Weighted arithmetic mean

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 with 20 students, one with 30 test grades in each class as follows:

: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):

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":

:<math>\bar{x} = \frac{(20\times80) + (30\times90)}{20 + 30} = 86.</math>

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 a convex combination.

Using the previous example, we would get the following weights:

:<math>\frac{20}{20 + 30} = 0.4</math>

:<math>\frac{30}{20 + 30} = 0.6</math>

Then, apply the weights like this:

:<math>\bar{x} = (0.4\times80) + (0.6\times90) = 86.</math>

Mathematical definition

Formally, the weighted mean of a non-empty finite tuple of data <math>\left( x_1, x_2, \dots , x_n \right)</math>,

with corresponding non-negative weights <math>\left( w_1, w_2, \dots , w_n \right)</math> is

:<math>\bar{x} = \frac{ \sum\limits_{i=1}^n w_i x_i}{\sum\limits_{i=1}^n w_i},</math>

which expands to:

:<math>\bar{x} = \frac{w_1 x_1 + w_2 x_2 + \cdots + w_n x_n}{w_1 + w_2 + \cdots + w_n}.</math>

Therefore, data elements with a high weight contribute more to the weighted mean than do elements with a low weight. The weights may not be negative in order for the equation to work. 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., <math display=inline>\sum\limits_{i=1}^n {w_i'} = 1</math>.

For such normalized weights, the weighted mean is equivalently:

:<math>\bar {x} = \sum\limits_{i=1}^n {w_i' x_i}</math>.

One can always normalize the weights by making the following transformation on the original weights:

:<math>w_i' = \frac{w_i}{\sum\limits_{j=1}^n{w_j</math>.

The ordinary mean <math display=inline>\frac {1}{n}\sum\limits_{i=1}^n {x_i}</math> is a special case of the weighted mean where all data have equal weights.

If the data elements are independent and identically distributed random variables with variance <math>\sigma^2</math>, the standard error of the weighted mean, <math>\sigma_{\bar{x</math>, can be shown via uncertainty propagation to be:

:<math display="inline">

\sigma_{\bar{x = \sigma \sqrt{ \sum\limits_{i=1}^n w_i'^2 }

</math>

Variance-defined weights

For the weighted mean of a list of data for which each element <math>x_i</math> potentially comes from a different probability distribution with known variance <math>\sigma_i^2</math>, all having the same mean, one possible choice for the weights is given by the reciprocal of variance:

:<math>w_i = \frac{1}{\sigma_i^2}.</math>

The weighted mean in this case is:

:<math>\bar{x} = \frac{ \sum_{i=1}^n \left( \dfrac{ x_i}{\sigma_i^{2 \right)}{\sum_{i=1}^n \dfrac{1}{\sigma_i^{2}

=\frac{\sum_{i=1}^n\left(x_i\cdot w_i\right)}{\sum_{i=1}^n w_i},</math>

and the standard error of the weighted mean (with inverse-variance weights) is:

:<math>\sigma_{\bar{x = \sqrt{\frac{ 1 }{\sum_{i=1}^n \sigma_i^{-2}

=\sqrt{\frac{1}{\sum_{i=1}^n w_i,</math>

Note this reduces to <math> \sigma_{\bar{x^2 = \sigma_0^2/n</math> when all <math>\sigma_i = \sigma_0</math>.

It is a special case of the general formula in previous section,

:<math> \sigma^2_{\bar x} = \sum_{i=1}^n {w_i'^2 \sigma^2_i} = \frac{ \sum_{i=1}^n {\sigma_i^{-4} \sigma^2_i} }{\left(\sum_{i=1}^n \sigma_i^{-2}\right)^2}.</math>

The equations above can be combined to obtain:

:<math>\bar{x} = \sigma_{\bar{x^2 \sum_{i=1}^n \frac{x_i}{\sigma_i^2}.</math>

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.

Statistical properties

Expectancy

The weighted sample mean, <math>\bar{x}</math>, 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

then the weighted sample mean has expectation

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

Variance

Simple i.i.d. case

When treating the weights as constants, and having a sample of n observations from uncorrelated random variables, all with the same variance and expectation (as is the case for i.i.d random variables), then the variance of the weighted mean can be estimated as the multiplication of the unweighted variance by Kish's design effect (see proof):

: <math>\operatorname{Var}(\bar y_w) = \hat \sigma_y^2 \frac{\overline{w^2{ \bar{w}^2 } </math>

With <math>\hat \sigma_y^2 = \frac{\sum_{i=1}^n (y_i - \bar y)^2}{n-1} </math>, <math>\bar{w} = \frac{\sum_{i=1}^n w_i}{n} </math>, and <math>\overline{w^2} = \frac{\sum_{i=1}^n w_i^2}{n} </math>

However, this estimation is rather limited due to the strong assumption about the y observations. This has led to the development of alternative, more general, estimators.

