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The weighted arithmetic mean is similar to an ordinary arithmetic mean (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. 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 430050=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 (20times 80)+(30times 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.

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 weights:

displaystyle frac 2020+30=0.4

displaystyle frac 3020+30=0.6

Then, apply the weights like this:

bar x=(0.4times 80)+(0.6times 90)=86.

Mathematical definition

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

x_1,x_2,dots ,x_n,

(where x represents a set of mean values)
with non-negative weights

bar x=frac sum limits _i=1^nw_ix_isum limits _i=1^nw_i,

which means:

bar x=frac w_1x_1+w_2x_2+cdots +w_nx_nw_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. $displaystyle sum _i=1^nw_i'=1$ . For such normalized weights the weighted mean is simply
$displaystyle bar x=sum _i=1^nw_i'x_i$ .
Note that one can always normalize the weights by making the following transformation on the original weights $w_i'=frac w_isum _j=1^nw_j$ . Using the normalized weight yields the same results as when using the original weights. Indeed,

beginalignedbar x&=sum _i=1^nw'_ix_i=sum _i=1^nfrac w_isum _j=1^nw_jx_i=frac sum _i=1^nw_ix_isum _j=1^nw_j\&=frac sum _i=1^nw_ix_isum _i=1^nw_i.endaligned

The common mean $frac 1nsum _i=1^nx_i$ is a special case of the weighted mean where all data have equal weights, $displaystyle w_i=1$ . When the weights are normalized then $w_i'=frac 1n.$

Similarly, the weighted average uncertainty, $barsigma$ can be shown to be
$displaystyle beginalignedbar sigma &=left(sqrt sum _i=1^nw_iright)^-1endaligned$

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

displaystyle E(X_i)=mu _i,

then the weighted sample mean has expectation

E(bar X)=sum _i=1^nw_imu _i.

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

displaystyle E(bar X)=mu .

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

displaystyle sigma _bar X^2=sum _i=1^nw_i^2sigma _i^2

whose square root $displaystyle sigma _bar X$ can be called the standard error of the weighted mean.

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

sigma _bar X^2=sigma _0^2sum _i=1^nw_i^2,

where $1/nleq sum _i=1^nw_i^2leq 1$ . The variance attains its maximum value, $sigma _0^2$ , 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 $displaystyle sigma _bar X=sigma _0/sqrt n$ , 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 $w_i$ by $w_i'=frac w_isum _i=1^nw_i$ .

Dealing with variance

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

w_i=frac 1sigma _i^2.

The weighted mean in this case is:

bar x=frac sum _i=1^nleft(x_isigma _i^-2right)sum _i=1^nsigma _i^-2,

and the uncertainty on the weighted mean is:

displaystyle sigma _bar x=sqrt frac 1sum _i=1^nsigma _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=sigma _bar x^2sum _i=1^nx_i/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 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 $chi ^2$ is too large. The correction that must be made is

displaystyle hat sigma _bar x^2=sigma _bar x^2chi _nu ^2

where $chi _nu ^2$ is the reduced chi-squared. This gives the scaled variance in the weighted mean as:

displaystyle hat sigma _bar x^2=frac 1sum _i=1^nsigma _i^-2times frac 1(n-1)sum _i=1^nfrac (x_i-bar x)^2sigma _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),

displaystyle sigma ^2=frac sum _i=1^n(x_i-bar x)^2n-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 $displaystyle hat sigma _mathrm w ^2$ is defined similarly to the normal biased sample variance $displaystyle hat sigma ^2$ :

displaystyle beginalignedhat sigma ^2 &=frac sum _i=1^Nleft(x_i-mu right)^2N\hat sigma _mathrm w ^2&=frac sum _i=1^Nw_ileft(x_i-mu ^*right)^2V_1endaligned

where $V_1=sum _i=1^Nw_i$ , which is 1 for normalized weights. If the weights are frequency weights (and thus are random variables), it can be shown that $displaystyle hat sigma _mathrm w ^2$ is the maximum likelihood estimator of $sigma ^2$ for 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, then the unbiased estimator is:

beginaligneds^2 &=frac sum _i=1^Nw_ileft(x_i-mu ^*right)^2V_1-1endaligned

This effectively applies Bessel's correction for frequency weights.

