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.

The term weighted average usually refers to a weighted arithmetic mean, but weighted versions of other means can also be calculated, such as the weighted geometric mean and the weighted harmonic mean.

Example[link]

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, and 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:

Failed to parse (Missing texvc executable; please see math/README to configure.): \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):

Failed to parse (Missing texvc executable; please see math/README to configure.): \bar{x} = \frac{(20)80 + (30)90}{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.

Mathematical definition[link]

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

Failed to parse (Missing texvc executable; please see math/README to configure.): \{x_1, x_2, \dots , x_n\},

with non-negative weights

Failed to parse (Missing texvc executable; please see math/README to configure.): \{w_1, w_2, \dots, w_n\},

is the quantity

Failed to parse (Missing texvc executable; please see math/README to configure.): \bar{x} = \frac{ \sum_{i=1}^n w_i x_i}{\sum_{i=1}^n w_i},

which means:

Failed to parse (Missing texvc executable; please see math/README to configure.): \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 Failed to parse (Missing texvc executable; please see math/README to configure.): 1 , i.e. Failed to parse (Missing texvc executable; please see math/README to configure.): \sum_{i=1}^n {w_i} = 1 . For such normalized weights the weighted mean is simply Failed to parse (Missing texvc executable; please see math/README to configure.): \bar {x} = \sum_{i=1}^n {w_i x_i} .

The common mean Failed to parse (Missing texvc executable; please see math/README to configure.): \frac {1}{n}\sum_{i=1}^n {x_i}

is a special case of the weighted mean where all data have equal weights, Failed to parse (Missing texvc executable; please see math/README to configure.): w_i=w

. When the weights are normalized then Failed to parse (Missing texvc executable; please see math/README to configure.): w_i=\frac{1}{n}.

Convex combination[link]

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:

Failed to parse (Missing texvc executable; please see math/README to configure.): \frac{20}{20 + 30} = 0.4\,

Failed to parse (Missing texvc executable; please see math/README to configure.): \frac{30}{20 + 30} = 0.6\,

Failed to parse (Missing texvc executable; please see math/README to configure.): \bar{x} = \frac{(0.4)80 + (0.6)90}{0.4 + 0.6} = 86.

This simplifies to:

Failed to parse (Missing texvc executable; please see math/README to configure.): \bar{x} = (0.4)80 + (0.6)90 = 86.

Statistical properties[link]

The weighted sample mean, Failed to parse (Missing texvc executable; please see math/README to configure.): \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

Failed to parse (Missing texvc executable; please see math/README to configure.): E(x_i )=\bar {x_i},

then the weighted sample mean has expectation

Failed to parse (Missing texvc executable; please see math/README to configure.): E(\bar{x}) = \sum_{i=1}^n {w_i \bar{x_i}}.

Particularly, if the expectations of all observations are equal, Failed to parse (Missing texvc executable; please see math/README to configure.): \bar {x_i}=c , then the expectation of the weighted sample mean will be the same,

Failed to parse (Missing texvc executable; please see math/README to configure.): E(\bar{x})= c. \,

For uncorrelated observations with standard deviations Failed to parse (Missing texvc executable; please see math/README to configure.): \sigma_i , the weighted sample mean has standard deviation

Failed to parse (Missing texvc executable; please see math/README to configure.): \sigma(\bar x)= \sqrt {\sum_{i=1}^n {w_i^2 \sigma^2_i}}.

Consequently, when the standard deviations of all observations are equal, Failed to parse (Missing texvc executable; please see math/README to configure.): \sigma_i=d , the weighted sample mean will have standard deviation Failed to parse (Missing texvc executable; please see math/README to configure.): \sigma(\bar x)= d \sqrt {V_2} . Here Failed to parse (Missing texvc executable; please see math/README to configure.): V_2

is the quantity

Failed to parse (Missing texvc executable; please see math/README to configure.): V_2=\sum_{i=1}^n {w_i^2},

such that Failed to parse (Missing texvc executable; please see math/README to configure.): 1/n \le V_2\le 1 . It attains its minimum value for equal weights, and its maximum when all weights except one are zero. In the former case we have Failed to parse (Missing texvc executable; please see math/README to configure.): \sigma(\bar x)=d/ \sqrt {n} , which is related to the central limit theorem.

