Simple measure of similarity for the market graph construction

Abstract

A simple measure of similarity for the construction of the market graph is proposed. The measure is based on the probability of the coincidence of the signs of the stock returns. This measure is robust, has a simple interpretation, is easy to calculate and can be used as measure of similarity between any number of random variables. For the case of pairwise similarity the connection of this measure with the sign correlation of Fechner is noted. The properties of the proposed measure of pairwise similarity in comparison with the classic Pearson correlation are studied. The simple measure of pairwise similarity is applied (in parallel with the classic correlation) for the study of Russian and Swedish market graphs. The new measure of similarity for more than two random variables is introduced and applied to the additional deeper analysis of Russian and Swedish markets. Some interesting phenomena for the cliques and independent sets of the obtained market graphs are observed.

Appendices

Appendix A: Derivation of the formulas for connection between the classical and the sign correlation coefficients for the normal distribution

Suppose that $X,Y$ are normally distributed random variables and the correlation coefficient between $X$ and $Y$ is $\rho (X,Y)=\rho$.

Consider the random variables $U,V$ having normal distribution with zero mean and variance equal to one, and such that $\rho (U,V)=0$. Let the random variable $X=\alpha U + \beta V$, the random variable $Y=\gamma U + \delta V$. To ensure that $\rho (X,Y)=\rho$ it is necessary to satisfy the following conditions:

$$\begin{aligned} \left\{ \begin{array}{l} \alpha ^2+\beta ^2=1 \\ \gamma ^2+\delta ^2=1 \\ \alpha \gamma +\beta \delta =\rho \end{array} \right. \end{aligned}$$

This equation system allows the following solution: $\alpha =\beta =\frac{1}{\sqrt{2}}$, $\gamma =\frac{1}{\sqrt{2}}(\rho +\sqrt{1-\rho ^2})$, $\delta =\frac{1}{\sqrt{2}}(\rho -\sqrt{1-\rho ^2})$. Thus, random variables $X = \frac{U}{\sqrt{2}} + \frac{V}{\sqrt{2}}$ and $Y = \frac{(\rho + \sqrt{1-\rho ^2})U}{\sqrt{2}} + \frac{(\rho - \sqrt{1-\rho ^2})V}{\sqrt{2}}$ have correlation coefficient $\rho$.

$$\begin{aligned}&P(\{sign(X)=sign(Y)\}) = P(X > 0, Y > 0) + P(X < 0, Y < 0)\nonumber \\&\quad =P(U + V > 0, (\rho + \sqrt{1 - \rho ^2})U + (\rho - \sqrt{1 - \rho ^2})V > 0) \nonumber \\&\quad +P(U + V < 0, (\rho + \sqrt{1 - \rho ^2})U + (\rho - \sqrt{1 - \rho ^2})V < 0) \end{aligned}$$

(17)

Let $(\rho +\sqrt{1-\rho ^2})>0$, then (17) looks like:

$$\begin{aligned}&P(\{sign(X)=sign(Y)\}) \nonumber \\&\quad =P\left( U > -V, U > -\frac{(\rho - \sqrt{1 - \rho ^2})}{(\rho + \sqrt{1 - \rho ^2})}V\right) + P\left( U < -V, U < -\frac{(\rho - \sqrt{1 - \rho ^2})}{(\rho + \sqrt{1 - \rho ^2})}V\right) \nonumber \\&\quad = P\left( U > -V, U > \frac{2\rho \sqrt{1 - \rho ^2} - 1}{2\rho ^2 - 1}V\right) + P\left( U < -V, U < \frac{2\rho \sqrt{1 \!-\! \rho ^2} \!-\! 1}{2\rho ^2 \!-\! 1}V\right) \qquad \end{aligned}$$

(18)

As the random variables $U,V \sim N(0,1)$ and are independent, then the probability of hitting the sector formed by two lines from $(0,0)$ at angle $\alpha$, equals to $\frac{\alpha }{2\pi }$.

Angle between lines $U=-V$ and $U=\frac{2\rho \sqrt{1 - \rho ^2}-1}{2\rho ^2 - 1}V$ forming the sector $\left( U > -V, U > \frac{2\rho \sqrt{1 - \rho ^2}-1}{2\rho ^2 - 1}V\right)$, equals to $\alpha =\frac{3\pi }{4}-arctg\left( \frac{2\rho \sqrt{1 - \rho ^2}-1}{2\rho ^2 - 1}\right)$.

