# Concept of “identically distributed copy” in Kendall's tau

$(X_1, X_2)$ are the log returns of companies $X_1$ and $X_2,$ respectively. Then "we create an identically distributed copy"...

$$(X_1, X_2) \underset{id}{=} (X_1', X_2')$$

to define Kendall's tau as

$$\tau (X_1, X_2) = \Pr \left((X_1 - X_1') (X_2 - X_2') > 0\right) - \Pr \left((X_1 - X_1') (X_2- X_2') < 0\right)$$

I get the concordant and discordant pairs explanation in Wikipedia, but I don't know what is meant by "identically distributed copy".

Is it a permutation of the sample, is it the sample sorted in decreasing or increasing order?

And how is this definition reconciled with the Wikipedia explanation?

• Here you define the "theoretical" Kendall's tau coefficient. On wiki this is the Kendall's tau coefficient for two paired samples, an estimate of the theoretical one. – Stéphane Laurent Nov 4 '17 at 16:12

X1 and X2 are random variables, not observed (sample) values; getting things with identical distributions doesn't involve doing anything to sample values. It means having a new pair of random variables with the same distribution as the first two.

The thing being discussed here is Kendall's tau as a population quantity (since a copula is a joint distribution, not a collection of samples). The thing in Wikipedia (presuming you mean the Wikipedia page on Kendall's tau) is the corresponding sample correlation measure.

It's the same as the distinction between $E(X)$ and $\bar{x}$.

• The 'log returns of companies' sound like a sample. What do you do at a practical level? Subtracting a pdf (or cdf) from itself (copy) would always be zero... – Antoni Parellada Nov 4 '17 at 16:03
• 1. It might do to review the notion of a random variable; and the distinction between a process and observations made on it. You could certainly have a random variable for something like log returns. 2. The distribution of a difference of two random variables is not the difference of the distributions (nor the densities) 3. The stuff you quote from the lecture is talking about facts about distributions (in particular, a measure of strength of a particular form of dependence that doesn't depend on the marginal distributions), ... – Glen_b Nov 4 '17 at 22:49
• ... in the same way we might establish a result about $E(X)$; it's not trying to say anything about how to estimate Kendall's tau from a sample any more than discussing Markov's inequality implies anything about estimating a tail probability. At a practical level, estimating tau depends on what you're trying to achieve and what you're prepared to assume as well as what's feasible for you to do. You can of course estimate tau in a direct way from a sample; but if you fit a copula via ML, say, then the tau will be implied by the parameter estimates in that fit. – Glen_b Nov 4 '17 at 22:49

An identically distributed copy (for short, a copy) of a random variable $X$ is a random variable $X'$ having the same distribution as $X$ and independent of $X$.

• How do you handle the X1 - X1', then? Subtracting the same pdf (of cdf) would be 0. Do you get random draws from the pdf? – Antoni Parellada Nov 4 '17 at 16:51
• @user183536 This is $X_1-X'_1$, a difference of random variables, there's no pdf or cdf to subtract. – Stéphane Laurent Nov 4 '17 at 16:52
• They are the same random variables, aren't they (copies)? When you try to get P(X1 - X1'), for instance, what do you do? – Antoni Parellada Nov 4 '17 at 16:56
• I expect Antoni understands this well by now but I worry others may have the same temporary confusion here (I've seen it occur many times). Imagine I plan to roll a six sided die and record the value on the uppermost face. That's $X_1$. Then I plan to roll it again and record the new value. That's $X_1'$. The random variables have the same distribution but are not copies of each other (I could get 6 then 3 for example) -- when I take their difference, I don't usually get 0. – Glen_b Oct 26 '18 at 1:00

In trying to illustrate pictorially the intuition behind the definition of the population Kendall's tau as

$$\tau (X_1, X_2) =\Pr\left[(X_1-X_1')\,(X_2 - X_2') >0 \right]- \Pr\left[(X_1-X_1')\,(X_2 - X_2') <0 \right]\tag 1$$

and the relationship both 1. Between the iid random vector $(X_1,X_2)$ and its copy $(X_1', X_2')$; as well as 2. Between these random vectors (2-tuples of dependent random variables) and their joint and marginal densities and distributions (pdf and cdf's), we can take a look at a somewhat illustrative value of $X_1$ in a bivariate normal distribution with a covariance $\mathrm{cov}(X_1,X_2)=0.2.$

In the example, $X_1\sim N(0,0.25)$ and $X_2 \sim N(0,0.25).$ The Pearson correlation is, therefore,

$$\rho(X_1,X_2)=\frac{\mathrm{cov}(X_1,X_2)}{\sigma_{X_1}\,\sigma_{X_2}}=\frac{0.2}{\sqrt{0.25\times0.25}}=0.8.$$

If we consider an illustrative draw from $X_1$ towards the left of the distribution, e.g. $x_1=-1,$ the conditional mean of $X_2$ given this value of $X_1$ will be given by

\begin{align} \mathbb E(X_2|X_1=x)=\mu_Y+\rho \dfrac{\sigma_Y}{\sigma_X}(x-\mu_X)= -0.8 \end{align}

and these conditional mean values will increase from left to right linearly as on the following plot:

and assuming constant variance, its value will be $\mathrm{var}(X_2\vert X_1=x)=\sigma^2_{X_2}\,(1-\rho^2)=0.09$

Now, looking at the first part of equation (1), $(X_1-X_1')$ will be negative whenever the independent draw from the identical rv $X_1'$ is less negative than $x_1,$ which is highly probable: $\Pr(x_2 > -1) = 0.977.$

Looking at second term of the multiplication, i.e. $(X_2-X_2'),$ we know that $\mathbb E[X_2\vert X_1=-1]$ is below the mean of the marginal distribution of $X_2,$ which was designed to be $\mu_{X_2}=0.$

Therefore, precisely because it is highly likely that $x_1' > x_1,$ rendering $(X_1-X_1')<0,$ it is also going to be more likely for $x_2'>x_2,$ since it will more probably come from a conditional normal distribution with $\mathbb E[X_2'\vert X_1'=x_1' ] > \mathbb E[X_2\vert X_1=-1 ],$ in which case $(X_2-X_2')$ will also be negative, rendering $(X_1-X_1')\,(X_2 - X_2') >0.$

The same argument (inverted) holds if we were to look at a value of $x_1 =+1.$

Therefore, the positive correlation imposed on this bivariate normal, would indeed result in a positive Kendall $\tau$ if we swept (integrate) from $-\infty$ to $+\infty,$ to the extent that $\Pr\left[(X_1-X_1')\,(X_2 - X_2') >0 \right]> \Pr\left[(X_1-X_1')\,(X_2 - X_2') <0 \right].$

The inverse would be easy to show if the correlation decided upon had been negative.