Dealing with variance[link]

For the weighted mean of a list of data for which each element Failed to parse (Missing texvc executable; please see math/README to configure.): x_i\,\!

comes from a different probability distribution with known variance Failed to parse (Missing texvc executable; please see math/README to configure.): {\sigma_i}^2\,

, one possible choice for the weights is given by:

Failed to parse (Missing texvc executable; please see math/README to configure.): w_i = \frac{1}{\sigma_i^2}.

The weighted mean in this case is:

Failed to parse (Missing texvc executable; please see math/README to configure.): \bar{x} = \frac{ \sum_{i=1}^n (x_i/{\sigma_i}^2)}{\sum_{i=1}^n (1/{\sigma_i}^2)},

and the variance of the weighted mean is:

Failed to parse (Missing texvc executable; please see math/README to configure.): \sigma_{\bar{x}}^2 = \frac{ 1 }{\sum_{i=1}^n (1/{\sigma_i}^2)},

which reduces to Failed to parse (Missing texvc executable; please see math/README to configure.): \sigma_{\bar{x}}^2 = \frac{ {\sigma_0}^2 }{n} , when all Failed to parse (Missing texvc executable; please see math/README to configure.): \sigma_i = \sigma_0.\,

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.

[edit] 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 Failed to parse (Missing texvc executable; please see math/README to configure.): \chi^2

is too large. The correction that must be made is

Failed to parse (Missing texvc executable; please see math/README to configure.): \sigma_{\bar{x}}^2 \rightarrow \sigma_{\bar{x}}^2 \chi^2_\nu \,

where Failed to parse (Missing texvc executable; please see math/README to configure.): \chi^2_\nu

is Failed to parse (Missing texvc executable; please see math/README to configure.): \chi^2
divided by the number of degrees of freedom, in this case n − 1. This gives the variance in the weighted mean as:

Failed to parse (Missing texvc executable; please see math/README to configure.): \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, Failed to parse (Missing texvc executable; please see math/README to configure.): \sigma_i = \sigma_0 , they cancel out in the weighted mean variance, Failed to parse (Missing texvc executable; please see math/README to configure.): \sigma_{\bar{x}}^2 , which then reduces to the standard error of the mean (squared), Failed to parse (Missing texvc executable; please see math/README to configure.): \sigma_{\bar{x}}^2 = \sigma^2/n , in terms of the sample standard deviation (squared), Failed to parse (Missing texvc executable; please see math/README to configure.): \sigma^2 = \sum_{i=1}^n (x_i - \bar{x} )^2 / (n-1) .

Weighted sample variance[link]

Typically when a mean is calculated it is important to know the variance and standard deviation about that mean. When a weighted mean Failed to parse (Missing texvc executable; please see math/README to configure.): \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:

Failed to parse (Missing texvc executable; please see math/README to configure.): \sigma^2\ = \frac{ \sum_{i=1}^N{\left(x_i - \mu\right)^2} }{ N }

Failed to parse (Missing texvc executable; please see math/README to configure.): \sigma^2_\mathrm{weighted} = \frac{\sum_{i=1}^N w_i \left(x_i - \mu^*\right)^2 }{V_1}

where Failed to parse (Missing texvc executable; please see math/README to configure.): 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. The unbiased estimator of a weighted population variance (assuming each Failed to parse (Missing texvc executable; please see math/README to configure.): x_i

is drawn from a Gaussian distribution with variance Failed to parse (Missing texvc executable; please see math/README to configure.): 1/w_i

) is given by ^[1]:

Failed to parse (Missing texvc executable; please see math/README to configure.): s^2\ = \frac {V_1} {V_1^2-V_2} \sum_{i=1}^N w_i \left(x_i - \mu^*\right)^2,

where Failed to parse (Missing texvc executable; please see math/README to configure.): V_2 = \sum_{i=1}^n {w_i^2}

as introduced previously.

