Derivative of the Joint Distribution Interpretation

Question

Given two continuous random variables $X$ and $Y$, the joint cumulative distribution function $F_{X,Y}$ is defined as:

$$F_{X,Y}(x,y)=\mathbb{P}(X\le x, Y\le y)=\displaystyle\int_{-\infty}^{x}\int_{-\infty}^{y} f_{X,Y}(t_1,t_2)\mathrm{d}t_1\mathrm{d}t_2,$$

where $f_{X,Y}$ is the joint probability density function of $X$ and $Y$.

The second partial derivative $\partial^2F_{X,Y}(x,y)/\partial x\partial y$ gives the joint probability density $f_{X,Y}(x,y)$. I am trying to find out what the following partial derivatives represent:

$$\begin{align} \dfrac{\partial}{\partial x}F_{X,Y}(x,y) &= \int_{-\infty}^{y} f_{X,Y}(x,t_2)\mathrm{d}t_2, \\[6pt] \dfrac{\partial}{\partial y}F_{X,Y}(x,y) &= \int_{-\infty}^{x} f_{X,Y}(t_1,y)\mathrm{d}t_1. \\[6pt] \end{align}$$

What do these partial derivatives represent? Do they have any particular interpretation?

Answer 1

The first-order partial derivatives of a multivariate joint distribution function can be considered as giving the density of the differentiated variable, jointly with the cumulative probability of the other variable(s). One simple way to see this interpretation is to convert the partial derivative to a density integral, integrated over the other dimensions. From the fundamental theorem of calculus we can write the partial derivative as:

$$\begin{aligned} \frac{\partial}{\partial x} F_{X,Y} (x, y) &= \int \limits_{- \infty}^y f_{X,Y} (x, t) dt \\[6pt] &= \int \limits_{- \infty}^y f_{Y|X} (t|x) f_X(x) dt \\[6pt] &= \int \limits_{- \infty}^y f_{Y|X} (t|x) dt \times f_X(x) \\[6pt] &= \mathbb{P} (Y \leqslant y | X = x) f_X (x). \end{aligned}$$

This shows that the partial derivative gives us the joint density over the line $Y \leqslant y, X = x$ (within the two-dimensional space of the two random variables). The partial derivative with respect to $y$ has an analogous interpretation.

Answer 2

If you take the joint CDF over xy and derive it over just one of the variables - you're left with marginal PDF for that same variable.

Let's prove using a simple joint distribution of two i.i.d. RVs X and Y ~Expo(1)

On one hand, we can get to the marginal PDF through the joint PDF: $$ F_{XY}(x,y) = \iint_{XY} f_{XY}(x,y) dxdy = \int_{0}^{\infty }\int_{0}^{\infty } e^{-x}e^{-y}dxdy \\ f_{XY}(x,y) = \frac{\partial^{2} }{\partial x \partial y} F_{XY}(x,y) = e^{-x}e^{-y} \\ f_{X}(x) = \int_{Y}f_{XY}(x,y)dy = e^{-x} \int_{0}^{\infty } e^{-y}dy = e^{-x} $$ Where last equation simplifies to e^(-x) because PDF of standalone y integrates to 1.

Alternatively we can go directly from joint CDF to marginal PDF: $$f_{X}(x) = \frac{\partial }{\partial x} F_{XY}(x,y) = e^{-x} \int_{0}^{\infty } e^{-y}dy = e^{-x}$$

Marginal PDF, if you're unfamiliar, is basically the PDF of X standalone, "freed up" from Y.