# Why the transformation of probability density function cannot forming probability density function? What my mistakes?

Let $$\theta$$ is a random variable which uniform distributed $$(-\pi,\pi)$$, with probability density function $$f(\theta)=\begin{cases}\dfrac{1}{2\pi}, -\pi<\theta<\pi\\ 0, \text{otherwise}\end{cases}.$$

Let $$Y=\cos\theta$$. Find the probability density function of $$Y$$.

To answer this question I'm using Jacobian transformation like this.

$$Y=\cos\theta$$ imply $$\theta=\arccos Y$$.

The Jacobian is $$\vert J\vert = \left\vert\dfrac{d\theta}{dy}\right\vert= \left\vert-\dfrac{1}{\sqrt{1-y^2}}\right\vert=\dfrac{1}{\sqrt{1-y^2}}.$$ So I have $$g_Y(y)=f(\theta)\vert J\vert=f(\arccos Y)\vert J\vert=\dfrac{1}{2\pi\sqrt{1-y^2}}.$$

To find the range of $$y$$, I try to plot the cosine graphics from $$\theta=-\pi$$ to $$\theta=\pi$$ like this.

We can see that the range of $$y$$ is $$-1. So I have

$$g_Y(y)=\begin{cases}\dfrac{1}{2\pi\sqrt{1-y^2}}, -1

Now I want to check that $$g(y)$$ is a probability density function with the property of pdf.

$$\begin{eqnarray} \int\limits_{-1}^{1} \dfrac{1}{2\pi\sqrt{1-y^2}} dy &=& \left[-\dfrac{1}{2\pi} \arccos y\right]_{-1}^1\\ &=& -\dfrac{1}{2\pi} \arccos (1)+\dfrac{1}{2\pi} \arccos (-1)\\ &=& -\dfrac{1}{2\pi} (0)+\dfrac{1}{2\pi} (\pi)\\ &=& \dfrac{1}{2}. \end{eqnarray}$$

If $$g(y)$$ is p.d.f. the integral should be $$1$$.

What is my mistake in my work? Did I make a mistake in determining the range of $$y$$ or the transformation?

• As what you plot the transformation, it is not a one-to-one transformation for the entire domain, but it is a one-to-one transformation on $[-\pi, 0]$ and $[0, \pi]$ individually if you split them. You may think about when does $Y = \cos \theta \Rightarrow \theta = \arccos Y$ holds - the principal domain. Essentially, you have done the transformation, for half of the domain, so you got $1/2$ in the process. The another half is the same, so you add them up and will get $1$. – BGM Jul 10 '19 at 12:44
• @BGM Yes, I understand what you mean. Thank you very much for your help. – Ongky Denny Wijaya Jul 10 '19 at 14:05

The reason for the discrepancy is the following: The connection between the $$\theta$$-domain and the $$y$$-domain is not bijective, but $$2:1$$.
I'd argue as follows: When $$Y=\cos\Theta$$ then we know that $$-1\leq Y\leq 1$$. Let $$G$$ be the cumulative distribution function of $$Y$$. Then $$G(y)=P(Y\leq y)=P(\cos\Theta\leq y)=P(\arccos y\leq|\Theta|\leq\pi)=2{\pi-\arccos y\over2\pi}\ .$$ This gives $$g_Y(y)=G'(y)={1\over\pi\sqrt{1-y^2}}\qquad(-1\leq y\leq1)\ .$$