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# Variance

## From Wikipedia, the free encyclopedia

Example of samples from two populations with the same mean but different variances. The red population has mean 100 and variance 100 (SD=10) while the blue population has mean 100 and variance 2500 (SD=50).

In probability theory and statistics, variance is the expectation of the squared deviation of a random variable from its mean. Informally, it measures how far a set of (random) numbers are spread out from their average value. Variance has a central role in statistics, where some ideas that use it include descriptive statistics, statistical inference, hypothesis testing, goodness of fit, and Monte Carlo sampling. Variance is an important tool in the sciences, where statistical analysis of data is common. The variance is the square of the standard deviation, the second central moment of a distribution, and the covariance of the random variable with itself, and it is often represented by ${\displaystyle \sigma ^{2}}$, ${\displaystyle s^{2}}$, or ${\displaystyle \operatorname {Var} (X)}$.

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• ✪ Range, variance and standard deviation as measures of dispersion | Khan Academy
• ✪ Sample variance | Descriptive statistics | Probability and Statistics | Khan Academy
• ✪ ANOVA 1: Calculating SST (total sum of squares) | Probability and Statistics | Khan Academy
• ✪ Statistics: Sample variance | Descriptive statistics | Probability and Statistics | Khan Academy
• ✪ Review and intuition why we divide by n-1 for the unbiased sample | Khan Academy

#### Transcription

In the last video we talked about different ways to represent the central tendency or the average of a data set. What we're going to do in this video is to expand that a little bit to understand how spread apart the data is as well. So let's just think about this a little bit. Let's say I have negative 10, 0, 10, 20 and 30. Let's say that's one data set right there. And let's say the other data set is 8, 9, 10, 11 and 12. Now let's calculate the arithmetic mean for both of these data sets. So let's calculate the mean. And when you go further on in statistics, you're going to understand the difference between a population and a sample. We're assuming that this is the entire population of our data. So we're going to be dealing with the population mean. We're going to be dealing with, as you see, the population measures of dispersion. I know these are all fancy words. In the future, you're not going to have all of the data. You're just going to have some samples of it, and you're going to try to estimate things for the entire population. So I don't want you to worry too much about that just now. But if you are going to go further in statistics, I just want to make that clarification. Now, the population mean, or the arithmetic mean of this data set right here, it is negative 10 plus 0 plus 10 plus 20 plus 30 over-- we have five data points-- over 5. And what is this equal to? That negative 10 cancels out with that 10, 20 plus 30 is 50 divided by 5, it's equal to 10. Now, what's the mean of this data set? 8 plus 9 plus 10 plus 11 plus 12, all of that over 5. And the way we could think about it, 8 plus 12 is 20, 9 plus 11 is another 20, so that's 40, and then we have a 50 there. Add another 10. So this, once again, is going to be 50 over 5. So this has the exact same population means. Or if you don't want to worry about the word population or sample and all of that, both of these data sets have the exact same arithmetic mean. When you average all these numbers and divide by 5 or when you take the sum of these numbers and divide by 5, you get 10, some of these numbers and divide by 5, you get 10 as well. But clearly, these sets of numbers are different. You know, if you just looked at this number, you'd say, oh, maybe these sets are very similar to each other. But when you look at these two data sets, one thing might pop out at you. All of these numbers are very close to 10. I mean, the furthest number here is two away from 10. 12 is only two away from 10. Here, these numbers are further away from 10. Even the closer ones are still 10 away and these guys are 20 away from 10. So this right here, this data set right here is more disperse, right? These guys are further away from our mean than these guys are from this mean. So let's think about different ways we can measure dispersion, or how far away we are from the center, on average. Now one way, this is kind of the most simple way, is the range. And you won't see it used too often, but it's kind of a very simple way of understanding how far is the spread between the largest and the smallest number. You literally take the largest number, which is 30 in our example, and from that, you subtract the smallest number. So 30 minus negative 10, which is equal to 40, which tells us that the difference between the largest and the smallest number is 40, so we have a range of 40 for this data set. Here, the range is the largest number, 12, minus the smallest number, which is 8, which is equal to 4. So here range is actually a pretty good measure of dispersion. We say, OK, both of these guys have a mean of 10. But when I look at the range, this guy has a much larger range, so that tells me this is a more disperse set. But range is always not going to tell you the whole picture. You might have two data sets with the exact same range where still, based on how things are bunched up, it could still have very different distributions of where the numbers lie. Now, the one that you'll see used most often is called the variance. Actually, we're going to see the standard deviation in this video. That's probably what's used most often, but it has a very close relationship to the variance. So the symbol for the variance-- and we're going to deal with the population variance. Once again, we're assuming that this is all of the data for our whole population, that we're not just sampling, taking a subset, of the data. So the variance, its symbol is literally this sigma, this Greek letter, squared. That is the symbol for variance. And we'll see that the sigma letter actually is the symbol for standard deviation. And that is for a reason. But anyway, the definition of a variance is you literally take each of these data points, find the difference between those data points and your mean, square them, and then take the average of those squares. I know that sounds very complicated, but when I actually calculate it, you're going to see it's not too bad. So remember, the mean here is 10. So I take the first data point. Let me do it over here. Let me scroll down a little bit. So I take the first data point. Negative 10. From that, I'm going to subtract our mean and I'm going to square that. So I just found the difference from that first data point to the mean and squared it. And that's essentially to make it positive. Plus the second data point, 0 minus 10, minus the mean-- this is the mean; this is that 10 right there-- squared plus 10 minus 10 squared-- that's the middle 10 right there-- plus 20 minus 10-- that's the 20-- squared plus 30 minus 10 squared. So this is the squared differences between each number and the mean. This is the mean right there. I'm finding the difference between every data point and the mean, squaring them, summing them up, and then dividing by that number of data points. So I'm taking the average of these numbers, of the squared distances. So when you say it kind of verbally, it sounds very complicated. But you're taking each number. What's the difference between that, the mean, square it, take the average of those. So I have 1, 2, 3, 4, 5, divided by 5. So what is this going to be equal to? Negative 10 minus 10 is negative 20. Negative 20 squared is 400. 0 minus 10 is negative 10 squared is 100, so plus 100. 10 minus 10 squared, that's just 0 squared, which is 0. Plus 20 minus 10 is 10 squared, is 100. Plus 30 minus 10, which is 20, squared is 400. All of that over 5. And what do we have here? 400 plus 100 is 500, plus another 500 is 1000. It's equal to 1000/5, which is equal to 200. So in this situation, our variance is going to be 200. That's our measure of dispersion there. And let's compare it to this data set over here. Let's compare it to the variance of this less-dispersed data set. So let me scroll over a little bit so we have some real estate, although I'm running out. Maybe I could scroll up here. There you go. Let me calculate the variance of this data set. So we already know its mean. So its variance of this data set is going to be equal to 8 minus 10 squared plus 9 minus 10 squared plus 10 minus 10 squared plus 11 minus 10-- let me scroll up a little bit-- squared plus 12 minus 10 squared. Remember, that 10 is just the mean that we calculated. You have to calculate the mean first. Divided by-- we have 1, 2, 3, 4, 5 squared differences. So this is going to be equal to-- 8 minus 10 is negative 2 squared, is positive 4. 9 minus 10 is negative 1 squared, is positive 1. 10 minus 10 is 0 squared. You still get 0. 11 minus 10 is 1. Square it, you get 1. 12 minus 10 is 2. Square it, you get 4. And what is this equal to? All of that over 5. This is 10/5. So this is going to be--all right, this is 10/5, which is equal to 2. So the variance here-- let me make sure I got that right. Yes, we have 10/5. So the variance of this less-dispersed data set is a lot smaller. The variance of this data set right here is only 2. So that gave you a sense. That tells you, look, this is definitely a less-dispersed data set then that there. Now, the problem with the variance is you're taking these numbers, you're taking the difference between them and the mean, then you're squaring it. It kind of gives you a bit of an arbitrary number, and if you're dealing with units, let's say if these are distances. So this is negative 10 meters, 0 meters, 10 meters, this is 8 meters, so on and so forth, then when you square it, you get your variance in terms of meters squared. It's kind of an odd set of units. So what people like to do is talk in terms of standard deviation, which is just the square root of the variance, or the square root of sigma squared. And the symbol for the standard deviation is just sigma. So now that we've figured out the variance, it's very easy to figure out the standard deviation of both of these characters. The standard deviation of this first one up here, of this first data set, is going to be the square root of 200. The square root of 200 is what? The square root of 2 times 100. This is equal to 10 square roots of 2. That's that first data set. Now the standard deviation of the second data set is just going to be the square root of its variance, which is just 2. So the second data set has 1/10 the standard deviation as this first data set. This is 10 roots of 2, this is just the root of 2. So this is 10 times the standard deviation. And this, hopefully, will make a little bit more sense. Let's think about it. This has 10 times more the standard deviation than this. And let's remember how we calculated it. Variance, we just took each data point, how far it was away from the mean, squared that, took the average of those. Then we took the square root, really just to make the units look nice, but the end result is we said that that first data set has 10 times the standard deviation as the second data set. So let's look at the two data sets. This has 10 times the standard deviation, which makes sense intuitively, right? I mean, they both have a 10 in here, but each of these guys, 9 is only one away from the 10, 0 is 10 away from the 10, 10 less. 8 is only two away. This guy is 20 away. So it's 10 times, on average, further away. So the standard deviation, at least in my sense, is giving a much better sense of how far away, on average, we are from the mean. Anyway, hopefully, you found that useful.

