I am trying to solve this math question, but I find myself getting stuck. I have a joint PDF of the form:
$f(x, y) = (\pi R^2)^{-1}$ for all $(x, y)$ in the circle with radius $R$, $0$ otherwise.
Now I have found the marginal PDF's in this question, but the question asks for me to find the marginal DF of each random variable $X$ and $Y$.
I think the Joint DF is $xyf(x, y)$, as taking the derivative with respect to $x$, then $y$ will give back the joint PDF, but I am unsure. If someone could help me out it would be appreciated. Thanks!
You have a uniform distribution within the disc of radius $R$.
The support is thus $\big\{(x,y)\mid x\in(-R;R), y\in(-\sqrt{R^2-x^2~};\sqrt{R^2-x^2~})\big\}$
Therefore the marginal pdf for $X$ is:
$$f_X(s) ~=~ \int\limits_{-\sqrt{R^2-s^2}}^{\sqrt{R^2-s^2}} (\pi R^2)^{-1}~\mathbf 1_{s\in (-R;R)}\operatorname d t $$
The marginal CDF for $X$ is then $$F_X(x) = \left(\int\limits_{-R}^x \int\limits_{-\sqrt{R^2-s^2}}^{\sqrt{R^2-s^2}} (\pi R^2)^{-1}\operatorname d t \operatorname d s\right) \mathbf 1_{x\in(-R;R)}+\mathbf 1_{x\in[R;\infty)}$$
And symmetrically for $Y$.
The Joint CDF is $$F_{X,Y}(x,y) ~=~\begin{cases}0 &:& \text{elsewhere} \\[1ex] \int\limits_{-R}^x \int\limits_{-\sqrt{R^2-s^2}}^{y} (\pi R^2)^{-1}\operatorname d t\operatorname d s &:& x\in(-R;R), y\in(-\sqrt{R^2-x^2~};\sqrt{R^2-x^2~}) \\ F_X(x) &:& x\in(-R;R), y\in[\sqrt{R^2-x^2~};\infty) \\ F_Y(y)&:& y\in(-R;R), x\in[\sqrt{R^2-y^2~};\infty) \\ 1 &:& (x,y)\in[R;\infty)^2\end{cases}$$