# How does the notation of second order derivative read in plain English?

How does the following formula read in the context of Random Variable and probability?

$$f_{XY}(x, y) = \frac{\partial^2 \cdot F_{XY}(x, y)}{\partial x \, \partial y}$$

Since, the plot of a continuous CDF ($$F_{XY}(x,y)$$) looks like the following:

Source: Joseph K. Blitzstein, Jessica Hwang, Introduction to Probability. Page-287.

I am trying to translate the above formula in layman's terms. As far as I can tell, it means:

The Joint Probability Density Function(PDF) of the random variables $$X,Y$$ would be the rate of change of slope of the function $$F_{XY}(x, y)$$ in the both of the directions of $$X$$ and $$Y.$$

Am I correct?

• I'm guessing the notation is $\partial,$ (\partial) not $\delta$ (\delta). It is pronounced "the mixed second partial derivative of the CDF" or "the second partial derivative of the CDF with respect to $x$ and $y$". – spaceisdarkgreen Sep 22 '18 at 20:01
• Also, that is only the plot of a CDF for one variable, not a joint CDF for two variables – spaceisdarkgreen Sep 22 '18 at 20:02

A probability density function, as its name implies, can be interpreted as the density of probability with respect to length - the local "probability per meter" (or whatever the units might be). In the same way, your joint probability density function is the "probability per area" density.

Specifically, take a small rectangle $$R$$ with bottom-left corner $$(x, y)$$ of width $$w$$ and height $$h$$, and consider the probability that $$(X, Y)$$ is in $$R$$:

\begin{align} P((X,Y)\in R)&=P(X\in[x,x+w],Y\in[y,y+h])\\ &=F(x+w,y+h)-F(x+w,y)-F(x,y+h)+F(x,y) \end{align}

In the second equation I'm writing $$F$$ for $$F_{X,Y}$$, and the formula may not be very obvious, but it's easy to see by drawing a diagram. Now divide by the area to get the "probability density":

$$\frac{P((X,Y)\in R)}{wh}=\frac1w\left(\frac{F(x+w,y+h)-F(x+w,y)}h-\frac{F(x,y+h)-F(x,y)}h\right)$$

The right hand side, viewed as a formula in $$(w, h)$$, has a limit at zero equal to $$\frac{\partial^2 \cdot F(x, y)}{\partial x \partial y}=f_{(X,Y)}(x, y)$$, at least if $$F$$ is $$C^2$$. In other words, in the limit of an arbitrarily small rectangle, we get the PDF.

• what is wrong with my description/statement? – user366312 Sep 22 '18 at 22:19
• @yahoo.com It's the rate of change in the direction of $y$ of the rate of change in the direction of $x$. – Jack M Sep 22 '18 at 22:31