19
Feb 22

Distribution of the estimator of the error variance

If you are reading the book by Dougherty: this post is about the distribution of the estimator  $s^2$ defined in Chapter 3.

Consider regression

(1) $y=X\beta +e$

where the deterministic matrix $X$ is of size $n\times k,$ satisfies $\det \left( X^{T}X\right) \neq 0$ (regressors are not collinear) and the error $e$ satisfies

(2) $Ee=0,Var(e)=\sigma ^{2}I$

$\beta$ is estimated by $\hat{\beta}=(X^{T}X)^{-1}X^{T}y.$ Denote $P=X(X^{T}X)^{-1}X^{T},$ $Q=I-P.$ Using (1) we see that $\hat{\beta}=\beta +(X^{T}X)^{-1}X^{T}e$ and the residual $r\equiv y-X\hat{\beta}=Qe.$ $\sigma^{2}$ is estimated by

(3) $s^{2}=\left\Vert r\right\Vert ^{2}/\left( n-k\right) =\left\Vert Qe\right\Vert ^{2}/\left( n-k\right) .$

$Q$ is a projector and has properties which are derived from those of $P$

(4) $Q^{T}=Q,$ $Q^{2}=Q.$

If $\lambda$ is an eigenvalue of $Q,$ then multiplying $Qx=\lambda x$ by $Q$ and using the fact that $x\neq 0$ we get $\lambda ^{2}=\lambda .$ Hence eigenvalues of $Q$ can be only $0$ or $1.$ The equation $tr\left( Q\right) =n-k$
tells us that the number of eigenvalues equal to 1 is $n-k$ and the remaining $k$ are zeros. Let $Q=U\Lambda U^{T}$ be the diagonal representation of $Q.$ Here $U$ is an orthogonal matrix,

(5) $U^{T}U=I,$

and $\Lambda$ is a diagonal matrix with eigenvalues of $Q$ on the main diagonal. We can assume that the first $n-k$ numbers on the diagonal of $Q$ are ones and the others are zeros.

Theorem. Let $e$ be normal. 1) $s^{2}\left( n-k\right) /\sigma ^{2}$ is distributed as $\chi _{n-k}^{2}.$ 2) The estimators $\hat{\beta}$ and $s^{2}$ are independent.

Proof. 1) We have by (4)

(6) $\left\Vert Qe\right\Vert ^{2}=\left( Qe\right) ^{T}Qe=\left( Q^{T}Qe\right) ^{T}e=\left( Qe\right) ^{T}e=\left( U\Lambda U^{T}e\right) ^{T}e=\left( \Lambda U^{T}e\right) ^{T}U^{T}e.$

Denote $S=U^{T}e.$ From (2) and (5)

$ES=0,$ $Var\left( S\right) =EU^{T}ee^{T}U=\sigma ^{2}U^{T}U=\sigma ^{2}I$

and $S$ is normal as a linear transformation of a normal vector. It follows that $S=\sigma z$ where $z$ is a standard normal vector with independent standard normal coordinates $z_{1},...,z_{n}.$ Hence, (6) implies

(7) $\left\Vert Qe\right\Vert ^{2}=\sigma ^{2}\left( \Lambda z\right) ^{T}z=\sigma ^{2}\left( z_{1}^{2}+...+z_{n-k}^{2}\right) =\sigma ^{2}\chi _{n-k}^{2}.$

(3) and (7) prove the first statement.

2) First we note that the vectors $Pe,Qe$ are independent. Since they are normal, their independence follows from

$cov(Pe,Qe)=EPee^{T}Q^{T}=\sigma ^{2}PQ=0.$

It's easy to see that $X^{T}P=X^{T}.$ This allows us to show that $\hat{\beta}$ is a function of $Pe$:

$\hat{\beta}=\beta +(X^{T}X)^{-1}X^{T}e=\beta +(X^{T}X)^{-1}X^{T}Pe.$

Independence of $Pe,Qe$ leads to independence of their functions $\hat{\beta}$ and $s^{2}.$

11
Nov 19

My presentation at Kazakh National University

Today's talk: “Analysis of variance in the central limit theorem”
The talk is about results, which are a combination of methods of the function theory, functional analysis and probability theory. The intuition underlying the central limit theorem will be described, and the history and place of the results of the author in modern theory will be highlighted.

