# linear regression, symmetry of model does not lead to symmetry of coefficients

Experiment: You are given a large population of real numbers. For simplicity take the whole numbers from -n to n. Take two independent random samples x and y of size k and sort them (each one individually). Compute the linear regression $y=a+bx$.

Question: what are the expected values for a and b? What does the distribution look like?

As the setting is symmetric in x and y I expected $a=0$ and $b=1$ but numerics show that this is not the case.

I first posted this question in the math section:

but did not get any useful answers. This question here

symmetry of linear regression

is also closely related.

Here is some R code that runs this experiment.

n_pop  <- 100

n_sample <- 20

draws <- 1000

M <- matrix(c(1:(2*draws)), nrow=draws)

for (j in 1:draws) {

x <- sort(sample(-n_pop:n_pop,n_sample,replace=FALSE))

y <- sort(sample(-n_pop:n_pop,n_sample,replace=FALSE))

xy_model <- lm(y~x)

M[j,] <- coef(xy_model)

}

mean(M[,1])

mean(M[,2])

sd(M[,1])

sd(M[,2])

hist(M[,1],breaks=20)

hist(M[,2], breaks=20)

t.test(M[,2], mu=1, conf.level=0.99)

t.test(M[,1], mu=0, conf.level=0.99)


For this population and sample size I get around $a=1$ and $b=0.98$, $a=0$ and $b=1$ can be excluded with high confidence.

I suspect this is related to the fact that OLS minizes distance only on the y-axis but I don't understand why this changes the coefficients. Numerically, if the sample size gets smaller, the cofficients move farther way from 0 and 1. If one uses a Deming regression which minimizes euclidean distance to the line, numerically $a=0$ and $b=1$ looks plausible.

edit: the code above gives a distribution for $a$ that is symmetric around 0, I had an error in my R code. The question for $b$ remains and the answer below gives an excellent explanation why $b$ is not equal to $1$.

• What are you really after here? When will it ever be interesting to sort X and Y independently? You destroy any trace of the actual relationship – Repmat Oct 19 '17 at 11:52
• Maybe my wording wasn't clear. I started by taking 2 random samples from the same base population. If you just take them as is, they would have no correlation. After you sort them, you have two increasing sequences of numbers. As the two samples have the same statistical properties, the slope in the regression is approximately 1. However, the expected value of the slope is not exactly 1 and I'm trying to understand why. – quarague Oct 19 '17 at 13:03
• I cannot reproduce your results for the estimates of $a$: both simulation and theory indicate its mean is zero. What exactly is the symmetry that suggests the expectation of $b$'s estimate should be $1$? The map $x\to -x$ of the population does not induce any apparent symmetry in the distribution of the estimate of $b$, because that map leaves the distributions of $x$ and $y$ unchanged and therefore has no effect at all on the distribution of the estimates. – whuber Oct 19 '17 at 14:35
• The symmetry is P(x,y) = P(y,x). Probably some variation of regression dilution will cause a $E(\hat{b})<1$. It may be a bit difficult to show a direct derivation, although maybe there is some elegant short solution (and hopefully intuitive as well). Reasons why $E(\hat{a})\neq0$ are not directly clear. But the opposite is not trivial either. Why would theory state that $E(\hat{a})=0$? – Sextus Empiricus Oct 19 '17 at 16:32
• The a coefficient numerically has expectation zero. I had a slightly different setup before and forgot to adjust my code properly, my mistake. – quarague Oct 20 '17 at 7:48

## 1

The regression $y = a + bx + \epsilon$ is not symmetric for replacing y and x, as you already suggested via the link (symmetry of linear regression).

However, it is not trivial whether the expectation values, $\hat{a}$ and $\hat{b}$, are going to be different from a symmetric $E(\hat{a}_{xy})=E(\hat{a}_{yx})=0$ and $E(\hat{b}_{xy})=1/E(\hat{b}_{yx})=1$. Possibly the difference in the regression $y \sim x$ vs $x \sim y$ cancel over all possibilities (we will see this is true for $a$). For this we need to "see" all the possibilities and how they average/cancel.

## 2

We will show that the intercept is symmetric

$$E(\hat{a})=0$$

however the slope is not, and will be smaller than 1

$$E(\hat{b})<1$$

We show this by comparing the non-symmetric regression with a symmetric regression.

## 3

As symmetric regression we use the line in between the two regression lines lines defined by the y~x and x~y.

In algebraic terms:

$$\hat{b}_{sym} = \frac{\hat{b}_{xy} + \hat{b}_{yx}^{-1}}{2}$$

and

$$\hat{a}_{sym} = \frac{\hat{a}_{xy} - \hat{a}_{yx}\hat{b}_{yx}}{2}$$

## 4

Note the following about the symmetric slope.