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 Failed to parse (Missing texvc executable; please see math/README to configure.): x_i

are drawn from the same distribution and the integer weights Failed to parse (Missing texvc executable; please see math/README to configure.): w_i
indicate frequency of occurrence in the sample, then the unbiased estimator of the weighted population variance is given by

Failed to parse (Missing texvc executable; please see math/README to configure.): s^2\ = \frac {1} {V_1 - 1} \sum_{i=1}^N w_i \left(x_i - \mu^*\right)^2,

If all Failed to parse (Missing texvc executable; please see math/README to configure.): x_i

are unique, then Failed to parse (Missing texvc executable; please see math/README to configure.): N
counts the number of unique values, and Failed to parse (Missing texvc executable; please see math/README to configure.): V_1
counts the number of samples.

For example, if values Failed to parse (Missing texvc executable; please see math/README to configure.): \{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 Failed to parse (Missing texvc executable; please see math/README to configure.): \{2, 4, 5\}
with corresponding weights Failed to parse (Missing texvc executable; please see math/README to configure.): \{2, 1, 3\}

, and we should get the same results.

Vector-valued estimates[link]

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 Failed to parse (Missing texvc executable; please see math/README to configure.): \sigma^2

by the covariance matrix[2]:

Failed to parse (Missing texvc executable; please see math/README to configure.): W_i = \Sigma_i^{-1}.

The weighted mean in this case is:

Failed to parse (Missing texvc executable; please see math/README to configure.): \bar{\mathbf{x}} = \left(\sum_{i=1}^n \Sigma_i^{-1}\right)^{-1}\left(\sum_{i=1}^n \Sigma_i^{-1} \mathbf{x}_i\right),

and the covariance of the weighted mean is:

Failed to parse (Missing texvc executable; please see math/README to configure.): \Sigma_{\bar{\mathbf{x}}} = \left(\sum_{i=1}^n \Sigma_i^{-1}\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

Failed to parse (Missing texvc executable; please see math/README to configure.): \mathbf{x}_1 := [1 0]^\top, \qquad \Sigma_1 := \begin{bmatrix}1 & 0\\ 0 & 100\end{bmatrix}

Failed to parse (Missing texvc executable; please see math/README to configure.): \mathbf{x}_2 := [0 1]^\top, \qquad \Sigma_2 := \begin{bmatrix}100 & 0\\ 0 & 1\end{bmatrix}

then the weighted mean is:

Failed to parse (Missing texvc executable; please see math/README to configure.): \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)

Failed to parse (Missing texvc executable; please see math/README to configure.): =\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[link]

In the general case, suppose that Failed to parse (Missing texvc executable; please see math/README to configure.): \mathbf{X}=[x_1,\dots,x_n] , Failed to parse (Missing texvc executable; please see math/README to configure.): \mathbf{C}

is the covariance matrix relating the quantities Failed to parse (Missing texvc executable; please see math/README to configure.): x_i

, Failed to parse (Missing texvc executable; please see math/README to configure.): \bar{x}

is the common mean to be estimated, and Failed to parse (Missing texvc executable; please see math/README to configure.): \mathbf{W}
is the design matrix [1, ..., 1] (of length Failed to parse (Missing texvc executable; please see math/README to configure.): n

). The Gauss–Markov theorem states that the estimate of the mean having minimum variance is given by:

Failed to parse (Missing texvc executable; please see math/README to configure.): \sigma^2_\bar{x}=(\mathbf{W}^T \mathbf{C}^{-1} \mathbf{W})^{-1},

and

Failed to parse (Missing texvc executable; please see math/README to configure.): \bar{x} = \sigma^2_\bar{x} (\mathbf{W}^T \mathbf{C}^{-1} \mathbf{X}).