22
Dec 18

Application: distribution of sigma squared estimator

For the formulation of multiple regression and classical conditions on its elements see Application: estimating sigma squared. There we proved unbiasedness of the OLS estimator of $\sigma^2.$ Here we do more: we characterize its distribution and obtain unbiasedness as a corollary.

Preliminaries

We need a summary of what we know about the residual $r=y-\hat{y}$ and the projector $Q=I-P$ where $P=X^T(X^TX)^{-1}X^T:$

(1) $\Vert r\Vert^2=e^TQe.$

$P$ has $k$ unities and $n-k$ zeros on the diagonal of its diagonal representation, where $k$ is the number of regressors. With $Q$ it's the opposite: it has $n-k$ unities and $k$ zeros on the diagonal of its diagonal representation. We can always assume that the unities come first, so in the diagonal representation

(2) $Q=UDU^{-1}$

the matrix $U$ is orthogonal and $D$ can be written as

(3) $D=\left(\begin{array}{cc}I_{n-k}&0\\0&0\end{array}\right)$

where $I_{n-k}$ is an identity matrix and the zeros are zero matrices of compatible dimensions.

Characterization of the distribution of $s^2$

Exercise 1. Suppose the error vector $e$ is normal: $e\sim N(0,\sigma^2I).$ Prove that the vector $\delta =U^{-1}e/\sigma$ is standard normal.

Proof. By the properties of orthogonal matrices

$Var(\delta)=E\delta\delta^T=U^{-1}Eee^TU/\sigma^2=U^{-1}U=I.$

This, together with the equation $E\delta =0$, proves that $\delta$ is standard normal.

Exercise 2. Prove that $\Vert r\Vert^2/\sigma^2$ is distributed as $\chi _{n-k}^2.$

Proof. From (1) and (2) we have

$\Vert r\Vert^2/\sigma^2=e^TUDU^{-1}e/\sigma^2=(U^{-1}e)^TD(U^{-1}e)/\sigma^2=\delta^TD\delta.$

Now (3) shows that $\Vert r\Vert^2/\sigma^2=\sum_{i=1}^{n-k}\delta_i^2$ which is the definition of $\chi _{n-k}^2.$

Exercise 3. Find the mean and variance of $s^2=\Vert r\Vert^2/(n-k)=\sigma^2\chi _{n-k}^2/(n-k).$

Solution. From Exercise 2 we obtain the result proved earlier in a different way:

$Es^2=\sigma^2E\chi _{n-k}^2/(n-k)=\sigma^2.$

Further, using the variance of a standard normal

$Var(s^2)=\frac{\sigma^4}{(n-k)^2}\sum_{i=1}^{n-k}Var(\delta_i^2)=\frac{2\sigma^4}{n-k}.$

10
Dec 18

Distributions derived from normal variables

In the one-dimensional case the economic way to define normal variables is this: define a standard normal variable and then a general normal variable as its linear transformation.

In case of many dimensions, we follow the same idea. Before doing that we state without proofs two useful facts about independence of random variables (real-valued, not vectors).

Theorem 1. Suppose variables $X_1,...,X_n$ have densities $p_1(x_1),...,p_n(x_n).$ Then they are independent if and only if their joint density $p(x_1,...,x_n)$ is a product of individual densities: $p(x_1,...,x_n)=p_1(x_1)...p_n(x_n).$

Theorem 2. If variables $X,Y$ are normal, then they are independent if and only if they are uncorrelated: $cov(X,Y)=0.$

The necessity part (independence implies uncorrelatedness) is trivial.

