$\Theta$ is a random variable, uniformly distributed in $[-\pi,\pi]$. What will be $f_\Theta(\theta)$ if $\Theta$ is restricted to $[-\pi/4,\pi/4]$?

Question

Given : $\Phi$ and $\Theta$ are two random variables, related as $\Phi=\sin(2\Theta)$, where $\Theta$ is uniformly distributed in the interval $[-\pi,\pi]$ and I have to find $f_\Phi(\phi)$ i,e. probability density function of $\Phi$.

I am getting difficulty in calculating $f_\Theta(\theta)$ in restricted domain (we have to restrict the domain of $\Theta$ to make the $\Phi=\sin(2\Theta)$ an invertible function)

My approach: $\sin^{-1}(t)$ is invertible when t is in the range of $[\frac{-\pi}{2},\frac{\pi}{2}]$. Here,
$$\frac{-\pi}{2} \leq2\Theta\leq\frac{\pi}{2}$$ $$\Rightarrow\frac{-\pi}{4}\leq\Theta\leq\frac{\pi}{4}$$
Now, since $\Theta$ is uniformly distributed in the interval $[\frac{-\pi}{4},\frac{\pi}{4}]$, so $f_\Theta(\theta)=\frac{4}{2\pi}$, where $\Theta\in[\frac{-\pi}{4},\frac{\pi}{4}]$.

Is this CORRECT?? Is there any simpler way to get $f_\Theta(\theta)$ in restricted domain ?

I saw a solution in which first they found $f_\Theta(\theta)=\frac{1}{2\pi}$, considering $\Theta$ as uniformly distributed in interval $[-\pi, \pi]$. And then for finding $f_\Theta(\theta)$ in restricted domain, they multiplied $f_\Theta(\theta)=\frac{1}{2\pi}$ by $4$ . I didn't get why they simply multiplied by 4??

Answer 1

without doing a lot of calculations, observe that the inverted transformation function (in the inverteble range) is the following

$$\theta=\frac{1}{2}\arcsin \phi$$

with derivative

$$\theta'=\frac{1}{2\sqrt{1-\phi^2}}$$

thus (being the density of $\Theta$ just a constant) in order to get the integral=1 the requested density is

$$f_{\Phi}(\phi)=\frac{1}{\pi\sqrt{1-\phi^2}}\cdot\mathbb{1}_{(-1;1)}(\phi)$$

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Remember that you cannot write $f_X(X)$ and the correct notation is $f_X(x)$

I amended your post

Answer 2

Let $X=\sin\Theta$, where $\Theta$ is uniform distributed in $(-\pi,\pi)$ (It is enough to consider the interval $(-\pi,pi)$ instead of $[-\pi,\pi]$ for the endpoints have measure (probability) $0$). Notice that $X$ takes values in $(-1,1)$. For any bounded measurable function $g$ on $(-1,1)$

$$\mathbb{E}[g(X)]=\frac{1}{2\pi}\int^\pi_{-\pi}g(\sin\theta)\,d\theta= \frac{1}{2\pi}\Big(\int^{\pi/2}_{-\pi/2}g(\sin\theta)\,d\theta + \int^{-\pi/2}_{-\pi}g(\sin\theta)\,d\theta +\int^\pi_{\pi/2}g(\sin\theta)\,d\theta\Big) $$

1. Since $\sin$ is strictly increasing on $[-\pi/2,\pi/2]$, the change of variable $x=\sin\theta$ yields $$\int^{\pi/2}_{-\pi/2}g(\sin\theta)\,d\theta=\int^1_{-1}g(x)\frac{1}{\sqrt{1-x^2}}\,dx$$
2. Since $\sin$ is strictly decreasing on $[-\pi,\pi/2]$ and $[\pi/2,\pi]$ the change of variables $x=\sin\theta$ yields $$\begin{align} \int^{-\pi/2}_{-\pi}g(\sin\theta)\,d\theta&=-\int^{-1}_0g(x)\frac{1}{\sqrt{1-x^2}}\,dx=\int^0_{-1}g(x)\frac{1}{\sqrt{1-x^2}}\,dx\\ \int^{\pi}_{\pi/2}g(\sin\theta)\,d\theta&=-\int^{0}_{1}g(x)\frac{1}{\sqrt{1-x^2}}\,dx=\int^1_0g(x)\frac{1}{\sqrt{1-x^2}}\,dx \end{align} $$

Putting things together, we obtain $$ \mathbb{E}[g(X)]=\frac{1}{\pi}\int^1_{-1}g(x)\frac{1}{\sqrt{1-x^2}}\,dx$$

That is, the law of $X$ is absolutely continuous with respect the Lebesgue measure, and its density is given by $$\frac{1}{\pi}\frac{1}{\sqrt{1-x^2}}\mathbb{1}_{(-1,1)}(x)$$