### Weighted 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.

## Contents

- 1 Examples
- 2 Mathematical definition
- 3 Statistical properties
- 4 Dealing with variance
- 5 Weighted sample variance
- 6 Weighted sample covariance
- 7 Vector-valued estimates
- 8 Accounting for correlations
- 9 Decreasing strength of interactions
- 10 Exponentially decreasing weights
- 11 Weighted averages of functions
- 12 See also
- 13 Notes
- 14 External links

## 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):

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

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} = \frac{(0.4\times80) + (0.6\times90)}{0.4 + 0.6} = 86.

## Mathematical definition

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

- $\backslash \{x\_1,\; x\_2,\; \backslash dots\; ,\; x\_n\backslash \},$

with non-negative weights

- $\backslash \{w\_1,\; w\_2,\; \backslash dots,\; w\_n\backslash \},$

is the quantity

- $\backslash bar\{x\}\; =\; \backslash frac\{\; \backslash sum\_\{i=1\}^n\; w\_i\; x\_i\}\{\backslash sum\_\{i=1\}^n\; w\_i\},$

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. $\backslash sum\_\{i=1\}^n\; \{w\_i\}\; =\; 1$. For such normalized weights the weighted mean is simply $\backslash bar\; \{x\}\; =\; \backslash sum\_\{i=1\}^n\; \{w\_i\; x\_i\}$.

Note that one can always normalize the weights by making the following transformation on the weights $w\_i\text{'}\; =\; \backslash frac\{w\_i\}\{\backslash sum\_\{j=1\}^n\{w\_j\}\}$. Using the normalized weight yields the same results as when using the original weights. Indeed,

- $\backslash bar\{x\}\; =\; \backslash sum\_\{i=1\}^n\; w\text{'}\_i\; x\_i=\; \backslash sum\_\{i=1\}^n\; \backslash frac\{w\_i\}\{\backslash sum\_\{j=1\}^n\; w\_j\}\; x\_i\; =\; \backslash frac\{\; \backslash sum\_\{i=1\}^n\; w\_i\; x\_i\}\{\backslash sum\_\{j=1\}^n\; w\_j\}\; =\; \backslash frac\{\; \backslash sum\_\{i=1\}^n\; w\_i\; x\_i\}\{\backslash sum\_\{i=1\}^n\; w\_i\}.$

The common mean $\backslash frac\; \{1\}\{n\}\backslash sum\_\{i=1\}^n\; \{x\_i\}$ is a special case of the weighted mean where all data have equal weights, $w\_i=w$. When the weights are normalized then $w\_i\text{'}=\backslash frac\{1\}\{n\}.$

## Statistical properties

The weighted sample mean, $\backslash bar\{x\}$, 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

- $E(x\_i\; )=\backslash bar\; \{x\_i\},$

then the weighted sample mean has expectation

- $E(\backslash bar\{x\})\; =\; \backslash sum\_\{i=1\}^n\; \{w\_i\; \backslash bar\{x\_i\}\}.$

Particularly, if the expectations of all observations are equal, $\backslash bar\; \{x\_i\}=c$, then the expectation of the weighted sample mean will be the same,

- $E(\backslash bar\{x\})=\; c.\; \backslash ,$

For uncorrelated observations with standard deviations $\backslash sigma\_i$, the weighted sample mean has standard deviation

- $\backslash sigma(\backslash bar\; x)=\; \backslash sqrt\; \{\backslash sum\_\{i=1\}^n\; \{w\_i^2\; \backslash sigma^2\_i\}\}.$

Consequently, when the standard deviations of all observations are equal, $\backslash sigma\_i=d$, the weighted sample mean will have standard deviation $\backslash sigma(\backslash bar\; x)=\; d\; \backslash sqrt\; \{V\_2\}$. Here $V\_2$ is the quantity

- $V\_2=\backslash sum\_\{i=1\}^n\; \{w\_i^2\},$

such that $1/n\; \backslash le\; V\_2\backslash le\; 1$. It attains its minimum value for equal weights, and its maximum when all weights except one are zero. In the former case we have $\backslash sigma(\backslash bar\; x)=d/\; \backslash sqrt\; \{n\}$, 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 $w\_i\text{'}\; =\; \backslash frac\{w\_i\}\{\backslash sum\_\{i=1\}^n\{w\_i\}\}$.

## Dealing with variance

**
****
**

For the weighted mean of a list of data for which each element $x\_i\backslash ,\backslash !$ comes from a different probability distribution with known variance $\{\backslash sigma\_i\}^2\backslash ,$, 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 (x_iw_i)}{\sum_{i=1}^n w_i},

and the variance of the weighted mean is:

- $$

\sigma_{\bar{x}}^2 = \frac{ 1 }{\sum_{i=1}^n w_i},

which reduces to $\backslash sigma\_\{\backslash bar\{x\}\}^2\; =\; \backslash frac\{\; \{\backslash sigma\_0\}^2\; \}\{n\}$, when all $\backslash sigma\_i\; =\; \backslash sigma\_0.\backslash ,$

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 $\backslash chi^2$ is too large. The correction that must be made is

- $\backslash sigma\_\{\backslash bar\{x\}\}^2\; \backslash rightarrow\; \backslash sigma\_\{\backslash bar\{x\}\}^2\; \backslash chi^2\_\backslash nu\; \backslash ,$

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

- $\backslash sigma\_\{\backslash bar\{x\}\}^2\; =\; \backslash frac\{\; 1\; \}\{\backslash sum\_\{i=1\}^n\; 1/\{\backslash sigma\_i\}^2\}\; \backslash times\; \backslash frac\{1\}\{(n-1)\}\; \backslash sum\_\{i=1\}^n\; \backslash frac\{\; (x\_i\; -\; \backslash bar\{x\}\; )^2\}\{\; \backslash sigma\_i^2\; \};$

when all data variances are equal, $\backslash sigma\_i\; =\; \backslash sigma\_0$, they cancel out in the weighted mean variance, $\backslash sigma\_\{\backslash bar\{x\}\}^2$, which then reduces to the standard error of the mean (squared), $\backslash sigma\_\{\backslash bar\{x\}\}^2\; =\; \backslash sigma^2/n$, in terms of the sample standard deviation (squared), $\backslash sigma^2\; =\; \backslash sum\_\{i=1\}^n\; (x\_i\; -\; \backslash bar\{x\}\; )^2\; /\; (n-1)$.

