From joint cdf to joint pdf

Question

We can get the joint pdf by differentiating the joint cdf, $\Pr(X\le x, Y\le y)$ with respect to x and y. However, sometimes it's easier to find $\Pr(X\ge x, Y\ge y)$. Notice that taking the complement doesn't give the joint CDF, so we can't just differentiate and flip signs. Is there still some simple rule to differentiate this and get the pdf?

Answer 1

A joint distribution has domain $(-\infty, \infty) \times (-\infty, \infty)$. If we partition each component of the cartesian product in two by selecting some value $x$ and some value $y$, then we get $4$ subsets,

$$(-\infty, x] \times (-\infty, y],\;\;(-\infty, x] \times [y,\infty),\\ [x, \infty) \times (-\infty, y],\;\;[x, \infty) \times [y,\infty)$$

made up of intersections of two events,

$$A = P(X\le x), \;\; B = P(Y\le y)$$

and their corresponding complements.

Then (as the OP noted in a commnent),

$$\Pr(X\ge x, Y\ge y) = P(A^c\cap B^c) = 1 - P(A\cup B)$$

$$=1-\big[P(A) + P(B) - P(A\cap B)\big]$$

So it appears that by taking the cross-partial derivative of $\Pr(X\ge x, Y\ge y)$ we should again get the joint density. Let's verify that:

$$\Pr(X\ge x, Y\ge y) = \int_x^{\infty}\int_y^{\infty}f(s,t)dtds$$

$$\frac {\partial \Pr(X\ge x, Y\ge y)}{\partial y} = \int_x^{\infty} \left(\frac{\partial}{\partial y}\int_y^{\infty}f(s,t)dt\right)ds $$

$$=\int_x^{\infty}-f(s,y) ds$$

$$\frac {\partial^2 \Pr(X\ge x, Y\ge y)}{\partial y\partial x} = \frac {\partial }{\partial x} \int_x^{\infty}-f(s,y) ds = -\left(-f(x,y)\right) = f(x,y)$$

The above also means that we can obtain the joint pdf from any of the four joint events indicated by the breakdown of the support -but in the other two cases, we should multiply by $-1$.

$$\begin{align} f(x,y) =& \frac {\partial^2 \Pr(X\le x, Y\le y)}{\partial y\partial x}\\ =&\frac {\partial^2 \Pr(X\ge x, Y\ge y)}{\partial y\partial x}\\ =&-\frac {\partial^2 \Pr(X\le x, Y\ge y)}{\partial y\partial x}\\ =&-\frac {\partial^2 \Pr(X\ge x, Y\le y)}{\partial y\partial x} \end{align}$$