# Cosine Transformation on Uniform Distribution

Given the standard-Arcsin probabilty function, called $Arcsin(0,1)$, looking like this:

$F_X(x) = \left\{ \begin{matrix} 0 && x < 0\\ \frac{2}{\pi}\arcsin(\sqrt{x}) && 0 \le x \le 1\\ 1 && x > 1 \end{matrix} \right.$

Now say that $W$ is a uniform distribution over the interval $[0, \pi]$. We do a translation called $V$, and this is defined as the following: $V = -\cos(W)$ I have to proof that this is an $Arcsin(-1, 1)$ distribution. To dit this, I would do this:

Let's say that $g(x) = -\cos(x)$, and this is our transformation function.

$f_V(y) = f_W(g^{-1}(y))$

Now we only have to calculate our inverse function right? But when I do this, I don't come near the given answer. $y = -\cos\left({\frac{x}{\pi}}\right) = -\left(\cos\left(2 \cdot \frac{x}{2\pi}\right)\right) = 2\sin^2\left(\frac{x}{2\pi}\right)-1$

$y + 1 = 2\sin^2\left( \frac{x}{2\pi} \right)$

$\sqrt{\frac{y+1}{2}} = \sin\left(\frac{x}{2\pi}\right)$

$\arcsin\left(\sqrt{\frac{y+1}{2}}\right) = \frac{ x}{2\pi}$

$2\pi \arcsin\left(\sqrt{\frac{y+1}{2}}\right) = x$

The resulting function is a function $(-1,1) \rightarrow [0,10]$ So it's a factor 10 too much. My question is, what steps are missing? Or is the question just wrong?

• To start, the transformation is: $$f_V(v)= f_W(g^{-1}(v))\cdot\lvert \frac{\mathrm d g^{-1}(v)}{\mathrm d v}\rvert$$ Second, if $g(x)=-\cos(x)$ then $g^{-1}(y)= \arccos(-y)$. Dec 5, 2016 at 13:09

$F_V(y) = F_W(g^{-1}(y))$
Next thing which comes to mind, is that I actually didn't use this equation. So when you got a formula for $x$, you have to plug this back into $F_W(x)$. The third thing I did wrong is $y \ne -\cos(\frac{x}{\pi})$, instead I should have used $y = \cos(x)$.
Filling this in using the equations I used, the answer will be: $2 \arcsin\left(\sqrt{\frac{y+1}{2}}\right) = x$. This is something that looks more like it.
This means that: $F_V(y) = F_W\left(2 \arcsin\left(\sqrt{\frac{y+1}{2}}\right)\right)$ And this results in the following:
$F_V(y) = \left\{ \begin{matrix} 0 && y < 0\\ \frac{2\arcsin\left(\sqrt{\frac{y+1}{2}}\right)}{\pi} && 0 \le -\cos(y) \le 1\\ 1 && y > 1 \end{matrix} \right. = \left\{ \begin{matrix} 0 && y < -1\\ \frac{2}{\pi}\arcsin\left(\sqrt{\frac{y+1}{2}}\right) && -1 \le y \le 1\\ 1 && y > 1 \end{matrix} \right.$