The weighted arithmetic mean is similar to an ordinary

GNU Scientific Library - Reference manual, Version 1.15

2011.

/ref> :$\backslash begin\; \backslash mathbf\; \&=\; \backslash frac\; \backslash sum\_^N\; w\_i\; \backslash left(\backslash mathbf\_i\; -\; \backslash mu^*\backslash right)^T\; \backslash left(\backslash mathbf\_i\; -\; \backslash mu^*\backslash right)\; \backslash \backslash \; \&=\; \backslash frac\; .\; \backslash end$ The reasoning here is the same as in the previous section. Since we are assuming the weights are normalized, then $V\_1\; =\; 1$ and this reduces to: : $\backslash mathbf=\backslash frac.$ If all weights are the same, i.e. $\backslash textstyle\; w\_\; /\; V\_1=1/N$, then the weighted mean and covariance reduce to the unweighted sample mean and covariance above.

First Systems of Weighted Differential and Integral Calculus''

, 1980.

Tool to calculate Weighted Average

{{DEFAULTSORT:Weighted Mean Means Mathematical analysis Summary statistics

arithmetic mean
In mathematics and statistics, the arithmetic mean (, stress on first and third syllables of "arithmetic"), or simply the mean or the average (when the context is clear), is the sum of a collection of numbers divided by the count of numbers in the ...

(the most common type of average), except that 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
In mathematics and statistics, the arithmetic mean (, stress on first and third syllables of "arithmetic"), or simply the mean or the average (when the context is clear), is the sum of a collection of numbers divided by the count of numbers in the ...

. 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 mean for the morning class is 80 and the mean of the afternoon class is 90. The unweighted mean of the two means is 85. 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): :$\backslash bar\; =\; \backslash frac\; =\; 86.$ Or, this can be accomplished by weighting the class means by the number of students in each class. The larger class is given more "weight": :$\backslash bar\; =\; \backslash frac\; =\; 86.$ Thus, the weighted mean makes it possible to find the mean average student grade without knowing each student's score. Only the class means and the number of students in each class are needed.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 aconvex combinationImage:Convex combination illustration.svg, Given three points x_1, x_2, x_3 in a plane as shown in the figure, the point P ''is'' a convex combination of the three points, while Q is ''not.'' (Q is however an affine combination of the three points, a ...

.
Using the previous example, we would get the following weights:
:$\backslash frac\; =\; 0.4$
:$\backslash frac\; =\; 0.6$
Then, apply the weights like this:
:$\backslash bar\; =\; (0.4\backslash times80)\; +\; (0.6\backslash times90)\; =\; 86.$
Mathematical definition

Formally, the weighted mean of a non-empty finitemultiset
In mathematics, a multiset (or bag, or mset) is a modification of the concept of a Set (mathematics), set that, unlike a set, allows for multiple instances for each of its Element (mathematics), elements. The number of instances given for each elem ...

of data $\backslash ,$
with corresponding non-negative weights $\backslash $ is
:$\backslash bar\; =\; \backslash frac,$
which expands to:
:$\backslash bar\; =\; \backslash frac.$
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.:
:$\backslash sum\_^n\; =\; 1$.
For such normalized weights the weighted mean is then:
:$\backslash bar\; =\; \backslash sum\_^n$.
Note that one can always normalize the weights by making the following transformation on the original weights:
:$w\_i\text{'}\; =\; \backslash frac$.
Using the normalized weight yields the same results as when using the original weights:
:$\backslash begin\; \backslash bar\; \&=\; \backslash sum\_^n\; w\text{'}\_i\; x\_i=\; \backslash sum\_^n\; \backslash frac\; x\_i\; =\; \backslash frac\; \backslash \backslash \; \&\; =\; \backslash frac.\; \backslash end$
The ordinary mean $\backslash frac\; \backslash sum\_^n$ is a special case of the weighted mean where all data have equal weights.
The ''standard error of the weighted mean (unit input variances)'', $\backslash sigma\_$ can be shown via uncertainty propagation to be:
:$\backslash sigma\_\; =\; \backslash left(\backslash sqrt\; \backslash right)^$
Statistical properties