Survey sampling perspective

From a model based perspective, we are interested in estimating the variance of the weighted mean when the different <math>y_i</math> are not i.i.d random variables. An alternative perspective for this problem is that of some arbitrary sampling design of the data in which units are selected with unequal probabilities (with replacement). I.e.: selecting some element will not influence the probability of drawing another element (this doesn't apply for things such as cluster sampling design).

Since each element (<math>y_i</math>) is fixed, and the randomness comes from it being included in the sample or not (<math>I_i</math>), we often talk about the multiplication of the two, which is a random variable. To avoid confusion in the following section, let's call this term: <math>y'_i = y_i I_i</math>. With the following expectancy: <math>E[y'_i] = y_i E[I_i] = y_i \pi_i</math>; and variance: <math>V[y'_i] = y_i^2 V[I_i] = y_i^2 \pi_i(1-\pi_i)</math>.

When each element of the sample is inflated by the inverse of its selection probability, it is termed the <math>\pi</math>-expanded y values, i.e.: <math>\check y_i = \frac{y_i}{\pi_i}</math>. A related quantity is <math>p</math>-expanded y values: <math>\frac{y_i}{p_i} = n \check y_i</math>.

We have (at least) two versions of variance for the weighted mean: one with known and one with unknown population size estimation. There is no uniformly better approach, but the literature presents several arguments to prefer using the population estimation version (even when the population size is known).

:<math>

\widehat{\sigma_{\bar{x}_w}^2} = \frac{n}{(n-1)(n \bar{w} )^2} \left[\sum (w_i x_i - \bar{w} \bar{x}_w)^2 -

2 \bar{x}_w \sum (w_i - \bar{w})(w_i x_i - \bar{w} \bar{x}_w)

+ \bar{x}_w^2 \sum (w_i - \bar{w})^2 \right]

</math>

where <math>\bar{w} = \frac{\sum w_i}{n}</math>. Further simplification leads to

:<math>\widehat{\sigma_{\bar{x^2} = \frac{n}{(n-1)(n \bar{w} )^2} \sum w_i^2(x_i - \bar{x}_w)^2</math>

Gatz et al. mention that the above formulation was published by Endlich et al. (1988) when treating the weighted mean as a combination of a weighted total estimator divided by an estimator of the population size, based on the formulation published by Cochran (1977), as an approximation to the ratio mean. However, Endlich et al. didn't seem to publish this derivation in their paper (even though they mention they used it), and Cochran's book includes a slightly different formulation. Still, it's almost identical to the formulations described in previous sections.

Replication-based estimators

Because there is no closed analytical form for the variance of the weighted mean, it was proposed in the literature to rely on replication methods such as the Jackknife and Bootstrapping.

where <math>\operatorname{E}[s^2_{\mathrm{w] = \sigma_{\text{actual^2</math>. The degrees of freedom of this 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 <math> \mathbf{x}_{i}</math> (each set of single observations on each of the K random variables) is assigned a weight <math>w_i \geq0</math>.

Then the weighted mean vector <math> \mathbf{\mu^*}</math> is given by

:<math> \mathbf{\mu^*}=\frac{\sum_{i=1}^N w_i \mathbf{x}_i}{\sum_{i=1}^N w_i}.</math>

And the weighted covariance matrix is given by:

:<math>\mathbf{C} = \frac {\sum_{i=1}^N w_i \left(\mathbf{x}_i - \mu^*\right)^T \left(\mathbf{x}_i - \mu^*\right)} {V_1}.</math>

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 <math>\textstyle \mathbf{C}</math>, with Bessel's correction, is given by:

:<math>

\begin{align}

\mathbf{C} &= \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) \\

&= \frac {\sum_{i=1}^N w_i \left(\mathbf{x}_i - \mu^*\right)^T \left(\mathbf{x}_i - \mu^*\right)} {V_1 - (V_2 / V_1)}.

\end{align}

</math>

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

Since we are assuming the weights are normalized, then <math>V_1 = 1</math> and this reduces to:

: <math>\mathbf{C}=\frac{\sum_{i=1}^N w_i \left(\mathbf{x}_i - \mu^*\right)^T \left(\mathbf{x}_i - \mu^*\right)}{1-V_2}.</math>

If all weights are the same, i.e. <math> w_{i} / V_1=1/N</math>, 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 <math>\sigma^2</math> by the covariance matrix <math>\mathbf{C}</math> and the arithmetic inverse by the matrix inverse (both denoted in the same way, via superscripts); the weight matrix then reads:

<math display="block"> \mathbf{W}_i = \mathbf{C}_i^{-1}.</math>

The weighted mean in this case is:

