I cannot understand why the range of integral of $x, y$ are from $0$ to $1$ when $x > 0 $ and $y < 1$.

Question

First of all, I am not native speak so sorry for my poor English.

I am studying the text book named "the probability and statistics for engineering and scientists" by myself.

But I am not good at this subject.

So I need help from experts.(Please help me)

Q) Let X and Y be random variables with joint density function.

$$\ f(x, y) = 4xy (0<x, y<1) $$

Find the expected value of $$\ Z = {\sqrt{(X^2 + Y^2)}} $$.

The problem is that I cannot understand why the range of integral of x, y are from 0 to 1.

Please explain that above..

Answer 1

The integrals are "technically" not on $(0,1)\times(0,1)$ only, but on $\mathbb{R}^2$: you do have $$ \mathbb{E}[Z] = \int_{\mathbb{R}^2} dxdy f(x,y) \sqrt{x^2+y^2} $$ as... expected. However, by definition, $f(x,y)$ is non-zero only if $(x,y)\in (0,1)\times(0,1)$, so we end up having $$ \mathbb{E}[Z] = \int_{\mathbb{R}^2} dxdy 4xy \mathbf{1}_{(0,1)\times(0,1)}(x,y) \sqrt{x^2+y^2} = \int_{(0,1)\times(0,1)} dxdy 4xy \sqrt{x^2+y^2} $$

This is because $\mathbf{1}_{(0,1)\times(0,1)}(x,y)$ is by definition zero when $(x,y)\in(0,1)\times(0,1)$, and $1$ otherwise (it is the indicator function of $(0,1)\times(0,1)$).

Answer 2

Sometimes people write “$0<x,y<1$” to mean “$0<x<1$ and $0<y<1$”, similar to how one might write “$x,y \in (0,1)$” as an abbreviation for “$x \in (0,1)$ and $y \in (0,1)$”. And I think this is what's meant here. But this notation is a bit dangerous, since it can be interpreted (as you seem to have done) as “$0<x$ and $y<1$”, so I would recommend avoiding it.