# Weighted arithmetic mean

(Redirected from Weighted mean)
"Weighted mean" redirects here. For the geometric mean, see weighted geometric mean. For the harmonic mean, see weighted harmonic mean.

The weighted mean is similar to an arithmetic mean (the most common type of average), where instead of each of the data points contributing equally to the final average, some data points contribute more than others. The notion of weighted mean plays a role in descriptive statistics and also occurs in a more general form in several other areas of mathematics.

If all the weights are equal, then the weighted mean is the same as the arithmetic mean. While weighted means generally behave in a similar fashion to arithmetic means, they do have a few counterintuitive properties, as captured for instance in Simpson's paradox.

## Examples

### Basic example

Given two school classes, one with 20 students, and one with 30 students, the grades in each class on a test were:

Morning class = 62, 67, 71, 74, 76, 77, 78, 79, 79, 80, 80, 81, 81, 82, 83, 84, 86, 89, 93, 98
Afternoon class = 81, 82, 83, 84, 85, 86, 87, 87, 88, 88, 89, 89, 89, 90, 90, 90, 90, 91, 91, 91, 92, 92, 93, 93, 94, 95, 96, 97, 98, 99

The straight average for the morning class is 80 and the straight average of the afternoon class is 90. The straight average of 80 and 90 is 85, the mean of the two class means. However, this does not account for the difference in number of students in each class (20 versus 30); hence the value of 85 does not reflect the average student grade (independent of class). The average student grade can be obtained by averaging all the grades, without regard to classes (add all the grades up and divide by the total number of students):

$\bar{x} = \frac{4300}{50} = 86.$

Or, this can be accomplished by weighting the class means by the number of students in each class (using a weighted mean of the class means):

$\bar{x} = \frac{(20\times80) + (30\times90)}{20 + 30} = 86.$

Thus, the weighted mean makes it possible to find the average student grade in the case where only the class means and the number of students in each class are available.

### Convex combination example

Since only the relative weights are relevant, any weighted mean can be expressed using coefficients that sum to one. Such a linear combination is called a convex combination.

Using the previous example, we would get the following:

$\frac{20}{20 + 30} = 0.4\,$
$\frac{30}{20 + 30} = 0.6\,$
$\bar{x} = (0.4\times80) + (0.6\times90) = 86.$

## Mathematical definition

Formally, the weighted mean of a non-empty set of data

$\{x_1, x_2, \dots , x_n\},$

with non-negative weights

$\{w_1, w_2, \dots, w_n\},$

is the quantity

$\bar{x} = \frac{ \sum_{i=1}^n w_i x_i}{\sum_{i=1}^n w_i},$

which means:

$\bar{x} = \frac{w_1 x_1 + w_2 x_2 + \cdots + w_n x_n}{w_1 + w_2 + \cdots + w_n}.$

Therefore data elements with a high weight contribute more to the weighted mean than do elements with a low weight. The weights cannot be negative. Some may be zero, but not all of them (since division by zero is not allowed).

The formulas are simplified when the weights are normalized such that they sum up to $1$, i.e. $\sum_{i=1}^n {w_i} = 1$. For such normalized weights the weighted mean is simply $\bar {x} = \sum_{i=1}^n {w_i x_i}$.

Note that one can always normalize the weights by making the following transformation on the weights $w_i' = \frac{w_i}{\sum_{j=1}^n{w_j}}$. Using the normalized weight yields the same results as when using the original weights. Indeed,

$\bar{x} = \sum_{i=1}^n w'_i x_i= \sum_{i=1}^n \frac{w_i}{\sum_{j=1}^n w_j} x_i = \frac{ \sum_{i=1}^n w_i x_i}{\sum_{j=1}^n w_j} = \frac{ \sum_{i=1}^n w_i x_i}{\sum_{i=1}^n w_i}.$