## Definition

The variance of a random variable ${\displaystyle X}$ is the expected value of the squared deviation from the mean of ${\displaystyle X}$, ${\displaystyle \mu =\operatorname {E} [X]}$:

${\displaystyle \operatorname {Var} (X)=\operatorname {E} \left[(X-\mu )^{2}\right].}$

This definition encompasses random variables that are generated by processes that are discrete, continuous, neither, or mixed. The variance can also be thought of as the covariance of a random variable with itself:

${\displaystyle \operatorname {Var} (X)=\operatorname {Cov} (X,X).}$

The variance is also equivalent to the second cumulant of a probability distribution that generates ${\displaystyle X}$. The variance is typically designated as ${\displaystyle \operatorname {Var} (X)}$, ${\displaystyle \sigma _{X}^{2}}$, or simply ${\displaystyle \sigma ^{2}}$ (pronounced "sigma squared"). The expression for the variance can be expanded:

{\displaystyle {\begin{aligned}\operatorname {Var} (X)&=\operatorname {E} \left[(X-\operatorname {E} [X])^{2}\right]\\[4pt]&=\operatorname {E} \left[X^{2}-2X\operatorname {E} [X]+\operatorname {E} [X]^{2}\right]\\[4pt]&=\operatorname {E} \left[X^{2}\right]-2\operatorname {E} [X]\operatorname {E} [X]+\operatorname {E} [X]^{2}\\[4pt]&=\operatorname {E} \left[X^{2}\right]-\operatorname {E} [X]^{2}\end{aligned}}}

In other words, the variance of X is equal to the mean of the square of X minus the square of the mean of X. This equation should not be used for computations using floating point arithmetic because it suffers from catastrophic cancellation if the two components of the equation are similar in magnitude. There exist numerically stable alternatives.

### Discrete random variable

If the generator of random variable ${\displaystyle X}$ is discrete with probability mass function ${\displaystyle x_{1}\mapsto p_{1},x_{2}\mapsto p_{2},\ldots ,x_{n}\mapsto p_{n}}$ then

${\displaystyle \operatorname {Var} (X)=\sum _{i=1}^{n}p_{i}\cdot (x_{i}-\mu )^{2},}$

or equivalently

${\displaystyle \operatorname {Var} (X)=\left(\sum _{i=1}^{n}p_{i}x_{i}^{2}\right)-\mu ^{2},}$

where ${\displaystyle \mu }$ is the expected value, i.e.

${\displaystyle \mu =\sum _{i=1}^{n}p_{i}x_{i}.}$

(When such a discrete weighted variance is specified by weights whose sum is not 1, then one divides by the sum of the weights.)

The variance of a set of ${\displaystyle n}$ equally likely values can be written as

${\displaystyle \operatorname {Var} (X)={\frac {1}{n}}\sum _{i=1}^{n}(x_{i}-\mu )^{2},}$

where ${\displaystyle \mu }$ is the average value, i.e.,

${\displaystyle \mu ={\frac {1}{n}}\sum _{i=1}^{n}x_{i}.}$

The variance of a set of ${\displaystyle n}$ equally likely values can be equivalently expressed, without directly referring to the mean, in terms of squared deviations of all points from each other:[1]

${\displaystyle \operatorname {Var} (X)={\frac {1}{n^{2}}}\sum _{i=1}^{n}\sum _{j=1}^{n}{\frac {1}{2}}(x_{i}-x_{j})^{2}={\frac {1}{n^{2}}}\sum _{i}\sum _{j>i}(x_{i}-x_{j})^{2}.}$

### Continuous random variable

If the random variable ${\displaystyle X}$ represents samples generated by a continuous distribution with probability density function ${\displaystyle f(x)}$, and ${\displaystyle F(x)}$ is the corresponding cumulative distribution function, then the population variance is given by

{\displaystyle {\begin{aligned}\operatorname {Var} (X)=\sigma ^{2}&=\int (x-\mu )^{2}f(x)\,dx\\[4pt]&=\int x^{2}f(x)\,dx-2\mu \int xf(x)\,dx+\int \mu ^{2}f(x)\,dx\\[4pt]&=\int x^{2}\,dF(x)-2\mu \int x\,dF(x)+\mu ^{2}\int \,dF(x)\\[4pt]&=\int x^{2}\,dF(x)-2\mu \cdot \mu +\mu ^{2}\cdot 1\\[4pt]&=\int x^{2}\,dF(x)-\mu ^{2},\end{aligned}}}

or equivalently and conventionally,

${\displaystyle \operatorname {Var} (X)=\int x^{2}f(x)\,dx-\mu ^{2},}$

where ${\displaystyle \mu }$ is the expected value of ${\displaystyle X}$ given by

${\displaystyle \mu =\int xf(x)\,dx=\int x\,dF(x),}$

with the integrals being definite integrals taken for ${\displaystyle x}$ ranging over the range of ${\displaystyle X.}$

If a continuous distribution does not have a finite expected value, as is the case for the Cauchy distribution, it does not have a variance either. Many other distributions for which the expected value does exist also do not have a finite variance because the integral in the variance definition diverges. An example is a Pareto distribution whose index ${\displaystyle k}$ satisfies ${\displaystyle 1

## Examples

### Normal distribution

The normal distribution with parameters ${\displaystyle \mu }$ and ${\displaystyle \sigma }$ is a continuous distribution (also known as gaussian distribution) whose probability density function is given by

${\displaystyle f(x)={\frac {1}{\sqrt {2\pi \sigma ^{2}}}}e^{-{\frac {(x-\mu )^{2}}{2\sigma ^{2}}}}.}$

In this distribution, ${\displaystyle \operatorname {E} [X]=\mu }$ and the variance ${\displaystyle \operatorname {Var} (X)}$ is related with ${\displaystyle \sigma }$ via

${\displaystyle \operatorname {Var} (X)=\int _{-\infty }^{\infty }{\frac {x^{2}}{\sqrt {2\pi \sigma ^{2}}}}e^{-{\frac {(x-\mu )^{2}}{2\sigma ^{2}}}}\,dx-\mu ^{2}=\sigma ^{2}.}$

The role of the normal distribution in the central limit theorem is in part responsible for the prevalence of the variance in probability and statistics.

