Finding the probability density function given the cumulative distribution function in multivariate case. Usually, the connection between CDF ($F(x,y)$) and PDF($f(x,y)$) (in 2 dimension for example) is:
$$F(x,y) = \int\limits_{-\infty}^{x}\int\limits_{-\infty}^{y} f(x,y) \, dxdy$$
($F(x,y)$ is absolutely continuous.)
So, we get PDF from CDF, as partial derivative:
$$ f(x,y) = \frac{\partial ^2 F(x,y)}{\partial x \, \partial y}$$
However, when F(x,y) is "tricky" (or should I say "Degenerate"?) sometimes, it unclear how to find PDF from CDF.
For example:
$$ F(x,y) = \begin{cases}
x + y, & \quad 0 \leqslant x,y \leqslant 1 , \quad x+y \leqslant 1 \\
1, & \quad 0 \leqslant x,y \leqslant 1 , \quad 1 < x+y \\
\end{cases} \\ 
(x,y \in [0,1] )$$
Is its PDF ?
$$ f(x,y) = \begin{cases}
\delta(x) + \delta(y), & \quad 0 \leqslant x,y \leqslant 1 , \quad x+y \leqslant 1 \\
0, & \quad \text{otherwise}
\end{cases}$$
where $\delta$ is Dirac delta function.
Any hint about how to deal with such "tricky" CDFs is of great grateful !
 A: When a cumulative distribution function $\ F:\mathbb{R}^n\rightarrow[0,1]\ $ has discontinuities, I don't believe it's standard to take the generalised function representation of $\ \frac{\partial^nF}{\partial x_1\partial x_2\dots\partial x_n}\ $ as a definition of $\ F$'s density function. When $\ F\ $ is not absolutely continuous (i.e. when there's no ordinary real-valued function $\ f:\mathbb{R}^n\rightarrow[0,\infty)\ $ which satisfies the identity
$$F\left(x_1,x_2,\dots,x_n\right)=\int_{-\infty}^{x_1}\int_{-\infty}^{x_2}\dots \int_{-\infty}^{x_n} f\left(y_1,y_2,\dots,y_n\right)dy_n\dots dy_1\ )
$$
I believe $\ F\ $ is generally regarded as not possessing a density function, rather than as having a density function that is a generalised function.  Of course this is simply a matter of terminology, and if you find it convenient to work with generalised functions, I see no reason why you couldn't treat them as providing a sort of generalised density function for those cumulative distribution functions which don't have one in the usual sense.
I believe the generalised density function you've given for your example is correct.
A: Make the integral as the sum of two integrals.  $\int\int\delta (x)dxdy=y$ and $\int\int\delta (y)dydx=x$  For both the inner integral of the delta function $=1$
