\left[ -x^2 e^{-\lambda x} \right]}_0^\infty + \int_0^\infty 2 x e^{-\lambda x} \,dx \\
   &= 0 + \frac{2}{\lambda}\operatorname{E}[X] \\
   &= \frac{2}{\lambda^2}.
     

Thus, the variance of is given by $$\backslash operatorname\{Var\}(X)\; =\; \backslash operatorname\{E\}\backslash leftX^2\backslash right\; -\; \backslash operatorname\{E\}X^2\; =\; \backslash frac\{2\}\{\backslash lambda^2\}\; -\; \backslash left(\backslash frac\{1\}\{\backslash lambda\}\backslash right)^2\; =\; \backslash frac\{1\}\{\backslash lambda^2\}.$$

 \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.
     

The general formula for the variance of the outcome, , of an die is $$\backslash begin\{align\}$$

 \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}.
     

 \frac{1}{b - a} & \text{for } a \le x \le b, \\[3pt]
               0 & \text{for } x < a \text{ or } x > b
 \end{cases}
     

The variance of a constant is zero. $$\backslash operatorname\{Var\}(a)\; =\; 0.$$

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

$$\backslash operatorname\{Var\}X\; =\; \backslash operatorname\{E\}(\backslash operatorname\{Var\}X\backslash mid)\; +\; \backslash operatorname\{Var\}(\backslash operatorname\{E\}X\backslash mid).$$

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

In particular, if $Y$ is a discrete random variable assuming possible values $y\_1,\; y\_2,\; y\_3\; \backslash ldots$ with corresponding probabilities $p\_1,\; p\_2,\; p\_3\; \backslash ldots,$, then in the formula for total variance, the first term on the right-hand side becomes

$$\backslash operatorname\{E\}(\backslash operatorname\{Var\}X)\; =\; \backslash sum\_i\; p\_i\; \backslash sigma^2\_i,$$

where $\backslash sigma^2\_i\; =\; \backslash operatorname\{Var\}X$. Similarly, the second term on the right-hand side becomes

$$\backslash operatorname\{Var\}(\backslash operatorname\{E\}X)\; =\; \backslash sum\_i\; p\_i\; \backslash mu\_i^2\; -\; \backslash left(\backslash sum\_i\; p\_i\; \backslash mu\_i\backslash right)^2\; =\; \backslash sum\_i\; p\_i\; \backslash mu\_i^2\; -\; \backslash mu^2,$$

where $\backslash mu\_i\; =\; \backslash operatorname\{E\}X$ and $\backslash mu\; =\; \backslash sum\_i\; p\_i\; \backslash mu\_i$. Thus the total variance is given by

$$\backslash operatorname\{Var\}X\; =\; \backslash sum\_i\; p\_i\; \backslash sigma^2\_i\; +\; \backslash left(\; \backslash sum\_i\; p\_i\; \backslash mu\_i^2\; -\; \backslash mu^2\; \backslash right).$$

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

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

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

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

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, $\backslash mathit\{SS\}$): $$\backslash mathit\{SS\}\_\backslash text\{total\}\; =\; \backslash mathit\{SS\}\_\backslash text\{between\}\; +\; \backslash mathit\{SS\}\_\backslash text\{within\},$$ $$\backslash mathit\{SS\}\_\backslash text\{total\}\; =\; \backslash mathit\{SS\}\_\backslash text\{regression\}\; +\; \backslash mathit\{SS\}\_\backslash text\{residual\}.$$

$$2\backslash int\_0^\backslash infty\; u(1\; -\; F(u))\backslash ,du\; -\; \{\backslash left\backslash int\_0^\backslash infty\}^2.$$

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

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 arising from measurement anomalies or an unduly heavy-tailed distribution.

