# Derivative of the Joint Distribution Interpretation

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 $\dfrac{\partial^2}{\partial x\partial y}F_{X,Y}(x,y)$ gives the joint probability density $f_{X,Y}(x,y)$.

But what does, say, partial derivatives $\dfrac{\partial}{\partial x}F_{X,Y}(x,y)=\int_{-\infty}^{y} f_{X,Y}(x,t_2)\mathrm{d}t_2$ and $\dfrac{\partial}{\partial y}F_{X,Y}(x,y)=\int_{-\infty}^{x} f_{X,Y}(t_1,y)\mathrm{d}t_1$ represent? Do they have any particular interpretation?

$$\frac{\partial}{\partial x} F_{X,Y} (x, y) = \int \limits_{- \infty}^y f_{X,Y} (x, t) dt = \mathbb{P} (Y \leqslant y | X = x) f_X (x).$$
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.
• We could simplify the last term into $\mathbb{P}(Y \leq y, X = x)$ this also makes it more intuitive, at least to me $$\frac{\partial}{\partial x} \mathbb{P}(Y \leq y, X \leq x) = \mathbb{P}(Y \leq y, X = x)$$ – Martijn Weterings Mar 2 '18 at 10:23
• That statement seems to me to be false, since $\mathbb{P}(X=x) = 0$ for any continuous random variable $X$, and so the joint probability would also be zero. I think you have to keep the density of $X$ separate to the probability mass of $Y$. – Ben Mar 2 '18 at 20:48
• That's right I should have used $$\mathbb{P}(Y \leq y, x-\frac{1}{2} dx \leq X \leq x+\frac{1}{2} dx)$$ – Martijn Weterings Mar 3 '18 at 11:01
• Still need to divide by $dx$, otherwise it's still zero. ;) – Ben Mar 3 '18 at 13:24