Suppose that $X$ and $Y$ are the co-ordinates of a point chosen uniformly at random from
the disc {(x, y) ∈ R : $x^2$+ $y^2$ ≤ 1}. Define random variables $R>0$ and $Θ∈[0, 2π)$ via $X=Rcosθ$ and $Y=Rsinθ$ and by considering the area of an appropriate region in the plane determine, for $0≤θ_1≤θ_2≤2π$ and $0≤r_1≤r_2≤1 $, the value of the probability
$P(r_1≤R≤r_2 $ and $ θ_1≤Θ≤ θ_2)$
From this deduce the density of joint distribution of $R$ and $Θ$
My attempt so far:
I drew a sketch which indicated that the area that I needed to find was infact one sector minus another, giving the answer of $(1/2$)$(θ_2-θ_1)$$((r_2^2-r_1^2)$ but I am unsure how to use this to deduce the joint probability density function.
Any help appreciated!
Best Answer
The expression $\frac12(\theta_2-\theta_1)(r_2^2-r_1^2)$ is an area, not a probability. To get a probability you need to divide by the whole circle's area of $\pi$ to get $\frac1{2\pi}(\theta_2-\theta_1)(r_2^2-r_1^2)$. Now fix $\theta_1=r_1=0$ to get $P(\Theta\le\theta,R\le r)=\frac1{2\pi}\theta r^2$ and differentiate with respect to both variables to get the joint density of $f_{R,\Theta}(r,\theta)=\frac r\pi$.