The weighted sample mean, $\backslash bar$, 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. For simplicity, we assume normalized weights (weights summing to one). If the observations have expected values : $E(x\_i\; )=,$ then the weighted sample mean has expectation : $E(\backslash bar)\; =\; \backslash sum\_^n\; .$ In particular, if the means are equal, $\backslash mu\_i=\backslash mu$, then the expectation of the weighted sample mean will be that value, : $E(\backslash bar)=\; \backslash mu.$ For uncorrelated observations with variances $\backslash sigma^2\_i$, the variance of the weighted sample mean is : $\backslash sigma^2\_\; =\; \backslash sum\_^n$ whose square root $\backslash sigma\_$ can be called the ''standard error of the weighted mean (general case)''. Consequently, if all the observations have equal variance, $\backslash sigma^2\_i=\; \backslash sigma^2\_0$, the weighted sample mean will have variance : $\backslash sigma^2\_\; =\; \backslash sigma^2\_0\; \backslash sum\_^n\; ,$ where $1/n\; \backslash le\; \backslash sum\_^n\; \backslash le\; 1$. The variance attains its maximum value, $\backslash sigma\_0^2$, when all weights except one are zero. Its minimum value is found when all weights are equal (i.e., unweighted mean), in which case we have $\backslash sigma\_\; =\; \backslash sigma\_0\; /\; \backslash sqrt$, i.e., it degenerates into the standard error of the mean, squared. Note that because 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 $w\_i\text{'}\; =\; \backslash frac$.Variance weights

For the weighted mean of a list of data for which each element $x\_i$ potentially comes from a differentprobability distribution
In probability theory and statistics
Statistics is the discipline that concerns the collection, organization, analysis, interpretation, and presentation of data. In applying statistics to a scientific, industrial, or social problem, it is ...

with known variance
In probability theory
Probability theory is the branch of mathematics concerned with probability. Although there are several different probability interpretations, probability theory treats the concept in a rigorous mathematical manner by expr ...

$\backslash sigma\_i^2$, one possible choice for the weights is given by the reciprocal of variance:
:$w\_i\; =\; \backslash frac.$
The weighted mean in this case is:
:$\backslash bar\; =\; \backslash frac,$
and the ''standard error of the weighted mean (with variance weights)'' is:
:$\backslash sigma\_\; =\; \backslash sqrt,$
Note this reduces to $\backslash sigma\_^2\; =\; \backslash sigma\_0^2/n$ when all $\backslash sigma\_i\; =\; \backslash sigma\_0$.
It is a special case of the general formula in previous section,
:$\backslash sigma^2\_\; =\; \backslash sum\_^n\; =\; \backslash frac.$
The equations above can be combined to obtain:
:$\backslash bar\; =\; \backslash sigma\_^2\; \backslash sum\_^n\; \backslash frac.$
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 historical 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 $\backslash chi^2$ is too large. The correction that must be made is :$\backslash hat\_^2\; =\; \backslash sigma\_^2\; \backslash chi^2\_\backslash nu$ where $\backslash chi^2\_\backslash nu$ is the reduced chi-squared: :$\backslash chi^2\_\backslash nu\; =\; \backslash frac\; \backslash sum\_^n\; \backslash frac;$ The square root $\backslash hat\_$ can be called the ''standard error of the weighted mean (variance weights, scale corrected)''. When all data variances are equal, $\backslash sigma\_i\; =\; \backslash sigma\_0$, they cancel out in the weighted mean variance, $\backslash sigma\_^2$, which again reduces to the standard error of the mean (squared), $\backslash sigma\_^2\; =\; \backslash sigma^2/n$, formulated in terms of the sample standard deviation (squared), :$\backslash sigma^2\; =\; \backslash frac\; .$Bootstrapping validation

