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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. enter image description here

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

$$ g_Y(y)=\begin{cases}\dfrac{1}{2\pi\sqrt{1-y^2}}, -1<y<1\\ 0, \text{otherwise}\end{cases}. $$

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?

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    $\begingroup$ 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 $1$. $\endgroup$ – BGM Jul 10 '19 at 12:44
  • $\begingroup$ @BGM Yes, I understand what you mean. Thank you very much for your help. $\endgroup$ – Ongky Denny Wijaya Jul 10 '19 at 14:05
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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 $$g_Y(y)=G'(y)={1\over\pi\sqrt{1-y^2}}\qquad(-1\leq y\leq1)\ .$$

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  • $\begingroup$ Thank you very much @Christian Blatter $\endgroup$ – Ongky Denny Wijaya Jul 10 '19 at 14:01

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