Consequently,

$$\begin{aligned} P\left( U > -V, U > (2\rho \sqrt{1 - \rho ^2}-1)V\right)&= \frac{3}{8}-\frac{arctg\left( \frac{2\rho \sqrt{1 - \rho ^2}-1}{2\rho ^2 - 1}\right) }{2\pi },\\ P\left( U < -V, U < (2\rho \sqrt{1 - \rho ^2}-1)V\right)&= \frac{3}{8}-\frac{arctg\left( \frac{2\rho \sqrt{1 - \rho ^2}-1}{2\rho ^2 - 1}\right) }{2\pi }. \end{aligned}$$

Thus,

$$\begin{aligned}&P(\{sign(X)=sign(Y)\})\\&\quad =2\left( \frac{3}{8}-\frac{arctg\left( \frac{2\rho \sqrt{1 - \rho ^2}-1}{2\rho ^2 - 1}\right) }{2\pi }\right) =\frac{3}{4}-\frac{arctg\left( \frac{2\rho \sqrt{1 - \rho ^2}-1}{2\rho ^2 - 1}\right) }{\pi },\rho _{XY}^S\\&\quad =\frac{3}{2}-\frac{2arctg\left( \frac{2\rho \sqrt{1 - \rho ^2}-1}{2\rho ^2 - 1}\right) }{\pi }-1=\frac{1}{2}-\frac{arctg\left( \frac{2\rho \sqrt{1 - \rho ^2}-1}{2\rho ^2 - 1}\right) }{\pi /2}. \end{aligned}$$

Let $(\rho +\sqrt{1-\rho ^2})<0$, then (17) looks like:

$$\begin{aligned}&P(\{sign(X)=sign(Y)\}) \nonumber \\&\quad = P\left( U > -V, U < -\frac{(\rho - \sqrt{1 - \rho ^2})}{(\rho + \sqrt{1 - \rho ^2})}V\right) + P\left( U < -V, U > -\frac{(\rho - \sqrt{1 - \rho ^2})}{(\rho + \sqrt{1 - \rho ^2})}V\right) \nonumber \\&\quad =P\left( U > -V, U < \frac{2\rho \sqrt{1 - \rho ^2}-1}{2\rho ^2 - 1}V\right) + P\left( U < -V, U > \frac{2\rho \sqrt{1 - \rho ^2}-1}{2\rho ^2 - 1}V\right) \nonumber \\ \end{aligned}$$

(19)

Angle between lines $U=-V$ and $U=\frac{2\rho \sqrt{1 - \rho ^2} - 1}{2\rho ^2 - 1}V$ forming the sector $\left( U > -V, U < \frac{2\rho \sqrt{1 - \rho ^2} - 1}{2\rho ^2 - 1}V\right)$, equals to $\alpha =\frac{\pi }{4}-arctg\left( \frac{2\rho \sqrt{1 - \rho ^2} - 1}{2\rho ^2 - 1}\right)$.

Consequently,

$$\begin{aligned} P\left( U > -V, U < \frac{2\rho \sqrt{1 - \rho ^2} - 1}{2\rho ^2 - 1}V\right)&= \frac{1}{8}+\frac{arctg\left( \frac{2\rho \sqrt{1 - \rho ^2}-1}{2\rho ^2 - 1}\right) }{2\pi },\\ P\left( U < -V, U > \frac{2\rho \sqrt{1 - \rho ^2} - 1}{2\rho ^2 - 1}V\right)&= \frac{1}{8}+\frac{arctg\left( \frac{2\rho \sqrt{1 - \rho ^2}-1}{2\rho ^2 - 1}\right) }{2\pi }. \end{aligned}$$

Thus,

$$\begin{aligned} P(\{sign(X)=sign(Y)\})&= \frac{1}{4}+\frac{arctg\left( \frac{2\rho \sqrt{1 - \rho ^2}-1}{2\rho ^2 - 1}\right) }{\pi },\\ \rho _{XY}^S&= \frac{arctg\left( \frac{2\rho \sqrt{1 - \rho ^2}-1}{2\rho ^2 - 1}\right) }{\pi /2}-\frac{1}{2}. \end{aligned}$$

If $(\rho +\sqrt{1-\rho ^2})=0$, or $\rho =-\frac{\sqrt{2}}{2}$, (17) looks like this:

$$\begin{aligned} P(\{sign(X)=sign(Y)\})&= P(U>-V,V<0)+P(U<-V,V>0)\nonumber \\&= \frac{1}{8} + \frac{1}{8} = \frac{1}{4}\\ \rho _{XY}^S&= -\frac{1}{2}. \end{aligned}$$

Thus, the sign correlation coefficient as a function of the classical correlation coefficient is given by:

$$\begin{aligned} \rho _{XY}^S = \left\{ \begin{array}{ll} \frac{1}{2}-\frac{arctg\left( \frac{2\rho \sqrt{1 - \rho ^2}-1}{2\rho ^2 - 1}\right) }{\pi /2}, &{} \rho > -\frac{\sqrt{2}}{2} \\ -\frac{1}{2}, &{} \rho = -\frac{\sqrt{2}}{2} \\ \frac{arctg\left( \frac{2\rho \sqrt{1 - \rho ^2}-1}{2\rho ^2 - 1}\right) }{\pi /2}-\frac{1}{2}, &{} \rho < -\frac{\sqrt{2}}{2} \\ \end{array} \right. \end{aligned}$$

(20)

Appendix B: Contents of maximum cliques and independent sets for Russian and Swedish market graphs

See Tables 8, 9, 10, 11.

Table 8 Maximum independent set, Russia

Table 9 Maximum independent set, Sweden

Table 10 Maximum clique, Russia

Table 11 Maximum clique, Sweden