Density of a function of random variables $Y=\cos (X)$

Question

I'm really struggling to solve this question: Let $X$ be a random variable with $X \sim Exp(1)~$, Find the the density of $Y=\cos (X)$. I try to solve by dividing the domain in two parts, $(i-1)\pi$ and $i\pi$ to obtain the invert functions but after that I'm totally clueless, I don't know what to do. I appreciate any help.

Answer 1

That is indeed the way to go.

Rather than $i$, we shall use $n \in \Bbb N_0$.

The points for $Y$ shall be a fold of all points where $X=(2n\pi+\arccos(Y))$ or $X=(2(n+1)\pi-\arccos(Y))$ where $\arccos:[-1..1]\mapsto[0..\pi]$

Noting that for $y\in[-1..1]$, we have $\dfrac{\mathrm d \arccos(y)}{\mathrm d y}=\dfrac{-1}{\surd(1-y^2)}$

Thus we shall use the Jacobian transformation:

$$\begin{align}f_{Y}(y) &=\dfrac{\mathbf 1_{y\in[-1..1]}}{\surd(1-y^2)} \sum_{n=0}^\infty \left(f_{X}(2n\pi+\arccos(y))+ f_X(2(n+1)\pi-\arccos(y))\right)\end{align}$$

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From here it is a matter of substituting with the probability density function for $X$, distributing out all factors not containing $n$, and expressing resulting the Geometric Series in its closed form.