Concept of "identically distributed copy" in Kendall's tau In this precise moment of this lecture:
$(X_1, X_2)$ 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?
 A: 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 $\bar{x}$. 
A: 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 $X$.
A: 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.