### Exponential distribution

The exponential distribution with parameter ${\displaystyle \lambda }$ is a continuous distribution whose support is the semi-infinite interval ${\displaystyle [0,\infty )}$. Its probability density function is given by

${\displaystyle f(x)=\lambda e^{-\lambda x}}$

and it has expected value ${\displaystyle \mu =\lambda ^{-1}}$. The variance is equal to

${\displaystyle \operatorname {Var} (X)=\int _{0}^{\infty }x^{2}\lambda e^{-\lambda x}\,dx-\mu ^{2}=\lambda ^{-2}.}$

So for an exponentially distributed random variable, ${\displaystyle \sigma ^{2}=\mu ^{2}.}$

### Poisson distribution

The Poisson distribution with parameter ${\displaystyle \lambda }$ is a discrete distribution for ${\displaystyle k=0,1,2,\ldots }$. Its probability mass function is given by

${\displaystyle p(k)={\frac {\lambda ^{k}}{k!}}e^{-\lambda },}$

and it has expected value ${\displaystyle \mu =\lambda }$. The variance is equal to

${\displaystyle \operatorname {Var} (X)=\left(\sum _{k=0}^{\infty }k^{2}{\frac {\lambda ^{k}}{k!}}e^{-\lambda }\right)-\mu ^{2}=\lambda ,}$

So for a Poisson-distributed random variable, ${\displaystyle \sigma ^{2}=\mu }$.

### Binomial distribution

The binomial distribution with parameters ${\displaystyle n}$ and ${\displaystyle p}$ is a discrete distribution for ${\displaystyle k=0,1,2,\ldots ,n}$. Its probability mass function is given by

${\displaystyle p(k)={n \choose k}p^{k}(1-p)^{n-k},}$

and it has expected value ${\displaystyle \mu =np}$. The variance is equal to

${\displaystyle \operatorname {Var} (X)=\left(\sum _{k=0}^{n}k^{2}{n \choose k}p^{k}(1-p)^{n-k}\right)-\mu ^{2}=np(1-p).}$

As a simple example, the binomial distribution with ${\displaystyle p=1/2}$ describes the probability of getting ${\displaystyle k}$ heads in ${\displaystyle n}$ tosses of a fair coin. Thus the expected value of the number of heads is ${\displaystyle n/2,}$ and the variance is ${\displaystyle n/4.}$

### Fair die

A fair six-sided die can be modeled as a discrete random variable, X, with outcomes 1 through 6, each with equal probability 1/6. The expected value of X is ${\displaystyle (1+2+3+4+5+6)/6=7/2.}$ Therefore, the variance of X is

{\displaystyle {\begin{aligned}\operatorname {Var} (X)&=\sum _{i=1}^{6}{\frac {1}{6}}\left(i-{\frac {7}{2}}\right)^{2}\\[5pt]&={\frac {1}{6}}\left((-5/2)^{2}+(-3/2)^{2}+(-1/2)^{2}+(1/2)^{2}+(3/2)^{2}+(5/2)^{2}\right)\\[5pt]&={\frac {35}{12}}\approx 2.92.\end{aligned}}}

The general formula for the variance of the outcome, X, of an n-sided die is

{\displaystyle {\begin{aligned}\operatorname {Var} (X)&=\operatorname {E} (X^{2})-(\operatorname {E} (X))^{2}\\[5pt]&={\frac {1}{n}}\sum _{i=1}^{n}i^{2}-\left({\frac {1}{n}}\sum _{i=1}^{n}i\right)^{2}\\[5pt]&={\frac {(n+1)(2n+1)}{6}}-\left({\frac {n+1}{2}}\right)^{2}\\[4pt]&={\frac {n^{2}-1}{12}}.\end{aligned}}}

## Properties

### Basic properties

Variance is non-negative because the squares are positive or zero:

${\displaystyle \operatorname {Var} (X)\geq 0.}$

The variance of a constant random variable is zero, and if the variance of a variable in a data set is 0, then all the entries have the same value:

${\displaystyle P(X=a)=1\iff \operatorname {Var} (X)=0.}$

Variance is invariant with respect to changes in a location parameter. That is, if a constant is added to all values of the variable, the variance is unchanged:

${\displaystyle \operatorname {Var} (X+a)=\operatorname {Var} (X).}$

If all values are scaled by a constant, the variance is scaled by the square of that constant:

${\displaystyle \operatorname {Var} (aX)=a^{2}\operatorname {Var} (X).}$

The variance of a sum of two random variables is given by

${\displaystyle \operatorname {Var} (aX+bY)=a^{2}\operatorname {Var} (X)+b^{2}\operatorname {Var} (Y)+2ab\,\operatorname {Cov} (X,Y),}$
${\displaystyle \operatorname {Var} (aX-bY)=a^{2}\operatorname {Var} (X)+b^{2}\operatorname {Var} (Y)-2ab\,\operatorname {Cov} (X,Y),}$

where Cov(⋅, ⋅) is the covariance. In general we have for the sum of ${\displaystyle N}$ random variables ${\displaystyle \{X_{1},\dots ,X_{N}\}}$:

${\displaystyle \operatorname {Var} \left(\sum _{i=1}^{N}X_{i}\right)=\sum _{i,j=1}^{N}\operatorname {Cov} (X_{i},X_{j})=\sum _{i=1}^{N}\operatorname {Var} (X_{i})+\sum _{i\neq j}\operatorname {Cov} (X_{i},X_{j}).}$

These results lead to the variance of a linear combination as:

{\displaystyle {\begin{aligned}\operatorname {Var} \left(\sum _{i=1}^{N}a_{i}X_{i}\right)&=\sum _{i,j=1}^{N}a_{i}a_{j}\operatorname {Cov} (X_{i},X_{j})\\&=\sum _{i=1}^{N}a_{i}^{2}\operatorname {Var} (X_{i})+\sum _{i\not =j}a_{i}a_{j}\operatorname {Cov} (X_{i},X_{j})\\&=\sum _{i=1}^{N}a_{i}^{2}\operatorname {Var} (X_{i})+2\sum _{1\leq i

If the random variables ${\displaystyle X_{1},\dots ,X_{N}}$ are such that

${\displaystyle \operatorname {Cov} (X_{i},X_{j})=0\ ,\ \forall \ (i\neq j),}$

they are said to be uncorrelated. It follows immediately from the expression given earlier that if the random variables ${\displaystyle X_{1},\dots ,X_{N}}$ are uncorrelated, then the variance of their sum is equal to the sum of their variances, or, expressed symbolically:

${\displaystyle \operatorname {Var} \left(\sum _{i=1}^{N}X_{i}\right)=\sum _{i=1}^{N}\operatorname {Var} (X_{i}).}$

Since independent random variables are always uncorrelated, the equation above holds in particular when the random variables ${\displaystyle X_{1},\dots ,X_{n}}$ are independent. Thus independence is sufficient but not necessary for the variance of the sum to equal the sum of the variances.

