How is the standard definition of independence applied to time-series?

Question

So, [Wikipedia says] that the standard definition of independence is:

$f_{X,Y}(x,y) = f_X(x) f_Y(y)$

How is this applied to timeseries? How do we calculate each side of the equation?

If we're assuming that the probability distributions of the two series are the same, then $X=Y$ and we can just use $f(x,y) = f(x) f(y)$, right? So then I geuss I am basically asking what is f(x,y), and how do we calculate it?

Answer 1

Dependence is what makes time series, which are really sequences, interesting. Otherwise they would be an unordered set of random variables. Returning to your question, the issue is whether the joint distribution of $X_{t_i}$ and $X_{t_j}$ is separable: $f_{X_{t_i}, X_{t_j}}(x, y) \overset{?}{\equiv} f_{X_{t_i}}(x) f_{X_{t_j}}(y)$

Answer 2

Independence means that the joint density function for a set of variables is a product of their individual marginal densities. This doesn't change in the context of time series. $f(x,y)$ is the joint density for $X$ at $x$ and $Y$ at $y$. If $X$ and $Y$ are independent then it is the product of $f_X(x)$ and $f_Y(y)$. Note that $X$ and $Y$ do not have to have the same density. That is why Wikipedia used the subscripts and you should have too. Now in general if the variables are dependent than you need to know the form of the density and the value of the parameters. For example the bivariate normal density looks as follows:

$$ f_{X,Y}(x, y) = c e^{-q(x,y)} $$,

where the normalizing constant is

$$c = \frac{1}{2π \sqrt{(1-ρ^2)} σ_x σ_y} $$

and

$$q(x,y) = \frac{ \left( \frac{(x-μ_x)^2}{σ_{x}^2} - \frac{2 ρ (x-μ_x)(y-μ_y)}{σ_x σ_y} + \frac{(y-μ_y)^2}{σ_{y}^2}\right) }{2(1- ρ^2)} $$

We have two variances a correlation and constants in the formula. Given the parameters $ρ$, $σ_x$, $σ_y$ and $μ_x$ and $μ_y$ the joint density can be calculated.