Disclaimer: This is a dummy post

The following is an excerpt from my Statistics homework:

We want to find \( \mathbb{P} \big (\sin ( X ) \in [0, \frac{1}{1000}]\big) \) , which we can re-write as \(\mathbb{P} \left ( 0 \leq \sin ( X ) \leq \frac{1}{1000} \right) = \int_{0}^{\frac{1}{1000}} f_{\sin(X)}(\tau ) \, d\tau.\)

So it will suffice to find the pdf of our new random variable and perform the integration above. Denote \( \sin ( X ) = Y \) and note that on the set \( [0, \frac{1}{1000}] \), \( \sin \) is a strictly monotone function. We can then apply the non-linear change of variables formula presented in lecture to get that

\[\begin{align*} f_{Y} ( \tau ) &= \frac{e^{\arcsin(\tau)}}{|\cos( \arcsin(\tau) ) |}\\ &= \frac{e^{\arcsin(\tau)}} {\sqrt{\cos^2 ( \arcsin(\tau) ) }} \\ &= \frac{e^{\arcsin(\tau)}} {\sqrt{ 1 - \sin^2 ( \arcsin(\tau) ) }}\\ &= \frac{e^{\arcsin(\tau)}} {\sqrt{1 - \tau^2}}. \end{align*}\]

This is good because

\[\frac{d}{d \tau } ( \arcsin(\tau) ) = \frac{1}{\sqrt{1 - \tau^2}}\]

and we can use u-substitution (aka change of variables) to integrate:

\[\begin{align*} \int_{0}^{\frac{1}{1000}} \frac{e^{\arcsin(\tau)}} {\sqrt{1 - \tau^2}} \, d \tau &= \int_{0}^{\frac{1}{1000}} e^{-u} \, du \\ &= -e^{-\arcsin(x)} |_{\tau = 0}^{\tau=.001}\\ &= 1 - e^{-\arcsin(\tau ) }\\ &\approx 0.0009995003331251665 \end{align*}\]