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?


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:

$$\frac{\partial}{\partial x} F_{X,Y} (x, y) = \int \limits_{- \infty}^y f_{X,Y} (x, t) dt = \mathbb{P} (Y \leqslant y | X = x) f_X (x).$$

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.

  • $\begingroup$ We could simplify the last term into $\mathbb{P}(Y \leq y, X = x)$ this also makes it more intuitive, at least to me $$\frac{\partial}{\partial x} \mathbb{P}(Y \leq y, X \leq x) = \mathbb{P}(Y \leq y, X = x)$$ $\endgroup$ – Sextus Empiricus Mar 2 '18 at 10:23
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    $\begingroup$ 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 $Y$. $\endgroup$ – Reinstate Monica Mar 2 '18 at 20:48
  • $\begingroup$ That's right I should have used $$\mathbb{P}(Y \leq y, x-\frac{1}{2} dx \leq X \leq x+\frac{1}{2} dx)$$ $\endgroup$ – Sextus Empiricus Mar 3 '18 at 11:01
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    $\begingroup$ Still need to divide by $dx$, otherwise it's still zero. ;) $\endgroup$ – Reinstate Monica Mar 3 '18 at 13:24

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