Decreasing strength of interactions[link]

Consider the time series of an independent variable Failed to parse (Missing texvc executable; please see math/README to configure.): x

and a dependent variable Failed to parse (Missing texvc executable; please see math/README to configure.): y

, with Failed to parse (Missing texvc executable; please see math/README to configure.): n

observations sampled at discrete times Failed to parse (Missing texvc executable; please see math/README to configure.): t_i

. In many common situations, the value of Failed to parse (Missing texvc executable; please see math/README to configure.): y

at time Failed to parse (Missing texvc executable; please see math/README to configure.): t_i
depends not only on Failed to parse (Missing texvc executable; please see math/README to configure.): 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 Failed to parse (Missing texvc executable; please see math/README to configure.): z
for a window size Failed to parse (Missing texvc executable; please see math/README to configure.): m

.

Failed to parse (Missing texvc executable; please see math/README to configure.): z_k=\sum_{i=1}^m w_i x_{k+1-i}.

Exponentially decreasing weights[link]

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 Failed to parse (Missing texvc executable; please see math/README to configure.): 0<\Delta<1

at each time step. Setting Failed to parse (Missing texvc executable; please see math/README to configure.): w=1-\Delta
we can define Failed to parse (Missing texvc executable; please see math/README to configure.): m
normalized weights by

Failed to parse (Missing texvc executable; please see math/README to configure.): w_i=\frac {w^{i-1}}{V_1}

, where Failed to parse (Missing texvc executable; please see math/README to configure.): V_1

is the sum of the unnormalized weights. In this case Failed to parse (Missing texvc executable; please see math/README to configure.): V_1
is simply

Failed to parse (Missing texvc executable; please see math/README to configure.): V_1=\sum_{i=1}^m{w^{i-1}} = \frac {1-w^{m}}{1-w}

, approaching Failed to parse (Missing texvc executable; please see math/README to configure.): V_1=1/(1-w)

for large values of Failed to parse (Missing texvc executable; please see math/README to configure.): m

.

The damping constant Failed to parse (Missing texvc executable; please see math/README to configure.): 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 Failed to parse (Missing texvc executable; please see math/README to configure.): (1-w)^{-1}

, the weight approximately equals Failed to parse (Missing texvc executable; please see math/README to configure.): {e^{-1}}(1-w)=0.39(1-w) , the tail area the value Failed to parse (Missing texvc executable; please see math/README to configure.): e^{-1} , the head area Failed to parse (Missing texvc executable; please see math/README to configure.): {1-e^{-1}}=0.61 . The tail area at step Failed to parse (Missing texvc executable; please see math/README to configure.): n

is Failed to parse (Missing texvc executable; please see math/README to configure.): \le {e^{-n(1-w)}}

. Where primarily the closest Failed to parse (Missing texvc executable; please see math/README to configure.): n

observations matter and the effect of the remaining observations can be ignored safely, then choose Failed to parse (Missing texvc executable; please see math/README to configure.): w
such that the tail area is sufficiently small.

Weighted averages of functions[link]

The concept of weighted average can be extended to functions.^[3] Weighted averages of functions play an important role in the systems of weighted differential and integral calculus.^[4]

See also[link]

• Average
• Mean
• Distance-weighted estimator
• Summary statistics
• Central tendency
• Weight function
• Weighted least squares
• Weighted average cost of capital
• Weighting
• Weighted geometric mean
• Weighted harmonic mean
• Standard Deviation

Notes[link]

Further reading[link]

• Bevington, Philip. Data Reduction and Error Analysis for the Physical Sciences.

External links[link]

• David Terr, "Weighted Mean" from MathWorld.
• Weighted Mean Calculation