### Sum of uncorrelated variables (Bienaymé formula)

One reason for the use of the variance in preference to other measures of dispersion is that the variance of the sum (or the difference) of uncorrelated random variables is the sum of their variances:

${\displaystyle \operatorname {Var} \left(\sum _{i=1}^{n}X_{i}\right)=\sum _{i=1}^{n}\operatorname {Var} (X_{i}).}$

This statement is called the Bienaymé formula[2] and was discovered in 1853.[3][4] It is often made with the stronger condition that the variables are independent, but being uncorrelated suffices. So if all the variables have the same variance σ2, then, since division by n is a linear transformation, this formula immediately implies that the variance of their mean is

${\displaystyle \operatorname {Var} \left({\overline {X}}\right)=\operatorname {Var} \left({\frac {1}{n}}\sum _{i=1}^{n}X_{i}\right)={\frac {1}{n^{2}}}\sum _{i=1}^{n}\operatorname {Var} \left(X_{i}\right)={\frac {1}{n^{2}}}n\sigma ^{2}={\frac {\sigma ^{2}}{n}}.}$

That is, the variance of the mean decreases when n increases. This formula for the variance of the mean is used in the definition of the standard error of the sample mean, which is used in the central limit theorem.

To prove the initial statement, it suffices to show that

${\displaystyle \operatorname {Var} (X+Y)=\operatorname {Var} (X)+\operatorname {Var} (Y).}$

The general result then follows by induction. Starting with the definition,

{\displaystyle {\begin{aligned}\operatorname {Var} (X+Y)&=\operatorname {E} [(X+Y)^{2}]-(\operatorname {E} [X+Y])^{2}\\[5pt]&=\operatorname {E} [X^{2}+2XY+Y^{2}]-(\operatorname {E} [X]+\operatorname {E} [Y])^{2}.\end{aligned}}}

Using the linearity of the expectation operator and the assumption of independence (or uncorrelatedness) of X and Y, this further simplifies as follows:

{\displaystyle {\begin{aligned}\operatorname {Var} (X+Y)&=\operatorname {E} [X^{2}]+2\operatorname {E} [XY]+\operatorname {E} [Y^{2}]-(\operatorname {E} [X]^{2}+2\operatorname {E} [X]\operatorname {E} [Y]+\operatorname {E} [Y]^{2})\\[5pt]&=\operatorname {E} [X^{2}]+\operatorname {E} [Y^{2}]-\operatorname {E} [X]^{2}-\operatorname {E} [Y]^{2}\\[5pt]&=\operatorname {Var} (X)+\operatorname {Var} (Y).\end{aligned}}}

### Sum of variables

In general the variance of the sum of n variables is the sum of their covariances:

${\displaystyle \operatorname {Var} \left(\sum _{i=1}^{n}X_{i}\right)=\sum _{i=1}^{n}\sum _{j=1}^{n}\operatorname {Cov} (X_{i},X_{j})=\sum _{i=1}^{n}\operatorname {Var} (X_{i})+2\sum _{1\leq i

(Note: The second equality comes from the fact that Cov(Xi,Xi) = Var(Xi).)

Here Cov(⋅, ⋅) is the covariance, which is zero for independent random variables (if it exists). The formula states that the variance of a sum is equal to the sum of all elements in the covariance matrix of the components. The next expression states equivalently that the variance of the sum is the sum of the diagonal of covariance matrix plus two times the sum of its upper triangular elements (or its lower triangular elements); this emphasizes that the covariance matrix is symmetric. This formula is used in the theory of Cronbach's alpha in classical test theory.

So if the variables have equal variance σ2 and the average correlation of distinct variables is ρ, then the variance of their mean is

${\displaystyle \operatorname {Var} ({\overline {X}})={\frac {\sigma ^{2}}{n}}+{\frac {n-1}{n}}\rho \sigma ^{2}.}$

This implies that the variance of the mean increases with the average of the correlations. In other words, additional correlated observations are not as effective as additional independent observations at reducing the uncertainty of the mean. Moreover, if the variables have unit variance, for example if they are standardized, then this simplifies to

${\displaystyle \operatorname {Var} ({\overline {X}})={\frac {1}{n}}+{\frac {n-1}{n}}\rho .}$

This formula is used in the Spearman–Brown prediction formula of classical test theory. This converges to ρ if n goes to infinity, provided that the average correlation remains constant or converges too. So for the variance of the mean of standardized variables with equal correlations or converging average correlation we have

${\displaystyle \lim _{n\to \infty }\operatorname {Var} ({\overline {X}})=\rho .}$

Therefore, the variance of the mean of a large number of standardized variables is approximately equal to their average correlation. This makes clear that the sample mean of correlated variables does not generally converge to the population mean, even though the law of large numbers states that the sample mean will converge for independent variables.

### Matrix notation for the variance of a linear combination

Define ${\displaystyle X}$ as a column vector of ${\displaystyle n}$ random variables ${\displaystyle X_{1},\ldots ,X_{n}}$, and ${\displaystyle c}$ as a column vector of ${\displaystyle n}$ scalars ${\displaystyle c_{1},\ldots ,c_{n}}$. Therefore, ${\displaystyle c^{T}X}$ is a linear combination of these random variables, where ${\displaystyle c^{T}}$ denotes the transpose of ${\displaystyle c}$. Also let ${\displaystyle \Sigma }$ be the covariance matrix of ${\displaystyle X}$. The variance of ${\displaystyle c^{T}X}$ is then given by:[5]

${\displaystyle \operatorname {Var} (c^{T}X)=c^{T}\Sigma c.}$

### Weighted sum of variables

The scaling property and the Bienaymé formula, along with the property of the covariance Cov(aXbY) = ab Cov(XY) jointly imply that

${\displaystyle \operatorname {Var} (aX\pm bY)=a^{2}\operatorname {Var} (X)+b^{2}\operatorname {Var} (Y)\pm 2ab\,\operatorname {Cov} (X,Y).}$

This implies that in a weighted sum of variables, the variable with the largest weight will have a disproportionally large weight in the variance of the total. For example, if X and Y are uncorrelated and the weight of X is two times the weight of Y, then the weight of the variance of X will be four times the weight of the variance of Y.