1. The slope will be symmetric around the angle of 45 degrees

ie, a probability for an angle with $45+\alpha$ is equal to a probabilty for an angle with $45-\alpha$, also the expectation value for its angle coefficient will be 1, since:

• $\hat{b}_{sym}(x,y)=\hat{b}_{sym}(y,x)$, thus $E(\hat{b}_{sym}(x,y))=E(\hat{b}_{sym}(y,x))$

(note that we do not have the same for the slope of the asymmetric case $\hat{b}_{xy}(z_1,z_2)=\hat{b}_{yx}(z_2,z_1)$ )

• yet we must also have the symmetry $E(\hat{b}_{sym}(x,y))=1/E(\hat{b}_{sym}(y,x))$, the line that is described by the expectation of the coefficient, if we change the labels xy than we expect a mirror image which has a different b coefficient (the inverse).
2. The slope $\hat{b}_{xy}$ will be smaller due to regression to the mean.

Therefore... the symetric slope will be $E(\hat{b}_{sym})=1$, however then since always $\hat{b}_{xy} \leq \hat{b}_{sym}$ we will have $E(\hat{b}_{xy}) \leq E(\hat{b}_{sym})=1$. And the equality is only true when there are no error terms.

See the following image to recall the regression lines $xy$, $yx$ (red curves) and the symmetric one in between (green curve), this image is from the example code below: If we plot the angle of the symmetric line (green) and the regression line for the y~x model (the red one that has smaller angel with the x-axis) then we see that the angle of the symmetric line distributes evenly around 45 degrees, however the angle of the non-symmetric line $y~x$, is always lower: ## 5

What is left is to show that $E(\hat{a})=0$. I do not have a direct proof for this. However, it should be sufficient to notice that the regression to the mean (which is zero) does not favor either larger or smaller $a$. Note that there is a point-symmetry, $(x,y) \rightarrow (-x,-y)$, for which the angle $b$ is invariant, but not the intercept $\hat{a}$, which means that it, $a$, should be zero.

You obtain values of $\hat{a}$ different form zero, but you should take into account it's sample variance. (The sample error for $\hat{a}$ is different from zero)

## 6

Below is adjustment to your code to examine the above described effects

n_pop  <- 100

n_sample <- 20

draws <- 1000

M <- matrix(c(1:(2*draws)), nrow=draws)
K <- matrix(c(1:(2*draws)), nrow=draws)
ma <- matrix(c(1:(2*draws)), nrow=draws)
meanx <- matrix(c(1:(1*draws)), nrow=draws)

for (j in 1:draws) {

x <- sort(sample(-n_pop:n_pop,n_sample,replace=FALSE))

y <- sort(sample(-n_pop:n_pop,n_sample,replace=FALSE))

xy_model <- lm(y~x)
yx_model <- lm(x~y)

M[j,] <- coef(xy_model)
K[j,] <- coef(yx_model)

# xy and yx model coefficients last loop
b1 <- coef(xy_model)
b2 <- 1/coef(yx_model)
a1 <- coef(xy_model)
a2 <- -coef(yx_model)*coef(yx_model)

ma[j,1] <- 0.5*sum(a1,a2)
ma[j,2] <- 0.5*sum(b1,b2)

meanx[j,1] <- mean(x)
}

# means
mean(M[,1])
mean(M[,2])
mean(ma[,1])
mean(ma[,2])

# plotting last for-loop x-y as example of regresion fits
plot(x,y)

xp <- -n_pop:n_pop
lines(xp,a1+b1*xp,col=2)
lines(xp,a2+b2*xp,col=2)
lines(xp,0.5*a2+0.5*a1+0.5*b2*xp+0.5*b1*xp,col=3)

# plotting coefficients symmetric versus xy
plot(180/pi*atan(ma[,2]),180/pi*atan(M[,2]), xlab="symmetric fit angle", ylab="xy-fit angle")
lines(c(0,90),c(0,90),col=2)

# plotting mean of x vs beta
plot(meanx,ma[,2]-M[,2], xlab="mu_x",ylab="beta_sym - beta_xy",pch=21,

• Thanks a lot for this extensive answer. I computed the symmetrized slope after your formula and when I then look at $atan(\hat{b}_{sym})$ I indeed get a distribution that looks symmetric around 45 degrees (or pi/4) and with 10000 runs a confidence interval that contains pi/4. However, neither a nor the symmetrized version you suggested has a numeric expectation of zero in my experiments. – quarague Oct 20 '17 at 7:42