The common mean $\frac {1}{n}\sum_{i=1}^n {x_i}$ is a special case of the weighted mean where all data have equal weights, $w_i=w$. When the weights are normalized then $w_i'=\frac{1}{n}.$

## Statistical properties

The weighted sample mean, $\bar{X}$, with normalized weights (weights summing to one) is itself a random variable. Its expected value and standard deviation are related to the expected values and standard deviations of the observations as follows,

If the observations have expected values

$E(X_i )=\bar {\mu_i},$

then the weighted sample mean has expectation

$E(\bar{X}) = \sum_{i=1}^n {w_i \mu_i}.$

In particular, if the means are equal, $\mu_i=\mu$, then the expectation of the weighted sample mean will be that value,

$E(\bar{X})= \mu. \,$

For uncorrelated observations with variances $\sigma^2_i$, the variance of the weighted sample mean is

$\sigma^2_{\bar X} = \sum_{i=1}^n {w_i^2 \sigma^2_i}.$

Consequently,if all the observations have equal variance, $\sigma^2_i= \sigma^2_0$, the weighted sample mean will have variance

$\sigma^2_{\bar X} = \sigma^2_0 \sum_{i=1}^n {w_i^2},$

such that $1/n \le \sum_{i=1}^n {w_i^2} \le 1$. It attains its minimum value when all weights are equal, and its maximum when all weights except one are zero. In the former case we have $\sigma^2_{\bar X} = \sigma / \sqrt {n}$, which is related to the central limit theorem.

Note that due to the fact that one can always transform non-normalized weights to normalized weights all formula in this section can be adapted to non-normalized weights by replacing all $w_i$ by $w_i' = \frac{w_i}{\sum_{i=1}^n{w_i}}$.

## Dealing with variance

For the weighted mean of a list of data for which each element $x_i\,\!$ comes from a different probability distribution with known variance $\sigma_i^2\,$, one possible choice for the weights is given by:

$w_i = \frac{1}{\sigma_i^2}.$

The weighted mean in this case is:

$\bar{x} = \frac{ \sum_{i=1}^n \left( x_i \sigma_i^{-2} \right)}{\sum_{i=1}^n \sigma_i^{-2}},$

and the variance of the weighted mean is:

$\sigma_{\bar{x}}^2 = \frac{ 1 }{\sum_{i=1}^n \sigma_i^{-2}},$

which reduces to $\sigma_{\bar{x}}^2 = \sigma_0^2/n$ when all $\sigma_i = \sigma_0$.

The two equations above can be combined to obtain:

$\bar{x} = \sum_{i=1}^n x_i \sigma_{\bar{x}}^2 / \sigma_i^2.$

The significance of this choice is that this weighted mean is the maximum likelihood estimator of the mean of the probability distributions under the assumption that they are independent and normally distributed with the same mean.

### Correcting for over- or under-dispersion

Weighted means are typically used to find the weighted mean of experimental data, rather than theoretically generated data. In this case, there will be some error in the variance of each data point. Typically experimental errors may be underestimated due to the experimenter not taking into account all sources of error in calculating the variance of each data point. In this event, the variance in the weighted mean must be corrected to account for the fact that $\chi^2$ is too large. The correction that must be made is

$\sigma_{\bar{x}}^2 \rightarrow \sigma_{\bar{x}}^2 \chi^2_\nu \,$

where $\chi^2_\nu$ is $\chi^2$ divided by the number of degrees of freedom, in this case n − 1. This gives the variance in the weighted mean as:

$\sigma_{\bar{x}}^2 = \frac{ 1 }{\sum_{i=1}^n 1/{\sigma_i}^2} \times \frac{1}{(n-1)} \sum_{i=1}^n \frac{ (x_i - \bar{x} )^2}{ \sigma_i^2 };$

when all data variances are equal, $\sigma_i = \sigma_0$, they cancel out in the weighted mean variance, $\sigma_{\bar{x}}^2$, which then reduces to the standard error of the mean (squared), $\sigma_{\bar{x}}^2 = \sigma^2/n$, in terms of the sample standard deviation (squared), $\sigma^2 = \sum_{i=1}^n (x_i - \bar{x} )^2 / (n-1)$.