The expression above can be extended to a weighted sum of multiple variables:

${\displaystyle \operatorname {Var} \left(\sum _{i}^{n}a_{i}X_{i}\right)=\sum _{i=1}^{n}a_{i}^{2}\operatorname {Var} (X_{i})+2\sum _{1\leq i}\sum _{

### Product of independent variables

If two variables X and Y are independent, the variance of their product is given by[6]

{\displaystyle {\begin{aligned}\operatorname {Var} (XY)&=[\operatorname {E} (X)]^{2}\operatorname {Var} (Y)+[\operatorname {E} (Y)]^{2}\operatorname {Var} (X)+\operatorname {Var} (X)\operatorname {Var} (Y).\end{aligned}}}

Equivalently, using the basic properties of expectation, it is given by

${\displaystyle \operatorname {Var} (XY)=\operatorname {E} (X^{2})\operatorname {E} (Y^{2})-[\operatorname {E} (X)]^{2}[\operatorname {E} (Y)]^{2}.}$

### Product of statistically dependent variables

In general, if two variables are statistically dependent, the variance of their product is given by:

{\displaystyle {\begin{aligned}\operatorname {Var} (XY)={}&\operatorname {E} [X^{2}Y^{2}]-[\operatorname {E} (XY)]^{2}\\[5pt]={}&\operatorname {Cov} (X^{2},Y^{2})+\operatorname {E} (X^{2})\operatorname {E} (Y^{2})-[\operatorname {E} (XY)]^{2}\\[5pt]={}&\operatorname {Cov} (X^{2},Y^{2})+(\operatorname {Var} (X)+[\operatorname {E} (X)]^{2})(\operatorname {Var} (Y)+[\operatorname {E} (Y)]^{2})\\[5pt]&{}-[\operatorname {Cov} (X,Y)+\operatorname {E} (X)\operatorname {E} (Y)]^{2}\end{aligned}}}

### Decomposition

The general formula for variance decomposition or the law of total variance is: If ${\displaystyle X}$ and ${\displaystyle Y}$ are two random variables, and the variance of ${\displaystyle X}$ exists, then

${\displaystyle \operatorname {Var} [X]=\operatorname {E} (\operatorname {Var} [X\mid Y])+\operatorname {Var} (\operatorname {E} [X\mid Y]).}$

The conditional expectation ${\displaystyle \operatorname {E} (X\mid Y)}$ of ${\displaystyle X}$ given ${\displaystyle Y}$, and the conditional variance ${\displaystyle \operatorname {Var} (X\mid Y)}$ may be understood as follows. Given any particular value y of the random variable Y, there is a conditional expectation ${\displaystyle \operatorname {E} (X\mid Y=y)}$ given the event Y = y. This quantity depends on the particular value y; it is a function ${\displaystyle g(y)=\operatorname {E} (X\mid Y=y)}$. That same function evaluated at the random variable Y is the conditional expectation ${\displaystyle \operatorname {E} (X\mid Y)=g(Y).}$

In particular, if ${\displaystyle Y}$ is a discrete random variable assuming possible values ${\displaystyle y_{1},y_{2},y_{3}\ldots }$ with corresponding probabilities ${\displaystyle p_{1},p_{2},p_{3}\ldots ,}$, then in the formula for total variance, the first term on the right-hand side becomes

${\displaystyle \operatorname {E} (\operatorname {Var} [X\mid Y])=\sum _{i}p_{i}\sigma _{i}^{2},}$

where ${\displaystyle \sigma _{i}^{2}=\operatorname {Var} [X\mid Y=y_{i}]}$. Similarly, the second term on the right-hand side becomes

${\displaystyle \operatorname {Var} (\operatorname {E} [X\mid Y])=\sum _{i}p_{i}\mu _{i}^{2}-\left(\sum _{i}p_{i}\mu _{i}\right)^{2}=\sum _{i}p_{i}\mu _{i}^{2}-\mu ^{2},}$

where ${\displaystyle \mu _{i}=\operatorname {E} [X\mid Y=y_{i}]}$ and ${\displaystyle \mu =\sum _{i}p_{i}\mu _{i}}$. Thus the total variance is given by

${\displaystyle \operatorname {Var} [X]=\sum _{i}p_{i}\sigma _{i}^{2}+\left(\sum _{i}p_{i}\mu _{i}^{2}-\mu ^{2}\right).}$

A similar formula is applied in analysis of variance, where the corresponding formula is

${\displaystyle {\mathit {MS}}_{\text{total}}={\mathit {MS}}_{\text{between}}+{\mathit {MS}}_{\text{within}};}$

here ${\displaystyle {\mathit {MS}}}$ refers to the Mean of the Squares. In linear regression analysis the corresponding formula is

${\displaystyle {\mathit {MS}}_{\text{total}}={\mathit {MS}}_{\text{regression}}+{\mathit {MS}}_{\text{residual}}.}$

This can also be derived from the additivity of variances, since the total (observed) score is the sum of the predicted score and the error score, where the latter two are uncorrelated.

Similar decompositions are possible for the sum of squared deviations (sum of squares, ${\displaystyle {\mathit {SS}}}$):

${\displaystyle {\mathit {SS}}_{\text{total}}={\mathit {SS}}_{\text{between}}+{\mathit {SS}}_{\text{within}},}$
${\displaystyle {\mathit {SS}}_{\text{total}}={\mathit {SS}}_{\text{regression}}+{\mathit {SS}}_{\text{residual}}.}$

### Formulae for the variance

A formula often used for deriving the variance of a theoretical distribution is as follows:

${\displaystyle \operatorname {Var} (X)=\operatorname {E} (X^{2})-(\operatorname {E} (X))^{2}.}$

This will be useful when it is possible to derive formulae for the expected value and for the expected value of the square.

This formula is also sometimes used in connection with the sample variance. While useful for hand calculations, it is not advised for computer calculations as it suffers from catastrophic cancellation if the two components of the equation are similar in magnitude and floating point arithmetic is used. This is discussed in the article Algorithms for calculating variance.

### Calculation from the CDF

The population variance for a non-negative random variable can be expressed in terms of the cumulative distribution function F using

${\displaystyle 2\int _{0}^{\infty }u(1-F(u))\,du-{\Big (}\int _{0}^{\infty }(1-F(u))\,du{\Big )}^{2}.}$

This expression can be used to calculate the variance in situations where the CDF, but not the density, can be conveniently expressed.

### Characteristic property

The second moment of a random variable attains the minimum value when taken around the first moment (i.e., mean) of the random variable, i.e. ${\displaystyle \mathrm {argmin} _{m}\,\mathrm {E} \left(\left(X-m\right)^{2}\right)=\mathrm {E} (X)}$. Conversely, if a continuous function ${\displaystyle \varphi }$ satisfies ${\displaystyle \mathrm {argmin} _{m}\,\mathrm {E} (\varphi (X-m))=\mathrm {E} (X)}$ for all random variables X, then it is necessarily of the form ${\displaystyle \varphi (x)=ax^{2}+b}$, where a > 0. This also holds in the multidimensional case.[7]

### Units of measurement

Unlike expected absolute deviation, the variance of a variable has units that are the square of the units of the variable itself. For example, a variable measured in meters will have a variance measured in meters squared. For this reason, describing data sets via their standard deviation or root mean square deviation is often preferred over using the variance. In the dice example the standard deviation is √2.9 ≈ 1.7, slightly larger than the expected absolute deviation of 1.5.

The standard deviation and the expected absolute deviation can both be used as an indicator of the "spread" of a distribution. The standard deviation is more amenable to algebraic manipulation than the expected absolute deviation, and, together with variance and its generalization covariance, is used frequently in theoretical statistics; however the expected absolute deviation tends to be more robust as it is less sensitive to outliers arising from measurement anomalies or an unduly heavy-tailed distribution.

## Approximating the variance of a function

The delta method uses second-order Taylor expansions to approximate the variance of a function of one or more random variables: see Taylor expansions for the moments of functions of random variables. For example, the approximate variance of a function of one variable is given by

${\displaystyle \operatorname {Var} \left[f(X)\right]\approx \left(f'(\operatorname {E} \left[X\right])\right)^{2}\operatorname {Var} \left[X\right]}$

provided that f is twice differentiable and that the mean and variance of X are finite.

