 # I'm given a random variable X n </msub> that is wrt the following measure: Michelle Mendoza 2022-07-13 Answered
I'm given a random variable ${X}_{n}$ that is wrt the following measure:
$\underset{n\to \mathrm{\infty }}{lim}\frac{\mathrm{#}\left\{{X}_{n}\in \left[a,b\right]\right\}}{n}={\int }_{a}^{b}{\mathrm{sin}}^{2}\theta d\theta .$
with an error term $O\left(\mathrm{log}n/n\right)$.
Now I'm trying to find the distribution of ${\mathrm{cos}}^{2}{X}_{n}$. The text I'm reading says this is ${\int }_{a}^{b}{\mathrm{cos}}^{2}\theta d\theta$, which I'm already not sure of, and I also want to figure out what the error term of this is.
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Please note that I have added a factor $k=\frac{2}{\pi }$. See later.
First of all, as the values taken by the LHS of (1) are between 0 and 1, the same must be true for the RHS of (1).
This is why I have added factor $k$ in order that
$\frac{2}{\pi }{\int }_{0}^{\pi }{\mathrm{sin}}^{2}\theta d\theta =1.$
Let ${f}_{X}$ and ${F}_{X}$ be the pdf and cdf resp. of random variable $X$.
The LHS of (1) is plainly ${\int }_{a}^{b}{f}_{X}\left(x\right)dx$ under the classical "subjectivist" assumptions of probability (I think that the error term $O\left(\mathrm{log}\left(n\right)/n\right)$ comes from a theorem of De Finetti, not sure, but we don't need it ; I would even say that I find its presence a little strange).
Therefore, if identity (1) is valid for any $a,b$, they define a same measure ; therefore:
${f}_{X}\left(x\right)=k{\mathrm{sin}}^{2}\left(x\right)=k\frac{1}{2}\left(1-\mathrm{cos}\left(2x\right)\right)=\frac{1}{\pi }\left(1-\mathrm{cos}\left(2x\right)\right)$
giving, by integration:

Let $Y:={\mathrm{cos}}^{2}X$.
Let us compute the cdf of $Y$ which can take values $y\in \left[0,1\right]$.
${F}_{Y}\left(y\right)=P\left(Y
(last equality by symmetry of the pdf of $X$ wrt $\pi /2$)
${F}_{Y}\left(y\right)=2P\left(X<\underset{c}{\underset{⏟}{{\mathrm{cos}}^{-1}\left(\sqrt{y}\right)}}\right)=2{F}_{X}\left(c\right)$
$\begin{array}{}\text{(2)}& {F}_{Y}\left(y\right)=2\frac{1}{\pi }\left[\frac{1}{2}c-\frac{1}{4}\mathrm{sin}\left(2c\right)\right]\end{array}$
It remains to express (2) in a simpler form, and finally to take the derivative of cdf ${F}_{Y}$ with respect to $y$ in order to get the pdf ${f}_{Y}$.

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