<math display="block">\bar{\mathbf{x = \mathbf{C}_{\bar{\mathbf{x} \left(\sum_{i=1}^n \mathbf{W}_i \mathbf{x}_i\right),</math>

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

<math display="block">\mathbf{C}_{\bar{\mathbf{x} = \left(\sum_{i=1}^n \mathbf{W}_i\right)^{-1},</math>

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

: <math>\mathbf{x}_1 := \begin{bmatrix}1 & 0\end{bmatrix}^\top, \qquad \mathbf{C}_1 := \begin{bmatrix}1 & 0\\ 0 & 100\end{bmatrix}</math>

: <math>\mathbf{x}_2 := \begin{bmatrix}0 & 1\end{bmatrix}^\top, \qquad \mathbf{C}_2 := \begin{bmatrix}100 & 0\\ 0 & 1\end{bmatrix}</math>

then the weighted mean is:

: <math>

\begin{align}

\bar{\mathbf{x & = \left(\mathbf{C}_1^{-1} + \mathbf{C}_2^{-1}\right)^{-1} \left(\mathbf{C}_1^{-1} \mathbf{x}_1 + \mathbf{C}_2^{-1} \mathbf{x}_2\right) \\[5pt]

& =\begin{bmatrix} 0.9901 &0\\ 0& 0.9901\end{bmatrix}\begin{bmatrix}1\\1\end{bmatrix} = \begin{bmatrix}0.9901 \\ 0.9901\end{bmatrix}

\end{align}

</math>

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

:<math>\sigma^2_\bar{x}=(\mathbf{J}^T \mathbf{W} \mathbf{J})^{-1},</math>

and

:<math>\bar{x} = \sigma^2_\bar{x} (\mathbf{J}^T \mathbf{W} \mathbf{X}),</math>

where:

:<math>\mathbf{W} = \mathbf{C}^{-1}.</math>

Decreasing strength of interactions

Consider the time series of an independent variable <math>x</math> and a dependent variable <math>y</math>, with <math>n</math> observations sampled at discrete times <math>t_i</math>. In many common situations, the value of <math>y</math> at time <math>t_i</math> depends not only on <math>x_i</math> 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 <math>z</math> for a window size <math>m</math>.

:<math>z_k=\sum_{i=1}^m w_i x_{k+1-i}.</math>

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 <math>0<\Delta<1</math> at each time step. Setting <math>w=1-\Delta</math> we can define <math>m</math> normalized weights by

: <math>w_i=\frac {w^{i-1{V_1},</math>

where <math>V_1</math> is the sum of the unnormalized weights. In this case <math>V_1</math> is simply

: <math>V_1=\sum_{i=1}^m{w^{i-1 = \frac {1-w^{m{1-w},</math>

approaching <math>V_1=1/(1-w)</math> for large values of <math>m</math>.

The damping constant <math>w</math> 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 <math>(1-w)^{-1}</math>, the weight approximately equals <math>{e^{-1(1-w)=0.39(1-w)</math>, the tail area the value <math>e^{-1}</math>, the head area <math>{1-e^{-1=0.61</math>. The tail area at step <math>n</math> is <math>\le {e^{-n(1-w)</math>. Where primarily the closest <math>n</math> observations matter and the effect of the remaining observations can be ignored safely, then choose <math>w</math> 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.

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 <math>\chi^2</math> is too large. The correction that must be made is

:<math>\hat{\sigma}_{\bar{x^2 = \sigma_{\bar{x^2 \chi^2_\nu </math>

where <math>\chi^2_\nu</math> is the reduced chi-squared:

:<math>\chi^2_\nu = \frac{1}{(n-1)} \sum_{i=1}^n \frac{ (x_i - \bar{x} )^2}{ \sigma_i^2 };</math>

The square root <math>\hat{\sigma}_{\bar{x</math> can be called the standard error of the weighted mean (variance weights, scale corrected).

When all data variances are equal, <math>\sigma_i = \sigma_0</math>, they cancel out in the weighted mean variance, <math>\sigma_{\bar{x^2</math>, which again reduces to the standard error of the mean (squared), <math>\sigma_{\bar{x^2 = \sigma^2/n</math>, formulated in terms of the sample standard deviation (squared),

Weighted arithmetic mean

Examples

Basic example

Convex combination example

Mathematical definition

Variance-defined weights

Statistical properties

Expectancy

Variance

Simple i.i.d. case

Survey sampling perspective

Replication-based estimators

Weighted sample covariance

Frequency weights

Vector-valued estimates

Accounting for correlations

Decreasing strength of interactions

Exponentially decreasing weights

Weighted averages of functions

Correcting for over- or under-dispersion

See also

Notes

References

Further reading

External links