## Weighted sample variance

Typically when a mean is calculated it is important to know the variance and standard deviation about that mean. When a weighted mean $\mu^*$ is used, the variance of the weighted sample is different from the variance of the unweighted sample. The biased weighted sample variance is defined similarly to the normal biased sample variance:

$\sigma^2\ = \frac{ \sum_{i=1}^N{\left(x_i - \mu\right)^2} }{ N }$
$\sigma^2_\mathrm{weighted} = \frac{\sum_{i=1}^N w_i \left(x_i - \mu^*\right)^2 }{V_1}$

where $V_1 = \sum_{i=1}^n w_i$, which is 1 for normalized weights.

For small samples, it is customary to use an unbiased estimator for the population variance. In normal unweighted samples, the N in the denominator (corresponding to the sample size) is changed to N − 1. While this is simple in unweighted samples, it is not straightforward when the sample is weighted.

If each $x_i$ is drawn from a Gaussian distribution with variance $1/w_i$, the unbiased estimator of a weighted population variance is given by:[1]

$s^2\ = \frac {V_1} {V_1^2-V_2} \sum_{i=1}^N w_i \left(x_i - \mu^*\right)^2,$

where $V_2 = \sum_{i=1}^n {w_i^2}$.

Note: If the weights are not integral frequencies (for instance, if they have been standardized to sum to 1 or if they represent the variance of each observation's measurement) as in this case, then all information is lost about the total sample size n, whence it is not possible to use an unbiased estimator because it is impossible to estimate the Bessel correction factor $\frac{n}{(n-1)}$.

The degrees of freedom of the weighted, unbiased sample variance vary accordingly from N − 1 down to 0.

The standard deviation is simply the square root of the variance above.

If all of the $x_i$ are drawn from the same distribution and the integer weights $w_i$ indicate the number of occurrences ("repeat") of an observation in the sample, then the unbiased estimator of the weighted population variance is given by

$s^2\ = \frac {1} {V_1 - 1} \sum_{i=1}^N w_i \left(x_i - \mu^*\right)^2 = \frac {1} {\sum_{i=1}^n w_i - 1} \sum_{i=1}^N w_i \left(x_i - \mu^*\right)^2,$

If all $x_i$ are unique, then $N$ counts the number of unique values, and $V_1$ counts the number of samples.

For example, if values $\{2, 2, 4, 5, 5, 5\}$ are drawn from the same distribution, then we can treat this set as an unweighted sample, or we can treat it as the weighted sample $\{2, 4, 5\}$ with corresponding weights $\{2, 1, 3\}$, and we should get the same results.

As a side note, other approaches have been described to compute the weighted sample variance.[2]

## Weighted sample covariance

In a weighted sample, each row vector $\textstyle \textbf{x}_{i}$ (each set of single observations on each of the K random variables) is assigned a weight $\textstyle w_i \geq0$. Without loss of generality, assume that the weights are normalized:

$\sum_{i=1}^{N}w_i = 1.$

If they are not, divide the weights by their sum:

$w_i' = \frac{w_i}{\sum_{i=1}^{N}w_i}$

Then the weighted mean vector $\textstyle \mathbf{\mu^*}$ is given by

$\mathbf{\mu^*}=\sum_{i=1}^N w_i \mathbf{x}_i.$

(if the weights are not normalized, an equivalent formula to compute the weighted mean is:)

$\mathbf{\mu^*}=\frac{\sum_{i=1}^N w_i \mathbf{x}_i}{\sum_{i=1}^N w_i}.$

and an approximation of the unbiased weighted covariance matrix $\textstyle \mathbf{\Sigma}$ (but without Bessel correction) is:[3]

$\Sigma=\frac{\sum_{i=1}^{N}w_i}{\left(\sum_{i=1}^{N}w_i\right)^2-\sum_{i=1}^{N}w_i^2} \sum_{i=1}^N w_i \left(\mathbf{x}_i - \mu^*\right)^T\left(\mathbf{x}_i - \mu^*\right).$

If all weights are the same, with $\textstyle w_{i}=1/N$, then the weighted mean and covariance reduce to the sample mean and covariance above.