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 get the same result either way.

Reliability weights

If the weights are instead non-random (reliability weights), we can determine a correction factor to yield an unbiased estimator. Taking expectations we have,

displaystyle beginalignedoperatorname E [hat sigma ^2]&=frac sum _i=1^Noperatorname E [(x_i-mu )^2]N\&=operatorname E [(X-operatorname E [X])^2]-frac 1Noperatorname E [(X-operatorname E [X])^2]\&=left(frac N-1Nright)sigma _textactual^2\operatorname E [hat sigma _mathrm w ^2]&=frac sum _i=1^Nw_ioperatorname E [(x_i-mu ^*)^2]V_1\&=operatorname E [(X-operatorname E [X])^2]-frac V_2V_1^2operatorname E [(X-operatorname E [X])^2]\&=left(1-frac V_2V_1^2right)sigma _textactual^2endaligned

where $V_2=sum _i=1^Nw_i^2$ . Therefore, the bias in our estimator is $left(1-frac V_2V_1^2right)$ , analogous to the $left(frac N-1Nright)$ bias in the unweighted estimator. This means that to unbias our estimator we need to pre-divide by $1-left(V_2/V_1^2right)$ , 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:

displaystyle beginaligneds^2 &=frac hat sigma _mathrm w ^21-(V_2/V_1^2)\&=frac sum _i=1^Nw_i(x_i-mu ^*)^2V_1-(V_2/V_1)endaligned

,^[1]

where $displaystyle operatorname E [s^2]=sigma _textactual^2$ .

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.^[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 $textstyle w_igeq 0$ .

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

mathbf mu ^* =frac sum _i=1^Nw_imathbf x _isum _i=1^Nw_i.

And the weighted covariance matrix is given by:^[3]

displaystyle beginalignedSigma &=frac sum _i=1^Nw_ileft(mathbf x _i-mu ^*right)^Tleft(mathbf x _i-mu ^*right)V_1.endaligned

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 $textstyle mathbf Sigma$ , with Bessel's correction, is given by:^[3]

displaystyle beginalignedSigma &=frac sum _i=1^Nw_ileft(mathbf x _i-mu ^*right)^Tleft(mathbf x _i-mu ^*right)V_1-1.endaligned

Reliability weights

In the case of reliability weights, the weights are normalized:

displaystyle V_1=sum _i=1^Nw_i=1.

(If they are not, divide the weights by their sum to normalize prior to calculating $V_1$ :

displaystyle w_i'=frac w_isum _i=1^Nw_i

Then the weighted mean vector $textstyle mathbf mu ^*$ can be simplified to

mathbf mu ^* =sum _i=1^Nw_imathbf x _i.

and the unbiased weighted estimate of the covariance matrix $textstyle mathbf Sigma$ is:^[4]

displaystyle beginalignedSigma &=frac sum _i=1^Nw_ileft(sum _i=1^Nw_iright)^2-sum _i=1^Nw_i^2sum _i=1^Nw_ileft(mathbf x _i-mu ^*right)^Tleft(mathbf x _i-mu ^*right)\&=frac sum _i=1^Nw_ileft(mathbf x _i-mu ^*right)^Tleft(mathbf x _i-mu ^*right)V_1-(V_2/V_1).endaligned

The reasoning here is the same as in the previous section.

Since we are assuming the weights are normalized, then $displaystyle V_1=1$ and this reduces to:

displaystyle Sigma =frac sum _i=1^Nw_ileft(mathbf x _i-mu ^*right)^Tleft(mathbf x _i-mu ^*right)1-V_2.

If all weights are the same, i.e. $textstyle w_i/V_1=1/N$ , 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 $sigma ^2$ 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]

textW_i=Sigma _i^-1.

