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If $\theta\sim Unif[-1,1]$, then what is the CDF of $U=\sin(\pi\theta)$?

Now, its easy to see that $$P_{U}(t) = P\left(\theta \leq\frac{\sin^{-1}(t)}{\pi}\right)$$

somehow the answer is equal to :

$$ \mathrm{if} \,\,\,\, \, 0 \leq t \leq 1 : P(U\leq t) = \frac{1+\frac{2 \,\sin^{-1}(t)}{\pi}}{1-(-1)}$$

similarly for the other part of the interval but with: $$\frac{1-\frac{2 \,\sin^{-1}(-t)}{\pi}}{1-(-1)}$$

The denominator is of course form the definition of the Unifrom dist as the length of the interval. But I cant get how we obtained that numerator value, why is there a $1-2..$ I tried drawing the graph but I cant get it still.

Any hints about the clarification are appreciated!

StubbornAtom
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Fred
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    Since $\pi\theta\sim U(-\pi,\pi)$ which is equivalent to $U(0,2\pi)$, see https://math.stackexchange.com/questions/1026366/distribution-of-sine-of-uniform-random-variable-on-0-2-pi?rq=1. – StubbornAtom Jun 20 '19 at 07:11

1 Answers1

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To help to calculate the probability $$P(\sin(\pi \theta)\leq u)$$ take a look at the following figure: enter image description here

In the upper figure, $u>0$ and if $\pi\theta$ falls in the thick red region then the event in question occurs. That is, for $1\geq u>0$ the probability (the value of the cdf) at $u$ is

$$\frac{\pi+2\sin^{-1}(u)}{2\pi}=\frac12+\frac{\sin^{-1}(u)}{\pi}.$$

In the lower figure, one can see when the event in question takes place; look at the thick blue line. So the probability (the value of the cdf) for $-1\leq u\leq0$ is

$$\frac{\pi-2|\sin^{-1}(u)|}{2\pi}=\frac12-\frac{|\sin^{-1}(u)|}{\pi}.$$

zoli
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