Normal vectors

Let $z_1,...,z_n$ be independent standard normal variables. A standard normal variable is defined by its density, so all of $z_i$ have the same density. We achieve independence, according to Theorem 1, by defining their joint density to be a product of individual densities.

Definition 1. A standard normal vector of dimension $n$ is defined by

$z=\left(\begin{array}{c}z_1 \\... \\z_n \\ \end{array}\right)$

Properties$Ez=0$ because all of $z_i$ have means zero. Further, $cov(z_i,z_j)=0$ for $i\neq j$by Theorem 2 and variance of a standard normal is 1. Therefore, from the expression for variance of a vector we see that $Var(z)=I.$

Definition 2. For a matrix $A$ and vector $\mu$ of compatible dimensions a normal vector is defined by $X=Az+\mu.$

Properties$EX=AEz+\mu=\mu$ and

$Var(X)=Var(Az)=E(Az)(Az)^T=AEzz^TA^T=AIA^T=AA^T$

(recall that variance of a vector is always nonnegative).

Distributions derived from normal variables

In the definitions of standard distributions (chi square, t distribution and F distribution) there is no reference to any sample data. Unlike statistics, which by definition are functions of sample data, these and other standard distributions are theoretical constructs. Statistics are developed in such a way as to have a distribution equal or asymptotically equal to one of standard distributions. This allows practitioners to use tables developed for standard distributions.

Exercise 1. Prove that $\chi_n^2/n$ converges to 1 in probability.

Proof. For a standard normal $z$ we have $Ez^2=1$ and $Var(z^2)=2$ (both properties can be verified in Mathematica). Hence, $E\chi_n^2/n=1$ and

$Var(\chi_n^2/n)=\sum_iVar(z_i^2)/n^2=2/n\rightarrow 0.$

Now the statement follows from the simple form of the law of large numbers.

Exercise 1 implies that for large $n$ the t distribution is close to a standard normal.

30
Nov 18

Application: estimating sigma squared

Consider multiple regression

(1) $y=X\beta +e$

where

(a) the regressors are assumed deterministic, (b) the number of regressors $k$ is smaller than the number of observations $n,$ (c) the regressors are linearly independent, $\det (X^TX)\neq 0,$ and (d) the errors are homoscedastic and uncorrelated,

(2) $Var(e)=\sigma^2I.$

Usually students remember that $\beta$ should be estimated and don't pay attention to estimation of $\sigma^2.$ Partly this is because $\sigma^2$ does not appear in the regression and partly because the result on estimation of error variance is more complex than the result on the OLS estimator of $\beta .$

Definition 1. Let $\hat{\beta}=(X^TX)^{-1}X^Ty$ be the OLS estimator of $\beta$. $\hat{y}=X\hat{\beta}$ is called the fitted value and $r=y-\hat{y}$ is called the residual.

Exercise 1. Using the projectors $P=X(X^TX)^{-1}X^T$ and $Q=I-P$ show that $\hat{y}=Py$ and $r=Qe.$

Proof. The first equation is obvious. From the model we have $r=X\beta+e-P(X\beta +e).$ Since $PX\beta=X\beta,$ we have further $r=e-Pe=Qe.$

Definition 2. The OLS estimator of $\sigma^2$ is defined by $s^2=\Vert r\Vert^2/(n-k).$

Exercise 2. Prove that $s^2$ is unbiased: $Es^2=\sigma^2.$

Proof. Using projector properties we have

$\Vert r\Vert^2=(Qe)^TQe=e^TQ^TQe=e^TQe.$

Expectations of type $Ee^Te$ and $Eee^T$ would be easy to find from (2). However, we need to find $Ee^TQe$ where there is an obstructing $Q.$ See how this difficulty is overcome in the next calculation.