There is no unbiased (with Bessel correction) equation to compute the weighted covariance matrix in this case because if, as in the case above, the weights are not integral frequencies (for instance, if they have been standardized to sum to 1 or if they represent the variance of each observation's measurement), then all information is lost about the total sample size n, whence it is impossible to estimate precisely the Bessel correction factor $\frac{n}{(n-1)}$.

Alternatively, if each weight $\textstyle w_i \geq0$ assigns a number of occurrences for one observation value, so $\textstyle \textbf{x}_{i}$ (sometimes called the number of "repeats") and is unnormalized so that $\textstyle \sum_{i=1}^{N}w_i=N^*$ with $N^*$ being the sample size (total number of observations), then the biased weighted sample covariance matrix is given by:[4]

$\Sigma=\frac{1}{\sum_{i=1}^{N}w_i}\sum_{i=1}^N w_i \left(x_i - \mu^*\right)^T\left(x_i - \mu^*\right),$

and the correctly unbiased weighted sample covariance matrix is given by applying the Bessel correction (since $\sum_{i=1}^{N}w_i = N^*$ which is the real sample size):

$\Sigma=\frac{1}{\sum_{i=1}^{N}w_i - 1}\sum_{i=1}^N w_i \left(x_i - \mu^*\right)^T\left(x_i - \mu^*\right).$

## Vector-valued estimates

The above generalizes easily to the case of taking the mean of vector-valued estimates. For example, estimates of position on a plane may have less certainty in one direction than another. As in the scalar case, the weighted mean of multiple estimates can provide a maximum likelihood estimate. We simply replace the variance $\sigma^2$ by the covariance matrix $\Sigma$ and the arithmetic inverse by the matrix inverse (both denoted in the same way, via superscripts); the weight matrix then reads:[5]

$\text{W}_i = \Sigma_i^{-1}.$

The weighted mean in this case is:

$\bar{\mathbf{x}} = \Sigma_{\bar{\mathbf{x}}} \left(\sum_{i=1}^n \text{W}_i \mathbf{x}_i\right),$

(where the order of the matrix-vector product is not commutative), in terms of the covariance of the weighted mean:

$\Sigma_{\bar{\mathbf{x}}} = \left(\sum_{i=1}^n \text{W}_i\right)^{-1},$

For example, consider the weighted mean of the point [1 0] with high variance in the second component and [0 1] with high variance in the first component. Then

$\mathbf{x}_1 := [1 0]^\top, \qquad \Sigma_1 := \begin{bmatrix}1 & 0\\ 0 & 100\end{bmatrix}$
$\mathbf{x}_2 := [0 1]^\top, \qquad \Sigma_2 := \begin{bmatrix}100 & 0\\ 0 & 1\end{bmatrix}$

then the weighted mean is:

$\bar{\mathbf{x}} = \left(\Sigma_1^{-1} + \Sigma_2^{-1}\right)^{-1} \left(\Sigma_1^{-1} \mathbf{x}_1 + \Sigma_2^{-1} \mathbf{x}_2\right)$
$=\begin{bmatrix} 0.9901 &0\\ 0& 0.9901\end{bmatrix}\begin{bmatrix}1\\1\end{bmatrix} = \begin{bmatrix}0.9901 \\ 0.9901\end{bmatrix}$

which makes sense: the [1 0] estimate is "compliant" in the second component and the [0 1] estimate is compliant in the first component, so the weighted mean is nearly [1 1].