If all values are scaled by a constant, the variance is scaled by the square of that constant: $$\backslash operatorname\{Var\}(aX)=a^2\backslash operatorname\{Var\}(X).$$

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

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

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

If the random variables $X\_1,\backslash dots,X\_N$ are such that $$\backslash operatorname\{Cov\}(X\_i,X\_j)=0\backslash \; ,\backslash \; \backslash forall\backslash \; (i\backslash ne\; j)\; ,$$ then they are said to be uncorrelated. It follows immediately from the expression given earlier that if the random variables $X\_1,\backslash dots,X\_N$ are uncorrelated, then the variance of their sum is equal to the sum of their variances, or, expressed symbolically:

$$\backslash operatorname\{Var\}\backslash left(\backslash sum\_\{i=1\}^N\; X\_i\backslash right)\; =\; \backslash sum\_\{i=1\}^N\backslash operatorname\{Var\}(X\_i).$$

Since independent random variables are always uncorrelated (see ), the equation above holds in particular when the random variables $X\_1,\backslash dots,X\_n$ are independent. Thus, independence is sufficient but not necessary for the variance of the sum to equal the sum of the variances.

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

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

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

$$\backslash operatorname\{Var\}\backslash left(\backslash sum\_\{i=1\}^n\; X\_i\backslash right)\; =\; \backslash sum\_\{i=1\}^n\; \backslash operatorname\{Var\}(X\_i).$$

This statement is called the Bienaymé formulaLoève, M. (1977) "Probability Theory", Graduate Texts in Mathematics, Volume 45, 4th edition, Springer-Verlag, p. 12. and was discovered in 1853.Bienaymé, I.-J. (1853) "Considérations à l'appui de la découverte de Laplace sur la loi de probabilité dans la méthode des moindres carrés", Comptes rendus de l'Académie des sciences Paris, 37, p. 309–317; digital copy available [1] Bienaymé, I.-J. (1867) "Considérations à l'appui de la découverte de Laplace sur la loi de probabilité dans la méthode des moindres carrés", Journal de Mathématiques Pures et Appliquées, Série 2, Tome 12, p. 158–167; digital copy available [2][3] 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

 \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

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

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

$$\backslash begin\{align\}$$

 \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.
     

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

$$\backslash begin\{align\}$$

  \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).
     

(Note: The second equality comes from the fact that .)

Here, $\backslash operatorname\{Cov\}(\backslash cdot,\backslash cdot)$ 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

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

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 standard error. Moreover, if the variables have unit variance, for example if they are standardized, then this simplifies to

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

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

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

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.

If has a Poisson distribution, then $\backslash operatorname\{E\}N=\backslash operatorname\{Var\}(N)$ with estimator = . So, the estimator of $\backslash operatorname\{Var\}\backslash left(\backslash sum\_\{i=1\}^\{n\}X\_i\backslash right)$ becomes $n\{S\_x\}^2+n\backslash bar\{X\}^2$, giving $\backslash operatorname\{SE\}(\backslash bar\{X\})=\backslash sqrt\{\backslash frac$ (see standard error of the sample mean).

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

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:

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

$$\backslash operatorname\{Var\}(XY)\; =\; \backslash operatorname\{E\}\backslash left(X^2\backslash right)\; \backslash operatorname\{E\}\backslash left(Y^2\backslash right)\; -\; \backslash operatorname\{E\}(X)^2\; \backslash operatorname\{E\}(Y)^2.$$

 \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
     

$$\backslash operatorname\{Var\}\backslash leftf(X)\backslash right\; \backslash approx\; \backslash left(f\text{'}(\backslash operatorname\{E\}\backslash leftX\backslash right)\backslash right)^2\backslash operatorname\{Var\}\backslash leftX\backslash right$$

provided that f is twice differentiable and that the mean and variance of X are finite.

The simplest estimators for population mean and population variance are simply the mean and variance of the sample, the sample mean and (uncorrected) sample variance – these are consistent estimators (they converge to the value of the whole population as the number of samples increases) but can be improved. Most simply, the sample variance is computed as the sum of squared deviations about the (sample) mean, divided by n as the number of samples . However, using values other than n improves the estimator in various ways. Four common values for the denominator are n, n − 1, n + 1, and n − 1.5: n is the simplest (the variance of the sample), n − 1 eliminates bias, n + 1 minimizes mean squared error for the normal distribution, and n − 1.5 mostly eliminates bias in unbiased estimation of standard deviation for the normal distribution.

Firstly, if the true population mean is unknown, then the sample variance (which uses the sample mean in place of the true mean) is a biased estimator: it underestimates the variance by a factor of ( n − 1) / n; correcting this factor, resulting in the sum of squared deviations about the sample mean divided by n -1 instead of n, is called Bessel's correction. The resulting estimator is unbiased and is called the (corrected) sample variance or unbiased sample variance. If the mean is determined in some other way than from the same samples used to estimate the variance, then this bias does not arise, and the variance can safely be estimated as that of the samples about the (independently known) mean.