$E\Vert r\Vert^2=Ee^TQe$ ($e^TQe$ is a scalar, so its trace is equal to itself)

$=Etr(e^TQe)$ (applying trace-commuting)

$=Etr(Qee^T)$ (the regressors and hence $Q$ are deterministic, so we can use linearity of $E$)

$=tr(QEee^T)$ (applying (2)) $=\sigma^2tr(Q).$

$tr(P)=k$ because this is the dimension of the image of $P.$ Therefore $tr(Q)=n-k.$ Thus, $E\Vert r\Vert^2=\sigma^2(n-k)$ and $Es^2=\sigma^2.$

18
Nov 18

Application: Ordinary Least Squares estimator

Generalized Pythagoras theorem

Exercise 1. Let $P$ be a projector and denote $Q=I-P.$ Then $\Vert x\Vert^2=\Vert Px\Vert^2+\Vert Qx\Vert^2.$

Proof. By the scalar product properties

$\Vert x\Vert^2=\Vert Px+Qx\Vert^2=\Vert Px\Vert^2+2(Px)\cdot (Qx)+\Vert Qx\Vert^2.$

$P$ is symmetric and idempotent, so

$(Px)\cdot (Qx)=(Px)\cdot[(I-P)x]=x\cdot[(P-P^2)x]=0.$

This proves the statement.

Ordinary Least Squares (OLS) estimator derivation

Problem statement. A vector $y\in R^n$ (the dependent vector) and vectors $x^{(1)},...,x^{(k)}\in R^n$ (independent vectors or regressors) are given. The OLS estimator is defined as that vector $\beta \in R^k$ which minimizes the total sum of squares $TSS=\sum_{i=1}^n(y_i-x^{(1)}\beta_1-...-x^{(k)}\beta_k)^2.$

Denoting $X=(x^{(1)},...,x^{(k)}),$ we see that $TSS=\Vert y-X\beta\Vert^2$ and that finding the OLS estimator means approximating $y$ with vectors from the image $\text{Img}X.$ $x^{(1)},...,x^{(k)}$ should be linearly independent, otherwise the solution will not be unique.

Assumption. $x^{(1)},...,x^{(k)}$ are linearly independent. This, in particular, implies that $k\leq n.$

Exercise 2. Show that the OLS estimator is

(2) $\hat{\beta}=(X^TX)^{-1}X^Ty.$

Proof. By Exercise 1 we can use $P=X(X^TX)^{-1}X^T.$ Since $X\beta$ belongs to the image of $P,$ $P$ doesn't change it: $X\beta=PX\beta.$ Denoting also $Q=I-P$ we have

$\Vert y-X\beta\Vert^2=\Vert y-Py+Py-X\beta\Vert^2$

$=\Vert Qy+P(y-X\beta)\Vert^2$ (by Exercise 1)

$=\Vert Qy\Vert^2+\Vert P(y-X\beta)\Vert^2.$

This shows that $\Vert Qy\Vert^2$ is a lower bound for $\Vert y-X\beta\Vert^2.$ This lower bound is achieved when the second term is made zero. From

$P(y-X\beta)=Py-X\beta =X(X^TX)^{-1}X^Ty-X\beta=X[(X^TX)^{-1}X^Ty-\beta]$

we see that the second term is zero if $\beta$ satisfies (2).

Usually the above derivation is applied to the dependent vector of the form $y=X\beta+e$ where $e$ is a random vector with mean zero. But it holds without this assumption. See also simplified derivation of the OLS estimator.

8
May 18

Different faces of vector variance: again visualization helps

In the previous post we defined variance of a column vector $X$ with $n$ components by

$V(X)=E(X-EX)(X-EX)^T.$

In terms of elements this is the same as:

(1) $V(X)=\left(\begin{array}{cccc}V(X_1)&Cov(X_1,X_2)&...&Cov(X_1,X_n) \\Cov(X_2,X_1)&V(X_2)&...&Cov(X_2,X_n) \\...&...&...&... \\Cov(X_n,X_1)&Cov(X_n,X_2)&...&V(X_n) \end{array}\right).$

So why knowing the structure of this matrix is so important?

