I know the correct solution to this, but I am trying to understand why my approach is wrong.
Let P be a point that is randomly chosen from circle with radius R. My approach is the following:
Point P is a distance $d$ from the origin and is at an angle $\theta$ with the horizontal, where $d$ is uniformly distributed on $(0,R)$ and $\theta$ is uniformly distributed on $(0,2\pi)$. $d$ and $\theta$ are independent.
What we're interested in is the distribution of $X = d*cos(\theta)$, where the densities of $d$ and $\theta$ are $f_d(d) = 1/R$ and $f_\theta(\theta) = 1/2\pi$
Let $Y = cos(\theta)$, $F_Y(y) = P(Y<y) = P(cos(\theta) <y) = P(\theta <cos^{-1}(y)) = cos^{-1}(y)$
Therefore the density $f_y(y) = dF_y(y)/dy = -1/\sqrt {1 – y^2}$
What we want is the density $f_X(x)$. Because $d$ and $\theta$ are independent in the expression $X = d*cos(\theta) = d*Y$, then you would expect $f_X(x) = f_d(d)*f_Y(y) = -1/(R*\sqrt {1 – y^2})$
But this is the wrong answer. Where's my mistake?
Best Answer
$d$ is not uniformly distributed over $(0,R)$. Indeed, $d\leq t$ is equivalent to point is in the sircle with radius $t$, and cdf of $d$ is $$ F_d(t)=\mathbb P(d\leq t) = \frac{\pi t^2}{\pi R^2}=\frac{t^2}{R^2}, 0<t<R. $$ So, pdf of $d$ is $$ f_d(t)=\frac{2t}{R^2}, 0<t<R. $$