## Population variance and sample variance

Real-world observations such as the measurements of yesterday's rain throughout the day typically cannot be complete sets of all possible observations that could be made. As such, the variance calculated from the finite set will in general not match the variance that would have been calculated from the full population of possible observations. This means that one estimates the mean and variance that would have been calculated from an omniscient set of observations by using an estimator equation. The estimator is a function of the sample of n observations drawn without observational bias from the whole population of potential observations. In this example that sample would be the set of actual measurements of yesterday's rainfall from available rain gauges within the geography of interest.

The simplest estimators for population mean and population variance are simply the mean and variance of the sample, the sample mean and (uncorrected) sample variance – these are consistent estimators (they converge to the correct value as the number of samples increases), but can be improved. Estimating the population variance by taking the sample's variance is close to optimal in general, but can be improved in two ways. Most simply, the sample variance is computed as an average of squared deviations about the (sample) mean, by dividing by n. However, using values other than n improves the estimator in various ways. Four common values for the denominator are n, n − 1, n + 1, and n − 1.5: n is the simplest (population variance of the sample), n − 1 eliminates bias, n + 1 minimizes mean squared error for the normal distribution, and n − 1.5 mostly eliminates bias in unbiased estimation of standard deviation for the normal distribution.

Firstly, if the omniscient mean is unknown (and is computed as the sample mean), then the sample variance is a biased estimator: it underestimates the variance by a factor of (n − 1) / n; correcting by this factor (dividing by n − 1 instead of n) is called Bessel's correction. The resulting estimator is unbiased, and is called the (corrected) sample variance or unbiased sample variance. For example, when n = 1 the variance of a single observation about the sample mean (itself) is obviously zero regardless of the population variance. If the mean is determined in some other way than from the same samples used to estimate the variance then this bias does not arise and the variance can safely be estimated as that of the samples about the (independently known) mean.

Secondly, the sample variance does not generally minimize mean squared error between sample variance and population variance. Correcting for bias often makes this worse: one can always choose a scale factor that performs better than the corrected sample variance, though the optimal scale factor depends on the excess kurtosis of the population (see mean squared error: variance), and introduces bias. This always consists of scaling down the unbiased estimator (dividing by a number larger than n − 1), and is a simple example of a shrinkage estimator: one "shrinks" the unbiased estimator towards zero. For the normal distribution, dividing by n + 1 (instead of n − 1 or n) minimizes mean squared error. The resulting estimator is biased, however, and is known as the biased sample variation.

### Population variance

In general, the population variance of a finite population of size N with values xi is given by

{\displaystyle {\begin{aligned}\sigma ^{2}&={\frac {1}{N}}\sum _{i=1}^{N}\left(x_{i}-\mu \right)^{2}={\frac {1}{N}}\sum _{i=1}^{N}\left(x_{i}^{2}-2\mu x_{i}+\mu ^{2}\right)\\[5pt]&=\left({\frac {1}{N}}\sum _{i=1}^{N}x_{i}^{2}\right)-2\mu \left({\frac {1}{N}}\sum _{i=1}^{N}x_{i}\right)+\mu ^{2}\\[5pt]&=\left({\frac {1}{N}}\sum _{i=1}^{N}x_{i}^{2}\right)-\mu ^{2}\end{aligned}}}

where the population mean is

${\displaystyle \mu ={\frac {1}{N}}\sum _{i=1}^{N}x_{i}.}$

The population variance can also be computed using

${\displaystyle \sigma ^{2}={\frac {1}{N^{2}}}\sum _{i

This is true because

{\displaystyle {\begin{aligned}{\frac {1}{2N^{2}}}\sum _{i,j=1}^{N}\left(x_{i}-x_{j}\right)^{2}&={\frac {1}{2N^{2}}}\sum _{i,j=1}^{N}\left(x_{i}^{2}-2x_{i}x_{j}+x_{j}^{2}\right)\\[5pt]&={\frac {1}{2N}}\sum _{j=1}^{N}\left({\frac {1}{N}}\sum _{i=1}^{N}x_{i}^{2}\right)-\left({\frac {1}{N}}\sum _{i=1}^{N}x_{i}\right)\left({\frac {1}{N}}\sum _{j=1}^{N}x_{j}\right)\\[5pt]&\quad +{\frac {1}{2N}}\sum _{i=1}^{N}\left({\frac {1}{N}}\sum _{j=1}^{N}x_{j}^{2}\right)\\[5pt]&={\frac {1}{2}}\left(\sigma ^{2}+\mu ^{2}\right)-\mu ^{2}+{\frac {1}{2}}\left(\sigma ^{2}+\mu ^{2}\right)\\[5pt]&=\sigma ^{2}\end{aligned}}}

The population variance matches the variance of the generating probability distribution. In this sense, the concept of population can be extended to continuous random variables with infinite populations.

### Sample variance

In many practical situations, the true variance of a population is not known a priori and must be computed somehow. When dealing with extremely large populations, it is not possible to count every object in the population, so the computation must be performed on a sample of the population.[8] Sample variance can also be applied to the estimation of the variance of a continuous distribution from a sample of that distribution.

We take a sample with replacement of n values Y1, ..., Yn from the population, where n < N, and estimate the variance on the basis of this sample.[9] Directly taking the variance of the sample data gives the average of the squared deviations:

${\displaystyle \sigma _{Y}^{2}={\frac {1}{n}}\sum _{i=1}^{n}\left(Y_{i}-{\overline {Y}}\right)^{2}=\left({\frac {1}{n}}\sum _{i=1}^{n}Y_{i}^{2}\right)-{\overline {Y}}^{2}={\frac {1}{n^{2}}}\sum _{i,j\,:\,i

Here, ${\displaystyle {\overline {Y}}}$ denotes the sample mean:

${\displaystyle {\overline {Y}}={\frac {1}{n}}\sum _{i=1}^{n}Y_{i}.}$

Since the Yi are selected randomly, both ${\displaystyle {\overline {Y}}}$ and ${\displaystyle \sigma _{Y}^{2}}$ are random variables. Their expected values can be evaluated by averaging over the ensemble of all possible samples {Yi} of size n from the population. For ${\displaystyle \sigma _{Y}^{2}}$ this gives:

{\displaystyle {\begin{aligned}\operatorname {E} [\sigma _{Y}^{2}]&=\operatorname {E} \left[{\frac {1}{n}}\sum _{i=1}^{n}\left(Y_{i}-{\frac {1}{n}}\sum _{j=1}^{n}Y_{j}\right)^{2}\right]\\[5pt]&={\frac {1}{n}}\sum _{i=1}^{n}\operatorname {E} \left[Y_{i}^{2}-{\frac {2}{n}}Y_{i}\sum _{j=1}^{n}Y_{j}+{\frac {1}{n^{2}}}\sum _{j=1}^{n}Y_{j}\sum _{k=1}^{n}Y_{k}\right]\\[5pt]&={\frac {1}{n}}\sum _{i=1}^{n}\left[{\frac {n-2}{n}}\operatorname {E} [Y_{i}^{2}]-{\frac {2}{n}}\sum _{j\neq i}\operatorname {E} [Y_{i}Y_{j}]+{\frac {1}{n^{2}}}\sum _{j=1}^{n}\sum _{k\neq j}^{n}\operatorname {E} [Y_{j}Y_{k}]+{\frac {1}{n^{2}}}\sum _{j=1}^{n}\operatorname {E} [Y_{j}^{2}]\right]\\[5pt]&={\frac {1}{n}}\sum _{i=1}^{n}\left[{\frac {n-2}{n}}(\sigma ^{2}+\mu ^{2})-{\frac {2}{n}}(n-1)\mu ^{2}+{\frac {1}{n^{2}}}n(n-1)\mu ^{2}+{\frac {1}{n}}(\sigma ^{2}+\mu ^{2})\right]\\[5pt]&={\frac {n-1}{n}}\sigma ^{2}.\end{aligned}}}