Let $X_1,...,X_n$ be random variables and let $a_1,...,a_n$ be numbers. In the derivation of the variance of the slope estimator for simple regression we have to deal with the expression of type

(2) $V\left(\sum_{i=1}^na_iX_i\right).$

Question 1. How do you multiply a sum by a sum? I mean, how do you use summation signs to find the product $\left(\sum_{i=1}^na_i\right)\left(\sum_{i=1}^nb_i\right)$?

Answer 1. Whenever you have problems with summation signs, try to do without them. The product

$\left(a_1+...+a_n\right)\left(b_1+...+b_n\right)=a_1b_1+...+a_1b_n+...+a_nb_1+...+a_nb_n$

should contain ALL products $a_ib_j.$ Again, a matrix visualization will help:

$\left(\begin{array}{ccc}a_1b_1&...&a_1b_n \\...&...&... \\a_nb_1&...&a_nb_n \end{array}\right).$

The product we are looking for should contain all elements of this matrix. So the answer is

(3) $\left(\sum_{i=1}^na_i\right)\left(\sum_{i=1}^nb_i\right)=\sum_{i=1}^n\sum_{j=1}^na_ib_j.$

Formally, we can write $\sum_{i=1}^nb_i=\sum_{j=1}^nb_j$ (the sum does not depend on the index of summation, this is another point many students don't understand) and then perform the multiplication in (3).

Question 2. What is the expression for (2) in terms of covariances of components?

Answer 2. If you understand Answer 1 and know the relationship between variances and covariances, it should be clear that

(4) $V\left(\sum_{i=1}^na_iX_i\right)=Cov(\sum_{i=1}^na_iX_i,\sum_{i=1}^na_iX_i)$

$=Cov(\sum_{i=1}^na_iX_i,\sum_{j=1}^na_jX_j)=\sum_{i=1}^n\sum_{j=1}^na_ia_jCov(X_i,X_j).$

Question 3. In light of (1), separate variances from covariances in (4).

Answer 3. When $i=j,$ we have $Cov(X_i,X_j)=V(X_i),$ which are diagonal elements of (1). Otherwise, for $i\neq j$ we get off-diagonal elements of (1). So the answer is

(5) $V\left(\sum_{i=1}^na_iX_i\right)=\sum_{i=1}^na_i^2V(X_i)+\sum_{i\neq j}a_ia_jCov(X_i,X_j).$

Once again, in the first sum on the right we have only variances. In the second sum, the indices $i,j$ are assumed to run from $1$ to $n$, excluding the diagonal $i=j.$

Corollary. If $X_{i}$ are uncorrelated, then the second sum in (5) disappears:

(6) $V\left(\sum_{i=1}^na_iX_i\right)=\sum_{i=1}^na_i^2V(X_i).$

This fact has been used (with a slightly different explanation) in the derivation of the variance of the slope estimator for simple regression.

Question 4. Note that the matrix (1) is symmetric (elements above the main diagonal equal their mirror siblings below that diagonal). This means that some terms in the second sum on the right of (5) are repeated twice. If you group equal terms in (5), what do you get?

Answer 4. The idea is to write

$a_ia_jCov(X_i,X_j)+a_ia_jCov(X_j,X_i)=2a_ia_jCov(X_i,X_j),$

that is, to join equal elements above and below the main diagonal in (1). For this, you need to figure out how to write a sum of the elements that are above the main diagonal. Make a bigger version of (1) (with more off-diagonal elements) to see that the elements that are above the main diagonal are listed in the sum $\sum_{i=1}^{n-1}\sum_{j=i+1}^n.$ This sum can also be written as $\sum_{1\leq i Hence, (5) is the same as

(7) $V\left(\sum_{i=1}^na_iX_i\right)=\sum_{i=1}^na_i^2V(X_i)+2\sum_{i=1}^{n-1}\sum_{j=i+1}^na_ia_jCov(X_i,X_j)$

$=\sum_{i=1}^na_i^2V(X_i)+2\sum_{1\leq i

Unlike (6), this equation is applicable when there is autocorrelation.