## Accounting for correlations

In the general case, suppose that $\mathbf{X}=[x_1,\dots,x_n]$, $\mathbf{C}$ is the covariance matrix relating the quantities $x_i$, $\bar{x}$ is the common mean to be estimated, and $\mathbf{W}$ is the design matrix [1, ..., 1] (of length $n$). The Gauss–Markov theorem states that the estimate of the mean having minimum variance is given by:

$\sigma^2_\bar{x}=(\mathbf{W}^T \mathbf{C}^{-1} \mathbf{W})^{-1},$

and

$\bar{x} = \sigma^2_\bar{x} (\mathbf{W}^T \mathbf{C}^{-1} \mathbf{X}).$

## Decreasing strength of interactions

Consider the time series of an independent variable $x$ and a dependent variable $y$, with $n$ observations sampled at discrete times $t_i$. In many common situations, the value of $y$ at time $t_i$ depends not only on $x_i$ but also on its past values. Commonly, the strength of this dependence decreases as the separation of observations in time increases. To model this situation, one may replace the independent variable by its sliding mean $z$ for a window size $m$.

$z_k=\sum_{i=1}^m w_i x_{k+1-i}.$
Range weighted mean interpretation
Range (1–5) Weighted mean equivalence
3.34–5.00 Strong
1.67–3.33 Satisfactory
0.00–1.66 Weak

## Exponentially decreasing weights

In the scenario described in the previous section, most frequently the decrease in interaction strength obeys a negative exponential law. If the observations are sampled at equidistant times, then exponential decrease is equivalent to decrease by a constant fraction $0<\Delta<1$ at each time step. Setting $w=1-\Delta$ we can define $m$ normalized weights by

$w_i=\frac {w^{i-1}}{V_1},$

where $V_1$ is the sum of the unnormalized weights. In this case $V_1$ is simply

$V_1=\sum_{i=1}^m{w^{i-1}} = \frac {1-w^{m}}{1-w},$

approaching $V_1=1/(1-w)$ for large values of $m$.

The damping constant $w$ must correspond to the actual decrease of interaction strength. If this cannot be determined from theoretical considerations, then the following properties of exponentially decreasing weights are useful in making a suitable choice: at step $(1-w)^{-1}$, the weight approximately equals ${e^{-1}}(1-w)=0.39(1-w)$, the tail area the value $e^{-1}$, the head area ${1-e^{-1}}=0.61$. The tail area at step $n$ is $\le {e^{-n(1-w)}}$. Where primarily the closest $n$ observations matter and the effect of the remaining observations can be ignored safely, then choose $w$ such that the tail area is sufficiently small.

## Weighted averages of functions

The concept of weighted average can be extended to functions.[6] Weighted averages of functions play an important role in the systems of weighted differential and integral calculus.[7]

## Notes

1. ^ http://www.gnu.org/software/gsl/manual/html_node/Weighted-Samples.html
2. ^ Weighted Standard Error and its Impact on Significance Testing (WinCross vs. Quantum & SPSS), Dr. Albert Madansky
3. ^ Mark Galassi, Jim Davies, James Theiler, Brian Gough, Gerard Jungman, Michael Booth, and Fabrice Rossi. GNU Scientific Library - Reference manual, Version 1.15, 2011. Sec. 21.7 Weighted Samples
4. ^ George R. Price, Ann. Hum. Genet., Lond, pp485-490, Extension of covariance selection mathematics, 1972.
5. ^ James, Frederick (2006). Statistical Methods in Experimental Physics (2nd ed.). Singapore: World Scientific. p. 324. ISBN 981-270-527-9.
6. ^ G. H. Hardy, J. E. Littlewood, and G. Pólya. Inequalities (2nd ed.), Cambridge University Press, ISBN 978-0-521-35880-4, 1988.
7. ^ Jane Grossman, Michael Grossman, Robert Katz. The First Systems of Weighted Differential and Integral Calculus, ISBN 0-9771170-1-4, 1980.