Let $X$ be a continuous random variable having uniform distribution on $[0, 2\pi]$. What distribution has the random variable $Y=\sin X$ ? I think, it is also uniform. Am I right?
Asked
Active
Viewed 1.6k times
10
Davide Giraudo
- 172,925
Nikita Martynov
- 101
-
1No, it is not. Loosely speking: Since the sine is "slow" near its extremes, if is more likely that $Y\approx 1$ than that $Y\approx 0$. – Hagen von Eitzen Nov 17 '14 at 18:14
-
https://math.stackexchange.com/q/118108/321264 – StubbornAtom Oct 10 '20 at 15:00
2 Answers
12
By the Jacobian formula, $$f_Y(y)=\frac{\mathbf 1_{|y|\lt1}}{\pi\sqrt{1-y^2}}.$$ This is the so-called Arcsine distribution, which famously appears in probability theory and in number theory, as shown by two mathematical giants from the 20th century:
$\qquad\qquad\qquad\qquad$
Did
- 279,727
9
The support of the distribution is of course $[-1,1]$. For $y\in[0,1]$, we have \begin{align} \Pr(Y\le y) & = \Pr(0\le X\le\arcsin y\text{ or }2\pi\ge X\ge \pi-\arcsin y) \\[10pt] & = \frac{\arcsin y + (2\pi-(\pi-\arcsin y))}{2\pi}. \end{align} The density is the derivative of that, and that is not a constant function, so it's not uniformly distributed.