7
May 18

Variance of a vector: motivation and visualization

I always show my students the definition of the variance of a vector, and they usually don't pay attention. You need to know what it is, already at the level of simple regression (to understand the derivation of the slope estimator variance), and even more so when you deal with time series. Since I know exactly where students usually stumble, this post is structured as a series of questions and answers.

Think about ideas: how would you define variance of a vector?

Question 1. We know that for a random variable $X$, its variance is defined by

(1) $V(X)=E(X-EX)^{2}.$

Now let

$X=\left(\begin{array}{c}X_{1} \\... \\X_{n}\end{array}\right)$

be a vector with $n$ components, each of which is a random variable. How would you define its variance?

The answer is not straightforward because we don't know how to square a vector. Let $X^T=(\begin{array}{ccc}X_1& ...&X_n\end{array})$ denote the transposed vector. There are two ways to multiply a vector by itself: $X^TX$ and $XX^T.$

Question 2. Find the dimensions of $X^TX$ and $XX^T$ and their expressions in terms of coordinates of $X.$

Answer 2. For a product of matrices there is a compatibility rule that I write in the form

(2) $A_{n\times m}B_{m\times k}=C_{n\times k}.$

Recall that $n\times m$ in the notation $A_{n\times m}$ means that the matrix $A$ has $n$ rows and $m$ columns. For example, $X$ is of size $n\times 1.$ Verbally, the above rule says that the number of columns of $A$ should be equal to the number of rows of $B.$ In the product that common number $m$ disappears and the unique numbers ($n$ and $k$) give, respectively, the number of rows and columns of $C.$ Isn't the the formula
easier to remember than the verbal statement? From (2) we see that $X_{1\times n}^TX_{n\times 1}$ is of dimension 1 (it is a scalar) and $X_{n\times 1}X_{1\times n}^T$ is an $n\times n$ matrix.

For actual multiplication of matrices I use the visualization

(3) $\left(\begin{array}{ccccc}&&&&\\&&&&\\a_{i1}&a_{i2}&...&a_{i,m-1}&a_{im}\\&&&&\\&&&&\end{array}\right) \left(\begin{array}{ccccc}&&b_{1j}&&\\&&b_{2j}&&\\&&...&&\\&&b_{m-1,j}&&\\&&b_{mj}&&\end{array}\right) =\left( \begin{array}{ccccc}&&&&\\&&&&\\&&c_{ij}&&\\&&&&\\&&&&\end{array}\right)$

Short formulation. Multiply rows from the first matrix by columns from the second one.

Long Formulation. To find the element $c_{ij}$ of $C,$ we find a scalar product of the $i$th row of $A$ and $j$th column of $B:$ $c_{ij}=a_{i1}b_{1j}+a_{i2}b_{2j}+...$ To find all elements in the $i$th row of $C,$ we fix the $i$th row in $A$ and move right the columns in $B.$ Alternatively, to find all elements in the $j$th column of $C,$ we fix the $j$th column in $B$ and move down the rows in $A$. Using this rule, we have

(4) $X^TX=X_1^2+...+X_n^2,$ $XX^T=\left(\begin{array}{ccc}X_1^2&...&X_1X_n \\...&...&... \\X_nX_1&...&X_n^2 \end{array}\right).$

Usually students have problems with the second equation.

Based on (1) and (4), we have two candidates to define variance:

(5) $V(X)=E(X-EX)^T(X-EX)$

and

(6) $V(X)=E(X-EX)(X-EX)^T.$

Answer 1. The second definition contains more information, in the sense to be explained below, so we define variance of a vector by (6).

Question 3. Find the elements of this matrix.

