Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
add comment

2 Answers 2

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 ${2 \over \pi \sqrt{1 - \alpha^2}}$.

share|improve this answer
    
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
add comment

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?

share|improve this answer
    
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
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.