Secondly, the sample variance does not generally minimize mean squared error between sample variance and population variance. Correcting for bias often makes this worse: one can always choose a scale factor that performs better than the corrected sample variance, though the optimal scale factor depends on the excess kurtosis of the population (see ) and introduces bias. This always consists of scaling down the unbiased estimator (dividing by a number larger than n − 1) and is a simple example of a shrinkage estimator: one "shrinks" the unbiased estimator towards zero. For the normal distribution, dividing by n + 1 (instead of n − 1 or n) minimizes mean squared error. The resulting estimator is biased, however, and is known as the biased sample variation.

 \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]
          &= \operatorname{E}[x_i^2] - \mu^2
     

where the population mean is $\backslash mu\; =\; \backslash operatorname\{E\}x\_i\; =\; \backslash frac\; 1N\; \backslash sum\_\{i=1\}^N\; x\_i$ and $\backslash operatorname\{E\}x\_i^2\; =\; \backslash left(\backslash frac\{1\}\{N\}\; \backslash sum\_\{i=1\}^N\; x\_i^2\backslash right)$, where $\backslash operatorname\{E\}$ is the Expected value operator.

The population variance can also be computed using

(The right side has duplicate terms in the sum while the middle side has only unique terms to sum.) This is true because $$\backslash begin\{align\}$$

     &\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.
     

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.

We take a sample with replacement of values from the population of size , where , and estimate the variance on the basis of this sample.Montgomery, D. C. and Runger, G. C. (1994) Applied statistics and probability for engineers, page 201. John Wiley & Sons New York Directly taking the variance of the sample data gives the average of the squared deviations:

$$\backslash tilde\{S\}\_Y^2\; =$$

 \frac{1}{n} \sum_{i=1}^n \left(Y_i - \overline{Y}\right)^2 = \left(\frac 1n \sum_{i=1}^n Y_i^2\right) - \overline{Y}^2 =
 \frac{1}{n^2} \sum_{i,j\,:\,i
     
     

(See the section Population variance for the derivation of this formula.) Here, $\backslash overline\{Y\}$ denotes the sample mean:
     $$\backslash overline\{Y\}\; =\; \backslash frac\{1\}\{n\}\; \backslash sum\_\{i=1\}^n\; Y\_i\; .$$
     

Since the  are selected randomly, both $\backslash overline\{Y\}$ and $\backslash tilde\{S\}\_Y^2$ are Random variable. Their expected values can be evaluated by averaging over the ensemble of all possible samples  of size  from the population. For $\backslash tilde\{S\}\_Y^2$ this gives:
     $$\backslash begin\{align\}$$
 \operatorname{E}[\tilde{S}_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 1n \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 1n \sum_{i=1}^n \left( \operatorname{E}\left[Y_i^2\right] - \frac{2}{n} \left( \sum_{j \neq i} \operatorname{E}\left[Y_i Y_j\right] + \operatorname{E}\left[Y_i^2\right] \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 1n \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 1n \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{align}
     

Here $\backslash sigma^2\; =\; \backslash operatorname\{E\}Y\_i^2\; -\; \backslash mu^2$ derived in the section is population variance and $\backslash operatorname\{E\}Y\_i\; =\; \backslash operatorname\{E\}Y\_i\; \backslash operatorname\{E\}Y\_j\; =\; \backslash mu^2$ due to independency of $Y\_i$ and $Y\_j$.
     

Hence $\backslash tilde\{S\}\_Y^2$ gives an estimate of the population variance $\backslash sigma^2$ that is biased by a factor of $\backslash frac\{n\; -\; 1\}\{n\}$ because the expectation value of $\backslash tilde\{S\}\_Y^2$ is smaller than the population variance (true variance) by that factor. For this reason, $\backslash tilde\{S\}\_Y^2$ is referred to as the  biased sample variance.
     

 

 
 
  

 
 

 
Correcting for this bias yields the  unbiased sample variance, denoted $S^2$:
     

     $$S^2\; =\; \backslash frac\{n\}\{n\; -\; 1\}\; \backslash tilde\{S\}\_Y^2\; =\; \backslash frac\{n\}\{n\; -\; 1\}\; \backslash left\; =\; \backslash frac\{1\}\{n\; -\; 1\}\; \backslash sum\_\{i=1\}^n\; \backslash left(Y\_i\; -\; \backslash overline\{Y\}\; \backslash right)^2$$
     

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  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  yields an almost unbiased estimator.
     