## 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 $\backslash mu^*$ 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 $V\_1\; =\; \backslash sum\_\{i=1\}^n\; w\_i$, 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 $V\_2\; =\; \backslash sum\_\{i=1\}^n\; \{w\_i^2\}$ as introduced previously.

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 $\backslash frac\{n\}\{(n-1)\}$.

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 $\backslash \{2,\; 2,\; 4,\; 5,\; 5,\; 5\backslash \}$ 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 $\backslash \{2,\; 4,\; 5\backslash \}$ with corresponding weights $\backslash \{2,\; 1,\; 3\backslash \}$, 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 $\backslash textstyle\; \backslash textbf\{x\}\_\{i\}$ (each set of single observations on each of the *K* random variables) is assigned a weight $\backslash textstyle\; w\_i\; \backslash geq0$. Without loss of generality, assume that the weights are normalized:

$\backslash sum\_\{i=1\}^\{N\}w\_i\; =\; 1.$

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

$w\_i\text{'}\; =\; \backslash frac\{w\_i\}\{\backslash sum\_\{i=1\}^\{N\}w\_i\}$

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

- $\backslash mathbf\{\backslash mu^*\}=\backslash sum\_\{i=1\}^N\; w\_i\; \backslash mathbf\{x\}\_i.$

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

- $\backslash mathbf\{\backslash mu^*\}=\backslash frac\{\backslash sum\_\{i=1\}^N\; w\_i\; \backslash mathbf\{x\}\_i\}\{\backslash sum\_\{i=1\}^N\; w\_i\}.$

and the unbiased weighted covariance matrix $\backslash textstyle\; \backslash mathbf\{\backslash Sigma\}$ is

^{[3]}

$\backslash Sigma=\backslash frac\{\backslash sum\_\{i=1\}^\{N\}w\_i\}\{\backslash left(\backslash sum\_\{i=1\}^\{N\}w\_i\backslash right)^2-\backslash sum\_\{i=1\}^\{N\}w\_i^2\}\; \backslash sum\_\{i=1\}^N\; w\_i\; \backslash left(\backslash mathbf\{x\}\_i\; -\; \backslash mu^*\backslash right)^T\backslash left(\backslash mathbf\{x\}\_i\; -\; \backslash mu^*\backslash right).$

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

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

$\backslash Sigma=\backslash frac\{1\}\{\backslash sum\_\{i=1\}^\{N\}w\_i\}\backslash sum\_\{i=1\}^N\; w\_i\; \backslash left(x\_i\; -\; \backslash mu^*\backslash right)^T\backslash left(x\_i\; -\; \backslash mu^*\backslash right),$

and the unbiased weighted sample covariance matrix is given by applying the Bessel correction (since $\backslash sum\_\{i=1\}^\{N\}w\_i\; =\; N^*$ which is the real sample size):

$\backslash Sigma=\backslash frac\{1\}\{\backslash sum\_\{i=1\}^\{N\}w\_i\; -\; 1\}\backslash sum\_\{i=1\}^N\; w\_i\; \backslash left(x\_i\; -\; \backslash mu^*\backslash right)^T\backslash left(x\_i\; -\; \backslash mu^*\backslash right).$

## 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 $\backslash sigma^2$ by the covariance matrix:^{[5]}

- $$

W_i = \Sigma_i^{-1}.

The weighted mean in this case is:

- $$

\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:

- $$

\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

- $\backslash mathbf\{x\}\_1:=\; [1\; 0]^\backslash top,\; \backslash qquad\; \backslash Sigma\_1:=\; \backslash begin\{bmatrix\}1\; 0\backslash \backslash \; 0\; 100\backslash end\{bmatrix\}$
- $\backslash mathbf\{x\}\_2:=\; [0\; 1]^\backslash top,\; \backslash qquad\; \backslash Sigma\_2:=\; \backslash begin\{bmatrix\}100\; 0\backslash \backslash \; 0\; 1\backslash end\{bmatrix\}$

then the weighted mean is:

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

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 $\backslash mathbf\{X\}=[x\_1,\backslash dots,x\_n]$, $\backslash mathbf\{C\}$ is the covariance matrix relating the quantities $x\_i$, $\backslash bar\{x\}$ is the common mean to be estimated, and $\backslash mathbf\{W\}$ 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:

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

and

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

## 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 (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 $0<\backslash Delta<1$ at each time step. Setting $w=1-\backslash Delta$ we can define $m$ normalized weights by

- $w\_i=\backslash frac\; \{w^\{i-1\}\}\{V\_1\},$

where $V\_1$ is the sum of the unnormalized weights. In this case $V\_1$ is simply

- $V\_1=\backslash sum\_\{i=1\}^m\{w^\{i-1\}\}\; =\; \backslash frac\; \{1-w^\{m\}\}\{1-w\},$

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 $\backslash le\; \{e^\{-n(1-w)\}\}$. 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]}

## See also

- 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
- Weighted median
- Standard deviation

## Notes

### Further reading

## External links

- David Terr, "MathWorld.
- Weighted Mean Calculation