It has been shown by bootstrapping methods that the following is an accurate estimation for the square of the standard error of the mean (general case): :$\backslash sigma\_^2\; =\; \backslash frac\; \backslash left;\; href="/html/ALL/s/sum\_(w\_i\_x\_i\_-\_w\_s\_\backslash bar)^2\_-\_\; 2\_\backslash bar\_\backslash sum\_(w\_i\_-\_w\_s)(w\_i\_x\_i\_-\_w\_s\backslash bar)\; +\_\backslash bar^2\_\backslash sum\_(w\_i\_-\_w\_s)^2\_\backslash right.html"\; ;"title="sum\; (w\_i\; x\_i\; -\; w\_s\; \backslash bar)^2\; -\; 2\; \backslash bar\; \backslash sum\; (w\_i\; -\; w\_s)(w\_i\; x\_i\; -\; w\_s\backslash bar)\; +\; \backslash bar^2\; \backslash sum\; (w\_i\; -\; w\_s)^2\; \backslash right">sum\; (w\_i\; x\_i\; -\; w\_s\; \backslash bar)^2\; -\; 2\; \backslash bar\; \backslash sum\; (w\_i\; -\; w\_s)(w\_i\; x\_i\; -\; w\_s\backslash bar)\; +\; \backslash bar^2\; \backslash sum\; (w\_i\; -\; w\_s)^2\; \backslash right$ where $w\_s\; =\; \backslash sum\; w\_i$. Further simplification leads to :$\backslash sigma\_^2\; =\; \backslash frac\; \backslash sum\; w\_i^2(x\_i\; -\; \backslash bar)^2$Weighted sample variance

Typically when a mean is calculated it is important to know thevariance
In probability theory
Probability theory is the branch of mathematics concerned with probability. Although there are several different probability interpretations, probability theory treats the concept in a rigorous mathematical manner by expr ...

and standard deviation about that mean. When a weighted mean $\backslash mu^*$ is used, the variance of the weighted sample is different from the variance of the unweighted sample.
The ''biased'' weighted sample variance $\backslash hat\; \backslash sigma^2\_\backslash mathrm$ is defined similarly to the normal ''biased'' sample variance $\backslash hat\; \backslash sigma^2$:
:$\backslash begin\; \backslash hat\; \backslash sigma^2\backslash \; \&=\; \backslash frac\; N\; \backslash \backslash \; \backslash hat\; \backslash sigma^2\_\backslash mathrm\; \&=\; \backslash frac\; \backslash end$
where $\backslash sum\_^N\; w\_i\; =\; 1$ for normalized weights. If the weights are ''frequency weights'' (and thus are random variables), it can be shown that $\backslash hat\; \backslash sigma^2\_\backslash mathrm$ is the maximum likelihood estimator of $\backslash sigma^2$ for Independent and identically distributed random variables, iid Gaussian observations.
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 (see Bessel's correction). In the weighted setting, there are actually two different unbiased estimators, one for the case of ''frequency weights'' and another for the case of ''reliability weights''.
Frequency weights

If the weights are ''frequency weights'' (where a weight equals the number of occurrences), then the unbiased estimator is: :$\backslash begin\; s^2\backslash \; \&=\; \backslash frac\; \backslash end$ This effectively applies Bessel's correction for frequency weights. For example, if values $\backslash $ 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 $\backslash $ with corresponding weights $\backslash $, and we get the same result either way. If the frequency weights $\backslash $ are normalized to 1, then the correct expression after Bessel's correction becomes :$\backslash begin\; s^2\backslash \; \&=\; \backslash frac\; \backslash sum\_^N\; w\_i\; \backslash left(x\_i\; -\; \backslash mu^*\backslash right)^2\; \backslash end$ where the total number of samples is $\backslash sum\_^N\; w\_i$ (not $N$). In any case, the information on total number of samples is necessary in order to obtain an unbiased correction, even if $w\_i$ has a different meaning other than frequency weight. Note that the estimator can be unbiased only if the weights are not Standard score, standardized nor Normalization (statistics), normalized, these processes changing the data's mean and variance and thus leading to a Base rate fallacy, loss of the base rate (the population count, which is a requirement for Bessel's correction).Reliability weights

