2
$\begingroup$

A random triangle is formed in some way, such that all pairs of angles have the same joint distribution. What is the correlation between two of the angles (assuming that the variance of the angles is nonzero)?

Is it true that, since all will have the same joint distribution, tha5t the correlation between two of the angles is 1? They are identical-- no?

$\endgroup$
  • $\begingroup$ I believe you are right. I'm not entirely sure; however, this seems logical. $\endgroup$ – Jay Oct 28 '17 at 21:43
4
$\begingroup$

From properties of covariance, we have that $\text{Cov}(X,Y) = \frac{1}{2} \text{Cov}(X,Y+Z) = \frac{1}{2}\text{Cov}(X, \pi -X) = -\frac{1}{2}\text{Var}(X)$. This is also enough to imply that $\text{Corr}(X,Y) = -\frac{1}{2}$

$\endgroup$
  • $\begingroup$ Did you mean to write $\text{Corr}(X,Y) = -1/2$ ? $\endgroup$ – Moss Murderer Oct 29 '17 at 0:54
  • $\begingroup$ Nice! Very simple, direct approach. $\endgroup$ – whuber Oct 29 '17 at 14:17
3
$\begingroup$

Let the angles be $A,B,C$. About the only fact about them that's useful is that they sum to a constant (assuming these are Euclidean triangles), which I will call $\pi$:

$$\pi=A+B+C.\tag{1}$$

Correlation is computed in terms of first and second moments. Since all pairs of angles have the same joint distribution, they all have the same bivariate moments. This enables us to simplify the notation: write

$$\mu_1 = E[A]=E[B]=E[C]$$

for their common expectation,

$$\mu_2 = E[A^2] = E[B^2] = E[C^2]$$

for their common second moment, and

$$\mu_{11} = E[AB] = E[BC] = E[CA]$$

for their common cross-moment.

From $(1)$, its square, and linearity of expectation, we may obtain information about these moments:

$$\pi = E[A+B+C] = E[A]+E[B]+E[C] = \mu_1+\mu_1+\mu_1 = 3\mu_1\tag{2}$$

and

$$\eqalign{ \pi^2 &= E[(A+B+C)^2] = E[A^2 + 2AB + \cdots] = E[A^2] + 2E[AB] + \cdots\\&= 3\mu_2 + 6\mu_{11}.\tag{3} }$$

Finally, the correlations must all be the same, too. They are fractions where the numerator is the common covariance

$$\operatorname{Cov}(A,B)=\operatorname{Cov}(B,C)=\operatorname{Cov}(C,A)=E[CA]-E[C]E[A]=\mu_{11} - \mu_1^2$$

and the denominator is the product of two standard deviations. But all the standard deviations are the same, whence the product of two of them is the common variance,

$$\operatorname{Var}(A)=\operatorname{Var}(B)=\operatorname{Var}(C)=E[C^2]-E[C]^2 = \mu_2 - \mu_1^2.$$

The common correlation therefore is

$$\rho = \rho_{AB}=\rho_{BC}=\rho_{CA} = \frac{\mu_{11} - \mu_1^2}{\mu_2 - \mu_1^2}.\tag{4}$$

The question has been reduced to this:

Can you compute $(4)$ given $(2)$ and $(3)$?

Comparing $(2)$ with $(3)$ suggests squaring $(2)$ to obtain another expression for $\pi^2$:

$$(3\mu_1)^2 = \pi^2 = 3\mu_2 + 6\mu_{11}.$$

Let's subtract off suitable multiples of $\mu_1^2$ to obtain a relationship between the expressions that appear in $(4)$:

$$0 = (3\mu_1)^2 - 9\mu_1^2 = 3(\mu_2 - \mu_1^2) + 6(\mu_{11} - \mu_1^2).$$

Consequently, dividing the right hand side by $6(\mu_2 - \mu_1^2)$ (which, being six times the common variance, is assumed nonzero), we learn that

$$0 = \frac{1}{2} + \rho.$$


To demonstrate that the angles do not need to be constant, an example of such a distribution is $\pi$ times a Dirichlet$(\alpha,\alpha,\alpha)$ distribution for any $\alpha\gt 0$. Here are 100 such triangles generated according to that distribution with $\alpha=1$, oriented with $A$ at the left, $B$ at the right, and $C$ at the top of each one. Each has unit perimeter.

Figure

Although the correlations among the angles in this particular sample are not quite $\rho = -1/2$, they are close (they range from $-0.40$ to $-0.57$): and by generating more (I produced 20,000 of them), the correlations become indistinguishable from $-1/2$.

$\endgroup$
  • 2
    $\begingroup$ Very nice answer. It is quite striking that the set of triangles look essentially random (to me) despite the constraint. $\endgroup$ – Moss Murderer Oct 28 '17 at 23:06
  • $\begingroup$ @Moss Yes, that's the point. But if you choose larger values of $\alpha$ for the Dirichlet distribution, the triangles will tend to be much closer to equilateral. Note that another way to obtain a set of random triangles meeting the requirements of this problem is to use any distribution of angles and follow that up by permuting the roles of the three angles: that guarantees exchangeability (which is a stronger condition implying equal joint marginals). $\endgroup$ – whuber Oct 29 '17 at 14:16

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.