Hence ${\displaystyle \sigma _{Y}^{2}}$ gives an estimate of the population variance that is biased by a factor of ${\displaystyle {\frac {n-1}{n}}}$. For this reason, ${\displaystyle \sigma _{Y}^{2}}$ is referred to as the biased sample variance. Correcting for this bias yields the unbiased sample variance:

${\displaystyle s^{2}={\frac {n}{n-1}}\sigma _{Y}^{2}={\frac {n}{n-1}}\left({\frac {1}{n}}\sum _{i=1}^{n}\left(Y_{i}-{\overline {Y}}\right)^{2}\right)={\frac {1}{n-1}}\sum _{i=1}^{n}\left(Y_{i}-{\overline {Y}}\right)^{2}}$

Either estimator may be simply referred to as the sample variance when the version can be determined by context. The same proof is also applicable for samples taken from a continuous probability distribution.

The use of the term n − 1 is called Bessel's correction, and it is also used in sample covariance and the sample standard deviation (the square root of variance). The square root is a concave function and thus introduces negative bias (by Jensen's inequality), which depends on the distribution, and thus the corrected sample standard deviation (using Bessel's correction) is biased. The unbiased estimation of standard deviation is a technically involved problem, though for the normal distribution using the term n − 1.5 yields an almost unbiased estimator.

The unbiased sample variance is a U-statistic for the function ƒ(y1y2) = (y1 − y2)2/2, meaning that it is obtained by averaging a 2-sample statistic over 2-element subsets of the population.

### Distribution of the sample variance

Distribution and cumulative distribution of S22, for various values of ν = n − 1, when the yi are independent normally distributed.

Being a function of random variables, the sample variance is itself a random variable, and it is natural to study its distribution. In the case that Yi are independent observations from a normal distribution, Cochran's theorem shows that S2 follows a scaled chi-squared distribution:[10]

${\displaystyle (n-1){\frac {S^{2}}{\sigma ^{2}}}\sim \chi _{n-1}^{2}.}$

As a direct consequence, it follows that

${\displaystyle \operatorname {E} (S^{2})=\operatorname {E} \left({\frac {\sigma ^{2}}{n-1}}\chi _{n-1}^{2}\right)=\sigma ^{2},}$

and[11]

${\displaystyle \operatorname {Var} [s^{2}]=\operatorname {Var} \left({\frac {\sigma ^{2}}{n-1}}\chi _{n-1}^{2}\right)={\frac {\sigma ^{4}}{(n-1)^{2}}}\operatorname {Var} \left(\chi _{n-1}^{2}\right)={\frac {2\sigma ^{4}}{n-1}}.}$

If the Yi are independent and identically distributed, but not necessarily normally distributed, then[12][13]

${\displaystyle \operatorname {E} [S^{2}]=\sigma ^{2},\quad \operatorname {Var} [S^{2}]={\frac {\sigma ^{4}}{n}}\left((\kappa -1)+{\frac {2}{n-1}}\right)={\frac {1}{n}}\left(\mu _{4}-{\frac {n-3}{n-1}}\sigma ^{4}\right),}$

where κ is the kurtosis of the distribution and μ4 is the fourth central moment.

If the conditions of the law of large numbers hold for the squared observations, s2 is a consistent estimator of σ2. One can see indeed that the variance of the estimator tends asymptotically to zero. An asymptotically equivalent formula was given in Kenney and Keeping (1951:164), Rose and Smith (2002:264), and Weisstein (n.d.).[14][15][16]

### Samuelson's inequality

Samuelson's inequality is a result that states bounds on the values that individual observations in a sample can take, given that the sample mean and (biased) variance have been calculated.[17] Values must lie within the limits ${\displaystyle {\bar {y}}\pm \sigma _{Y}(n-1)^{1/2}.}$

### Relations with the harmonic and arithmetic means

It has been shown[18] that for a sample {yi} of real numbers,

${\displaystyle \sigma _{y}^{2}\leq 2y_{\max }(A-H),}$

where ymax is the maximum of the sample, A is the arithmetic mean, H is the harmonic mean of the sample and ${\displaystyle \sigma _{y}^{2}}$ is the (biased) variance of the sample.

This bound has been improved, and it is known that variance is bounded by

${\displaystyle \sigma _{y}^{2}\leq {\frac {y_{\max }(A-H)(y_{\max }-A)}{y_{\max }-H}},}$
${\displaystyle \sigma _{y}^{2}\geq {\frac {y_{\min }(A-H)(A-y_{\min })}{H-y_{\min }}},}$

where ymin is the minimum of the sample.[19]

## Tests of equality of variances

Testing for the equality of two or more variances is difficult. The F test and chi square tests are both adversely affected by non-normality and are not recommended for this purpose.

Several non parametric tests have been proposed: these include the Barton–David–Ansari–Freund–Siegel–Tukey test, the Capon test, Mood test, the Klotz test and the Sukhatme test. The Sukhatme test applies to two variances and requires that both medians be known and equal to zero. The Mood, Klotz, Capon and Barton–David–Ansari–Freund–Siegel–Tukey tests also apply to two variances. They allow the median to be unknown but do require that the two medians are equal.

The Lehmann test is a parametric test of two variances. Of this test there are several variants known. Other tests of the equality of variances include the Box test, the Box–Anderson test and the Moses test.

Resampling methods, which include the bootstrap and the jackknife, may be used to test the equality of variances.

## History

The term variance was first introduced by Ronald Fisher in his 1918 paper The Correlation Between Relatives on the Supposition of Mendelian Inheritance:[20]

The great body of available statistics show us that the deviations of a human measurement from its mean follow very closely the Normal Law of Errors, and, therefore, that the variability may be uniformly measured by the standard deviation corresponding to the square root of the mean square error. When there are two independent causes of variability capable of producing in an otherwise uniform population distributions with standard deviations ${\displaystyle \sigma _{1}}$ and ${\displaystyle \sigma _{2}}$, it is found that the distribution, when both causes act together, has a standard deviation ${\displaystyle {\sqrt {\sigma _{1}^{2}+\sigma _{2}^{2}}}}$. It is therefore desirable in analysing the causes of variability to deal with the square of the standard deviation as the measure of variability. We shall term this quantity the Variance...

Geometric visualisation of the variance of an arbitrary distribution (2, 4, 4, 4, 5, 5, 7, 9):
1. A frequency distribution is constructed.
2. The centroid of the distribution gives its mean.
3. A square with sides equal to the difference of each value from the mean is formed for each value.
4. Arranging the squares into a rectangle with one side equal to the number of values, n, results in the other side being the distribution's variance, σ².