Answer 3. Variance of a vector has variances of its components on the main diagonal and covariances outside it:

(7) $V(X)=\left(\begin{array}{cccc}V(X_1)&Cov(X_1,X_2)&...&Cov(X_1,X_n) \\Cov(X_2,X_1)&V(X_2)&...&Cov(X_2,X_n) \\...&...&...&... \\Cov(X_n,X_1)&Cov(X_n,X_2)&...&V(X_n) \end{array}\right).$

If you can't get this on your own, go back to Answer 2.

There is a matrix operation called trace and denoted $tr$. It is defined only for square matrices and gives the sum of diagonal elements of a matrix.

Exercise 1. Show that $tr(V(X))=E(X-EX)^T(X-EX).$ In this sense definition (6) is more informative than (5).

Exercise 2. Show that if $EX_1=...=EX_n=0$, then (7) becomes

$V(X)=\left(\begin{array}{cccc}EX^2_1&EX_1X_2&...&EX_1X_n \\EX_2X_1&EX^2_2&...&EX_2X_n \\...&...&...&... \\EX_nX_1&EX_nX_2&...&EX^2_n \end{array}\right).$

6
Oct 17

Significance level and power of test

In this post we discuss several interrelated concepts: null and alternative hypotheses, type I and type II errors and their probabilities. Review the definitions of a sample space and elementary events and that of a conditional probability.

Type I and Type II errors

Regarding the true state of nature we assume two mutually exclusive possibilities: the null hypothesis (like the suspect is guilty) and alternative hypothesis (the suspect is innocent). It's up to us what to call the null and what to call the alternative. However, the statistical procedures are not symmetric: it's easier to measure the probability of rejecting the null when it is true than other involved probabilities. This is why what is desirable to prove is usually designated as the alternative.

Usually in books you can see the following table.

 Decision taken Fail to reject null Reject null State of nature Null is true Correct decision Type I error Null is false Type II error Correct decision

This table is not good enough because there is no link to probabilities. The next video does fill in the blanks.

Video. Significance level and power of test

Significance level and power of test

The conclusion from the video is that

$\frac{P(T\bigcap R)}{P(T)}=P(R|T)=P\text{(Type I error)=significance level}$

$\frac{P(F\bigcap R)}{P(F)}=P(R|F)=P\text{(Correctly rejecting false null)=Power}$

11
Aug 17

Violations of classical assumptions

This will be a simple post explaining the common observation that "in Economics, variability of many variables is proportional to those variables". Make sure to review the assumptions; they tend to slip from memory. We consider the simple regression

(1) $y_i=a+bx_i+e_i.$

One of classical assumptions is

Homoscedasticity. All errors have the same variances$Var(e_i)=\sigma^2$ for all $i$.

We discuss its opposite, which is

Heteroscedasticity. Not all errors have the same variance. It would be wrong to write it as $Var(e_i)\ne\sigma^2$ for all $i$ (which means that all errors have variance different from $\sigma^2$). You can write that not all $Var(e_i)$ are the same but it's better to use the verbal definition.

Remark about Video 1. The dashed lines can represent mean consumption. Then the fact that variation of a variable grows with its level becomes more obvious.

Video 1. Case for heteroscedasticity

Figure 1. Illustration from Dougherty: as x increases, variance of the error term increases

Homoscedasticity was used in the derivation of the OLS estimator variance; under heteroscedasticity that expression is no longer valid. There are other implications, which will be discussed later.

Companies example. The Samsung Galaxy Note 7 battery fires and explosions that caused two recalls cost the smartphone maker at least $5 billion. There is no way a small company could have such losses. GDP example. The error in measuring US GDP is on the order of$200 bln, which is comparable to the Kazakhstan GDP. However, the standard deviation of the ratio error/GDP seems to be about the same across countries, if the underground economy is not too big. Often the assumption that the standard deviation of the regression error is proportional to one of regressors is plausible.

To see if the regression error is heteroscedastic, you can look at the graph of the residuals or use statistical tests.