Problem understanding join CDF of two independent random variables

Question

I have the following problem, which was asked in a similar question but it doesn't help me.

A dart is equally likely to land at any point inside a circular target of unit radius. Let $r$ and $\phi$ be the radius and the angle of the point

(a) Find the joint cdf of r and $\phi$

(b) Find the marginal cdf of R and $\phi$.

I have the solution (always for $0\leq r \leq 1$ and $0 \leq\phi \leq 2\pi$), which is $$F(R, \phi)=\frac{r^2\phi}{2\pi}$$

Since it is equal to the area of the pie slice divided between the area of the circle and $\frac{\pi}{\pi}=1$. I, however, don't understand this.

The angle is a uniform random variable between $0$ and $2\pi$, so it makes sense for its cdf to be $\frac{\phi}{2\pi}$. However, the radius is a uniform between $0$ and $1$, so, as far as I know, its cdf should be just $r$, not $r^2$. I can understand the area of the pie divided between the total area being the join cdf, but mathematically, there is something I am missing to justify the $r^2$. Can someone help me with this?

Also, my solution states that $F(r)=r^2$, so the problem is in the second bulletpoint too.

Answer 1

$$P(R\leq r)=\frac{\text{area of disc with radius }r}{\text{area of disc with radius }1}=\frac{\pi r^2}{\pi}=r^2$$

This is based on the fact that we are dealing with a unifom distribution on a disc with radius $1$.

Then for every suitable subset $A$ of the disc we have:$$P(\text{dart lands in }A)=\frac{\text{area of }A}{\text{area of disc with radius }1}$$

Answer 2

reading your question, the joint distribution is UNIFORM on the unit disk, thus the joint pdf is simply the reciprocal of the area thus

$$f_{XY}(x,y)=\frac{1}{\pi}$$

in order to find the pdf of the vector $(R;\Theta)$ indicating radius and angle, respectively, you simply can pass in polars obtainig

$$f_{P \Theta}(\rho;\theta)=\frac{\rho}{\pi}$$

now integrating you get your desired joint cdf

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to find marginals, integrate the opposite rv:

$$f_{P}(\rho)=2\rho$$

with CDF

$$F_{P}(\rho)=\int_0^\rho 2t dt=\rho^2$$

and

$$f_{\Theta}(\theta)=\int_0^1 \frac{\rho}{\pi}d\rho=\frac{1}{2\pi}$$

which is uniform in $\theta \in (0;2\pi)$