## Moment of inertia

The variance of a probability distribution is analogous to the moment of inertia in classical mechanics of a corresponding mass distribution along a line, with respect to rotation about its center of mass.[citation needed] It is because of this analogy that such things as the variance are called moments of probability distributions.[citation needed] The covariance matrix is related to the moment of inertia tensor for multivariate distributions. The moment of inertia of a cloud of n points with a covariance matrix of ${\displaystyle \Sigma }$ is given by[citation needed]

${\displaystyle I=n(\mathbf {1} _{3\times 3}\operatorname {tr} (\Sigma )-\Sigma ).}$

This difference between moment of inertia in physics and in statistics is clear for points that are gathered along a line. Suppose many points are close to the x axis and distributed along it. The covariance matrix might look like

${\displaystyle \Sigma ={\begin{bmatrix}10&0&0\\0&0.1&0\\0&0&0.1\end{bmatrix}}.}$

That is, there is the most variance in the x direction. Physicists would consider this to have a low moment about the x axis so the moment-of-inertia tensor is

${\displaystyle I=n{\begin{bmatrix}0.2&0&0\\0&10.1&0\\0&0&10.1\end{bmatrix}}.}$

## Semivariance

The semivariance is calculated in the same manner as the variance but only those observations that fall below the mean are included in the calculation. It is sometimes described as a measure of downside risk in an investments context. For skewed distributions, the semivariance can provide additional information that a variance does not.[citation needed]

For inequalities associated with the semivariance, see Chebyshev's inequality § Semivariances.

## Generalizations

### For complex variables

If ${\displaystyle x}$ is a scalar complex-valued random variable, with values in ${\displaystyle \mathbb {C} ,}$ then its variance is ${\displaystyle \operatorname {E} \left[(x-\mu )(x-\mu )^{*}\right],}$ where ${\displaystyle x^{*}}$ is the complex conjugate of ${\displaystyle x.}$ This variance is a real scalar.

### For vector-valued random variables

#### As a matrix

If ${\displaystyle X}$ is a vector-valued random variable, with values in ${\displaystyle \mathbb {R} ^{n},}$ and thought of as a column vector, then a natural generalization of variance is ${\displaystyle \operatorname {E} \left[(X-\mu )(X-\mu )^{\operatorname {T} }\right],}$ where ${\displaystyle \mu =\operatorname {E} (X)}$ and ${\displaystyle X^{\operatorname {T} }}$ is the transpose of ${\displaystyle X,}$ and so is a row vector. The result is a positive semi-definite square matrix, commonly referred to as the variance-covariance matrix (or simply as the covariance matrix).

If ${\displaystyle X}$ is a vector- and complex-valued random variable, with values in ${\displaystyle \mathbb {C} ^{n},}$ then the covariance matrix is ${\displaystyle \operatorname {E} \left[(X-\mu )(X-\mu )^{\dagger }\right],}$ where ${\displaystyle X^{\dagger }}$ is the conjugate transpose of ${\displaystyle X.}$[citation needed] This matrix is also positive semi-definite and square.

#### As a scalar

Another natural generalization of variance for such vector-valued random variables ${\displaystyle X,}$ which results in a scalar value rather than in a matrix, is obtained by interpreting the deviation between the random variable and its mean as the Euclidean distance. This results in ${\displaystyle \operatorname {E} \left[(X-\mu )^{\operatorname {T} }(X-\mu )\right]=\operatorname {tr} (C),}$ which is the trace of the covariance matrix.

## References

1. ^ Yuli Zhang, Huaiyu Wu, Lei Cheng (June 2012). Some new deformation formulas about variance and covariance. Proceedings of 4th International Conference on Modelling, Identification and Control(ICMIC2012). pp. 987–992.CS1 maint: Uses authors parameter (link)
2. ^ Loève, M. (1977) "Probability Theory", Graduate Texts in Mathematics, Volume 45, 4th edition, Springer-Verlag, p. 12.
3. ^ Bienaymé, I.-J. (1853) "Considérations à l'appui de la découverte de Laplace sur la loi de probabilité dans la méthode des moindres carrés", Comptes rendus de l'Académie des sciences Paris, 37, p. 309–317; digital copy available [1]
4. ^ Bienaymé, I.-J. (1867) "Considérations à l'appui de la découverte de Laplace sur la loi de probabilité dans la méthode des moindres carrés", Journal de Mathématiques Pures et Appliquées, Série 2, Tome 12, p. 158–167; digital copy available [2][3]
5. ^ Johnson, Richard; Wichern, Dean (2001). Applied Multivariate Statistical Analysis. Prentice Hall. p. 76. ISBN 0-13-187715-1
6. ^ Goodman, Leo A. (December 1960). "On the Exact Variance of Products". Journal of the American Statistical Association. 55 (292): 708. doi:10.2307/2281592. JSTOR 2281592.
7. ^ Kagan, A.; Shepp, L. A. (1998). "Why the variance?". Statistics & Probability Letters. 38 (4): 329–333. doi:10.1016/S0167-7152(98)00041-8.
8. ^ Navidi, William (2006) Statistics for Engineers and Scientists, McGraw-Hill, pg 14.
9. ^ Montgomery, D. C. and Runger, G. C. (1994) Applied statistics and probability for engineers, page 201. John Wiley & Sons New York
10. ^ Knight K. (2000), Mathematical Statistics, Chapman and Hall, New York. (proposition 2.11)
11. ^ Casella and Berger (2002) Statistical Inference, Example 7.3.3, p. 331[full citation needed]
12. ^ Cho, Eungchun; Cho, Moon Jung; Eltinge, John (2005) The Variance of Sample Variance From a Finite Population. International Journal of Pure and Applied Mathematics 21 (3): 387-394. http://www.ijpam.eu/contents/2005-21-3/10/10.pdf
13. ^ Cho, Eungchun; Cho, Moon Jung (2009) Variance of Sample Variance With Replacement. International Journal of Pure and Applied Mathematics 52 (1): 43–47. http://www.ijpam.eu/contents/2009-52-1/5/5.pdf
14. ^ Kenney, John F.; Keeping, E.S. (1951) Mathematics of Statistics. Part Two. 2nd ed. D. Van Nostrand Company, Inc. Princeton: New Jersey. http://krishikosh.egranth.ac.in/bitstream/1/2025521/1/G2257.pdf
15. ^ Rose, Colin; Smith, Murray D. (2002) Mathematical Statistics with Mathematica. Springer-Verlag, New York. http://www.mathstatica.com/book/Mathematical_Statistics_with_Mathematica.pdf
16. ^ Weisstein, Eric W. (n.d.) Sample Variance Distribution. MathWorld—A Wolfram Web Resource. http://mathworld.wolfram.com/SampleVarianceDistribution.html
17. ^ Samuelson, Paul (1968). "How Deviant Can You Be?". Journal of the American Statistical Association. 63 (324): 1522–1525. doi:10.1080/01621459.1968.10480944. JSTOR 2285901.
18. ^ Mercer, A. McD. (2000). "Bounds for A–G, A–H, G–H, and a family of inequalities of Ky Fan's type, using a general method". J. Math. Anal. Appl. 243 (1): 163–173. doi:10.1006/jmaa.1999.6688.
19. ^ Sharma, R. (2008). "Some more inequalities for arithmetic mean, harmonic mean and variance". J. Math. Inequalities. 2 (1): 109–114. CiteSeerX 10.1.1.551.9397. doi:10.7153/jmi-02-11.
20. ^
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