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For example, if $f(x) = \sin x$ and $x$ is uniformly distributed on $[0, \pi]$, how is the equation found that satisfies the probability distribution function of $f(x)$? I imagine the distribution function will be greater when the derivative of $f(x)$ is closer to zero, but this is just a guess.

I apologize if this question is vague or not advanced enough, but I can't find the answer anywhere.

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up vote 3 down vote accepted

Note that $\sin(x)$ increases from $x = 0$ to $x = {\pi \over 2}$, then decreases from ${\pi \over 2}$ to $\pi$, in a way symmetric about ${\pi \over 2}$. So for a given $0 \leq \alpha \leq 1$, the $x \in [0,\pi]$ for which $\sin(x) \leq \alpha$ consists of two segments, $[0,\beta]$ and $[\pi - \beta, \pi]$, where $\beta$ is the number for which $\sin(\beta) = \alpha$. In other words $\beta = \arcsin(\alpha)$.

Since $x$ is uniformly distributed on $[0,\pi]$, the probability $x$ is in $[0,\beta]$ is ${\beta \over \pi}$, and the probability $x$ is in $[\pi - \beta, \pi]$ is also ${\beta \over \pi}$. So the chance that $x$ is in one of these two segments is $2{\beta \over \pi}$. This means the probability $\sin(x) \leq \alpha$ is $2{\beta \over \pi}$, or ${2 \over \pi} \arcsin(\alpha)$. Thus this gives the distribution function of $\sin(x)$. The density function is obtained by differentiating with respect to $\alpha$; the result is ${2 \over \pi \sqrt{1 - \alpha^2}}$.

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Sorry for my late reply, but thank you for your answer! I finally understand this concept, and your answer has helped greatly. – Vortico Dec 14 '10 at 4:11

in searching the solution for a nearly problem, I have been falling in your interesting discussion. But there's something I doubt about Zarrax's answer. If the probablity density function is like your result, we can find out easily there's some values of alpha which makes the distribution function greater than 1, is it physically correct?

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Note that if $0 \leq \alpha \leq 1$, then $0 \leq (2/\pi)\arcsin(\alpha) \leq 1$. – Shai Covo Dec 30 '10 at 8:28

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