If the weights are instead non-random (''reliability weights''), we can determine a correction factor to yield an unbiased estimator. Assuming each random variable is sampled from the same distribution with mean $\backslash mu$ and actual variance $\backslash sigma\_^2$, taking expectations we have, :$\backslash begin\; \backslash operatorname\; [\backslash hat\; \backslash sigma^2]\; \&=\; \backslash frac\; N\; \backslash \backslash \; \&=\; \backslash operatorname\; [(X\; -\; \backslash operatorname[X])^2]\; -\; \backslash frac\; \backslash operatorname\; [(X\; -\; \backslash operatorname[X])^2]\; \backslash \backslash \; \&=\; \backslash left(\; \backslash frac\; N\; \backslash right)\; \backslash sigma\_^2\; \backslash \backslash \; \backslash operatorname\; [\backslash hat\; \backslash sigma^2\_\backslash mathrm]\; \&=\; \backslash frac\; \backslash \backslash \; \&=\; \backslash operatorname[(X\; -\; \backslash operatorname[X])^2]\; -\; \backslash frac\; \backslash operatorname[(X\; -\; \backslash operatorname[X])^2]\; \backslash \backslash \; \&=\; \backslash left(1\; -\; \backslash frac\backslash right)\; \backslash sigma\_^2\; \backslash end$ where $V\_1\; =\; \backslash sum\_^N\; w\_i$ and $V\_2\; =\; \backslash sum\_^N\; w\_i^2$. Therefore, the bias in our estimator is $\backslash left(1\; -\; \backslash frac\backslash right)$, analogous to the $\backslash left(\; \backslash frac\; \backslash right)$ bias in the unweighted estimator (also notice that $\backslash \; V\_1^2\; /\; V\_2\; =\; N\_$ is the effective sample size#weighted samples, effective sample size). This means that to unbias our estimator we need to pre-divide by $1\; -\; \backslash left(V\_2\; /\; V\_1^2\backslash right)$, ensuring that the expected value of the estimated variance equals the actual variance of the sampling distribution. The final unbiased estimate of sample variance is: :$\backslash begin\; s^2\_\backslash \; \&=\; \backslash frac\; \backslash \backslash \; \&=\; \backslash frac\; \backslash end$, where $\backslash operatorname[s^2\_]\; =\; \backslash sigma\_^2$. 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. As a side note, other approaches have been described to compute the weighted sample variance.Weighted sample covariance

In a weighted sample, each row vector $\backslash textstyle\; \backslash textbf\_$ (each set of single observations on each of the ''K'' random variables) is assigned a weight $\backslash textstyle\; w\_i\; \backslash geq0$. Then the weighted mean vector $\backslash textstyle\; \backslash mathbf$ is given by :$\backslash mathbf=\backslash frac.$ And the weighted covariance matrix is given by: :$\backslash begin\; \backslash mathbf\; \&=\; \backslash frac\; .\; \backslash end$ Similarly to weighted sample variance, there are two different unbiased estimators depending on the type of the weights.Frequency weights

If the weights are ''frequency weights'', the ''unbiased'' weighted estimate of the covariance matrix $\backslash textstyle\; \backslash mathbf$, with Bessel's correction, is given by: :$\backslash begin\; \backslash mathbf\; \&=\; \backslash frac\; .\; \backslash end$ Note that this estimator can be unbiased only if the weights are not Standard score, standardized nor Normalization (statistics), normalized, these processes changing the data's mean and variance and thus leading to a Base rate fallacy, loss of the base rate (the population count, which is a requirement for Bessel's correction).Reliability weights

In the case of ''reliability weights'', the weights are Normalizing constant, normalized: : $V\_1\; =\; \backslash sum\_^N\; w\_i\; =\; 1.$ (If they are not, divide the weights by their sum to normalize prior to calculating $V\_1$: : $w\_i\text{'}\; =\; \backslash frac$ Then the weighted mean vector $\backslash textstyle\; \backslash mathbf$ can be simplified to :$\backslash mathbf=\backslash sum\_^N\; w\_i\; \backslash mathbf\_i.$ and the ''unbiased'' weighted estimate of the covariance matrix $\backslash textstyle\; \backslash mathbf$ is:Mark Galassi, Jim Davies, James Theiler, Brian Gough, Gerard Jungman, Michael Booth, and Fabrice RossiGNU Scientific Library - Reference manual, Version 1.15

2011.

/ref> :$\backslash begin\; \backslash mathbf\; \&=\; \backslash frac\; \backslash sum\_^N\; w\_i\; \backslash left(\backslash mathbf\_i\; -\; \backslash mu^*\backslash right)^T\; \backslash left(\backslash mathbf\_i\; -\; \backslash mu^*\backslash right)\; \backslash \backslash \; \&=\; \backslash frac\; .\; \backslash end$ The reasoning here is the same as in the previous section. Since we are assuming the weights are normalized, then $V\_1\; =\; 1$ and this reduces to: : $\backslash mathbf=\backslash frac.$ If all weights are the same, i.e. $\backslash textstyle\; w\_\; /\; V\_1=1/N$, then the weighted mean and covariance reduce to the unweighted sample mean and covariance above.

