Calculate the joint CDF of two random variables

[docnet]

Homework Statement

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Relevant Equations

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$$f_{XY}=1$$
$$dzdy=2xdxdy⇒\frac{1}{2\sqrt{z}}dzdy=dxdy$$
$$f_{ZY}=\frac{1}{2\sqrt{z}}\quad \text{on some region S}$$
$$F_{ZY}=\int^y_{g}\int^x_{h}\frac{1}{2\sqrt{z}}dzdy\quad\text{for some}\quad g(x,y),h(x,y)$$
im learning how to find the region S using a change-of variables technique

 

[Orodruin]

The integration region for the joint CDF $f_{ZY}$ is always $z<a$ and $y<b$ by definition.

However, there is no real need to make the change of variables here in the first place. You can just integrate in the xy variables. (This is where you need to be careful with your integration region!)

Edit: Another complication in the change of variables is that your change of variables is not a bijection. There are two x values corresponding to the same z value for all $x \neq 0$.

 

[docnet]

because i get confused easily, i shamelessly copied the change of variables technique on this link i found on Goggle.

$$f_{ZY}=\frac{1}{2\sqrt{z}}\cdot f_{XY}\quad \text{on}\quad 0<z<(1-y)^2, 0<y<1$$

where $\frac{1}{2\sqrt{z}}$ is the absolute value of the Jacobian determinant and $f_{XY}=1$. the region was determined using the inverse transforms $x=\sqrt{z}$ and $y=y$.

is this okay to find the joint CDF by integrating over a square although the region S is not a square? $$F_{ZY}=\int_0^x\int_0^y\frac{1}{2\sqrt{z}}dzdy$$

or is it more like this because I'm supposed to include the boundary information somewhere in the integral?

$$F_{ZY}=\int_0^y\int^x_{(1-y)^2}\frac{1}{2\sqrt{z}}dzdy$$