The weighted mean in this case is:

bar mathbf x =Sigma _bar mathbf x left(sum _i=1^ntextW_imathbf x _iright),

(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^ntextW_iright)^-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

displaystyle mathbf x _1:=beginbmatrix1&0endbmatrix^top ,qquad Sigma _1:=beginbmatrix1&0\0&100endbmatrix

displaystyle mathbf x _2:=beginbmatrix0&1endbmatrix^top ,qquad Sigma _2:=beginbmatrix100&0\0&1endbmatrix

then the weighted mean is:

displaystyle beginalignedbar mathbf x &=left(Sigma _1^-1+Sigma _2^-1right)^-1left(Sigma _1^-1mathbf x _1+Sigma _2^-1mathbf x _2right)\[5pt]&=beginbmatrix0.9901&0\0&0.9901endbmatrixbeginbmatrix1\1endbmatrix=beginbmatrix0.9901\0.9901endbmatrixendaligned

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 $displaystyle mathbf X =[x_1,dots ,x_n]^T$ , $mathbfSigma$ is the covariance matrix relating the quantities $x_i$ , $bar x$ is the common mean to be estimated, and $mathbf J$ is a design matrix equal to a vector of ones $displaystyle [1,...,1]^T$ (of length $n$ ). The Gauss–Markov theorem states that the estimate of the mean having minimum variance is given by:

displaystyle sigma _bar x^2=(mathbf J ^Tmathbf W mathbf J )^-1,

and

displaystyle bar x=sigma _bar x^2(mathbf J ^Tmathbf W mathbf X ),

where:

displaystyle mathbf W =mathbf Sigma ^-1.

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^mw_ix_k+1-i.

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-1V_1,

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

V_1=sum _i=1^mw^i-1=frac 1-w^m1-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 $leq e^-n(1-w)$ . 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]

References

^ "GNU Scientific Library – Reference Manual: Weighted Samples". Gnu.org. Retrieved 22 December 2017.

^ "Weighted Standard Error and its Impact on Significance Testing (WinCross vs. Quantum & SPSS), Dr. Albert Madansky" (PDF). Analyticalgroup.com. Retrieved 22 December 2017.

^ ^a^b George R. Price (1972). "Ann. Hum. Genet., Lond, pp. 485-490, Extension of covariance selection mathematics" (PDF). Dynamics.org. Retrieved 22 December 2017.

^ 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

^ James, Frederick (2006). Statistical Methods in Experimental Physics (2nd ed.). Singapore: World Scientific. p. 324. ISBN 981-270-527-9.

^ G. H. Hardy, J. E. Littlewood, and G. Pólya. Inequalities (2nd ed.), Cambridge University Press, ISBN 978-0-521-35880-4, 1988.

^ Jane Grossman, Michael Grossman, Robert Katz. The First Systems of Weighted Differential and Integral Calculus, ISBN 0-9771170-1-4, 1980.

External links

David Terr. "Weighted Mean". MathWorld.

[1] "GNU Scientific Library – Reference Manual: Weighted Samples". Gnu.org. Retrieved 22 December 2017.

[2] "Weighted Standard Error and its Impact on Significance Testing (WinCross vs. Quantum & SPSS), Dr. Albert Madansky" (PDF). Analyticalgroup.com. Retrieved 22 December 2017.

[PRICE-1972-3] George R. Price (1972). "Ann. Hum. Genet., Lond, pp. 485-490, Extension of covariance selection mathematics" (PDF). Dynamics.org. Retrieved 22 December 2017.

[Galassi-2007-GSL-4] 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

[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.

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YTjnti

Weighted arithmetic mean

Contents

Examples

Basic example

Convex combination example

Mathematical definition

Statistical properties

Dealing with variance

Correcting for over- or under-dispersion

Weighted sample variance

Frequency weights

Reliability weights

Weighted sample covariance

Frequency weights

Reliability weights

Vector-valued estimates

Accounting for correlations

Decreasing strength of interactions

Exponentially decreasing weights

Weighted averages of functions

See also

References

Further reading

External links

Comments

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Weighted arithmetic mean

Contents

Examples

Basic example

Convex combination example

Mathematical definition

Statistical properties

Dealing with variance

Correcting for over- or under-dispersion

Weighted sample variance

Frequency weights

Reliability weights

Weighted sample covariance

Frequency weights

Reliability weights

Vector-valued estimates

Accounting for correlations

Decreasing strength of interactions

Exponentially decreasing weights

Weighted averages of functions

See also

References

Further reading

External links

Comments

Post a Comment

Popular posts from this blog

Executable numpy error

Trying to Print Gridster Items to PDF without overlapping contents

Mass disable jenkins jobs