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 $\backslash sigma^2$ by the covariance matrix $\backslash mathbf$ and the arithmetic inverse by the matrix inverse (both denoted in the same way, via superscripts); the weight matrix then reads: :$\backslash mathbf\_i\; =\; \backslash mathbf\_i^.$ The weighted mean in this case is: :$\backslash bar\; =\; \backslash mathbf\_\; \backslash left(\backslash sum\_^n\; \backslash mathbf\_i\; \backslash mathbf\_i\backslash right),$ (where the order of the matrix-vector product is not commutative), in terms of the covariance of the weighted mean: :$\backslash mathbf\_\; =\; \backslash left(\backslash sum\_^n\; \backslash mathbf\_i\backslash right)^,$ 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 : $\backslash mathbf\_1\; :=\; \backslash begin1\; \&\; 0\backslash end^\backslash top,\; \backslash qquad\; \backslash mathbf\_1\; :=\; \backslash begin1\; \&\; 0\backslash \backslash \; 0\; \&\; 100\backslash end$ : $\backslash mathbf\_2\; :=\; \backslash begin0\; \&\; 1\backslash end^\backslash top,\; \backslash qquad\; \backslash mathbf\_2\; :=\; \backslash begin100\; \&\; 0\backslash \backslash \; 0\; \&\; 1\backslash end$ then the weighted mean is: : $\backslash begin\; \backslash bar\; \&\; =\; \backslash left(\backslash mathbf\_1^\; +\; \backslash mathbf\_2^\backslash right)^\; \backslash left(\backslash mathbf\_1^\; \backslash mathbf\_1\; +\; \backslash mathbf\_2^\; \backslash mathbf\_2\backslash right)\; \backslash \backslash [5pt]\; \&\; =\backslash begin\; 0.9901\; \&0\backslash \backslash \; 0\&\; 0.9901\backslash end\backslash begin1\backslash \backslash 1\backslash end\; =\; \backslash begin0.9901\; \backslash \backslash \; 0.9901\backslash end\; \backslash end$ 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 $\backslash mathbf=[x\_1,\backslash dots,x\_n]^T$, $\backslash mathbf$ is the covariance matrix relating the quantities $x\_i$, $\backslash bar$ is the common mean to be estimated, and $\backslash mathbf$ is a design matrix equal to a vector of ones $[1,\; ...,\; 1]^T$ (of length $n$). The Gaussâ€“Markov theorem states that the estimate of the mean having minimum variance is given by: :$\backslash sigma^2\_\backslash bar=(\backslash mathbf^T\; \backslash mathbf\; \backslash mathbf)^,$ and :$\backslash bar\; =\; \backslash sigma^2\_\backslash bar\; (\backslash mathbf^T\; \backslash mathbf\; \backslash mathbf),$ where: :$\backslash mathbf\; =\; \backslash mathbf^.$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 $m$. :$z\_k=\backslash sum\_^m\; w\_i\; x\_.$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 $0<\backslash Delta<1$ at each time step. Setting $w=1-\backslash Delta$ we can define $m$ normalized weights by : $w\_i=\backslash frac\; ,$ where $V\_1$ is the sum of the unnormalized weights. In this case $V\_1$ is simply : $V\_1=\backslash sum\_^m\; =\; \backslash frac\; ,$ 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 $(1-w)^$, the weight approximately equals $(1-w)=0.39(1-w)$, the tail area the value $e^$, the head area $=0.61$. The tail area at step $n$ is $\backslash le$. 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. Weighted averages of functions play an important role in the systems of weighted differential and integral calculus.Jane Grossman, Michael Grossman, Robert KatzFirst Systems of Weighted Differential and Integral Calculus''

, 1980.

See also

* Average * Central tendency * Mean * Standard deviation * Summary statistics * Weight function * Weighted average cost of capital * Weighted geometric mean * Weighted harmonic mean * Weighted least squares * Weighted median * Weighting * Binomial_proportion_confidence_interval#Standard_error_of_a_proportion_estimation_when_using_weighted_data , Standard error of a proportion estimation when using weighted dataReferences

Further reading

* *External links

*Tool to calculate Weighted Average

{{DEFAULTSORT:Weighted Mean Means Mathematical analysis Summary statistics