How do I find the PDF from a multidimensional CDF with indicator functions? I have what I'm sure is a very stupid question.
When I have a two-dimensional random variable $\tilde{X}=(X_1,X_2)$ with the cdf $F(x_1,x_2)=(kx_1^2I_{(0,1)}(x_1)+I_{[1,\infty)}(x_1))(kx_2^2I_{(0,1)}(x_2)+I_{[1,\infty)}(x_2))$, how can I calculate the pdf from that? I get that I have to differentiate somehow, but what do I do with the indicator functions?
 A: Alternative way to describe the CDF without indicator functions
It is common to split up the CDF into cases like for the uniform distribution between $a$ and $b$ you have: $$F_X(x) = \begin{cases} 0 &:& x < a \\
\frac{x-a}{b-a} &:& a\leq x\leq b\\
1 &:& x>b
\end{cases}$$
and the density is the derivative $f_X(x) = {F_X}^\prime(x)$
$$f_X(x) = \begin{cases} 0 &:& x < a \\
\frac{1}{b-a} &:& a\leq x\leq b\\
0 &:& x>b
\end{cases}$$
You can get rid of the indicator in your function by describing it in the format above (only now you have to write the cases based on two variables together).

The derivative of a 2D CDF to obtain a PDF
The CDF is differently defined in a 2D case. One way is like
$$F_{X,Y}(x,y) = P(X\leq x, Y \leq y) = \int_{-\infty}^x\int_{-\infty}^y f_{X,Y}(t,s) ds dt$$
So you need to compute the derivative for both variables
$$\frac{\partial^2}{\partial  x\partial y} F_{X,Y}(x,y) = f_{X,Y}(x,y)$$
A: When in doubt, deal with indicator functions (and other awkward creatures like absolute values) by breaking the problem into cases.
Here we have
$$F(x_1,x_2) =
\begin{cases}
(kx_1^2)(kx_2^2)  & \text{if } x_1 \in (0,1) \text{ and } x_2 \in (0,1)\\
kx_1^2 &  \text{if } x_1 \in (0,1) \text{ and } x_2 \gt 1 \\
kx_2^2  & \text{if } x_1 \gt 1 \text{ and } x_2 \in (0,1)\\
1  & \text{if } x_1 \gt 1 \text{ and } x_2 \gt 1 \\
\end{cases}$$
Each of these is straightforward to differentiate.
Sometimes you'll be able to consolidate the cases afterwards. Here for instance, $\frac{\partial^2 F}{\partial x \partial y}$ is zero everywhere outside the unit square $0 \le x_1,x_2 \le 1$