The unbiased sample variance is a U-statistic for the function , meaning that it is obtained by averaging a 2-sample statistic over 2-element subsets of the population.
     

 

 
 
  

  Example
 
 

 
For a set of numbers {10, 15, 30, 45, 57, 52, 63, 72, 81, 93, 102, 105}, if this set is the whole data population for some measurement, then variance is the population variance 932.743 as the sum of the squared deviations about the mean of this set, divided by 12 as the number of the set members. If the set is a sample from the whole population, then the unbiased sample variance can be calculated as 1017.538 that is the sum of the squared deviations about the mean of the sample, divided by 11 instead of 12. A function VAR.S in Microsoft Excel gives the unbiased sample variance while VAR.P is for population variance.
     

 

 
 
  

  Distribution of the sample variance
 
 

 
 
 
Being a function of , 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 the unbiased sample variance  S2 follows a scaled chi-squared distribution (see also: asymptotic properties and an elementary proof):
     $$(n\; -\; 1)\; \backslash frac\{S^2\}\{\backslash sigma^2\}\; \backslash sim\; \backslash chi^2\_\{n-1\}$$
     

where  is the population variance. As a direct consequence, it follows that
     $$\backslash operatorname\{E\}\backslash left(S^2\backslash right)\; =\; \backslash operatorname\{E\}\backslash left(\backslash frac\{\backslash sigma^2\}\{n\; -\; 1\}\; \backslash chi^2\_\{n-1\}\backslash right)\; =\; \backslash sigma^2\; ,$$
     

and
 (2025). 9780534243128 ISBN 9780534243128
     

     $$$$
\operatorname{Var}\left[S^2\right] = \operatorname{Var}\left(\frac{\sigma^2}{n - 1} \chi^2_{n-1}\right) = \frac{\sigma^4}\right], where $\backslash mu\; =\; \backslash operatorname\{E\}(X)$ and $X^\{\backslash mathsf\{T\}\}$ is the transpose of , 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 $X$ is a vector- and complex-valued random variable, with values in $\backslash mathbb\{C\}^n,$ then the covariance matrix is $\backslash operatorname\{E\}\backslash left(X,$ where $X^\backslash dagger$ is the conjugate transpose of $X.$  This matrix is also positive semi-definite and square.
     

 

 
 
  

  As a scalar
 
 

 
Another generalization of variance for vector-valued random variables $X$, which results in a scalar value rather than in a matrix, is the generalized variance $\backslash det(C)$, the determinant of the covariance matrix. The generalized variance can be shown to be related to the multidimensional scatter of points around their mean.
 (2025). 9780471667193, Wiley Online Library. ISBN 9780471667193
     

A different generalization is obtained by considering the equation for the scalar variance, $\backslash operatorname\{Var\}(X)\; =\; \backslash operatorname\{E\}\backslash left(X$, and reinterpreting $(X\; -\; \backslash mu)^2$ as the squared Euclidean distance between the random variable and its mean, or, simply as the scalar product of the vector $X\; -\; \backslash mu$ with itself. This results in $\backslash operatorname\{E\}\backslash left(X\; =\; \backslash operatorname\{tr\}(C),$ which is the trace of the covariance matrix.
     

 

 
 
  

  See also
 
 

 
 
 
 

  

Bhatia–Davis inequality
  
  

Coefficient of variation
  
  

     Homoscedasticity
  
  

Least-squares spectral analysis for computing a frequency spectrum with spectral magnitudes in % of variance or in Decibel
  
  

Modern portfolio theory
  
  

Popoviciu's inequality on variances
  
  

Measures for statistical dispersion
  
  

Variance-stabilizing transformation
  
 

 

 
 
  

  Types of variance
 
 

 
 

  

     Correlation
  
  

Distance variance
  
  

Explained variance
  
  

     Pooled variance
  
  

     Pseudo-variance
  
 

 
 
 
 
 
 
 
 
 
Categories: Moments, Statistical Deviation And Dispersion, Articles Containing Proofs