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I have a question on how to derive the (marginal) pdf for the $k$-th random variable $X_k$, where $X_k = \cos(0.2\pi k + \theta)$, and $\theta$ is a random variable uniformly distributed over the interval $[0, 2\pi)$. Could anyone show me how to derive this?

utobi
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ryjing99
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1 Answers1

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I am not sure $k$ matters here

Indeed, running (Python):

import numpy as np 
import numpy.random as npr
k_arr=[1, 3, 5, 6, 10, 100]; theta=2*np.pi*npr.rand(50000); V=-0.2
[np.mean(np.cos(0.2*k+theta)<V) for k in k_arr]

Gives same results for different $k$. So if we would construct CDF for your variable:

$$ F\left(X_k\right)=Prob\left[\cos\left(\dots\right)<X_k\right]=\frac{1}{2\pi}\int_0^{2\pi}d\theta\,I_{X_k}\left(\cos\left(\alpha k+\theta\right)\right) $$

where $I_X\left(u\right)=1$ for $u<X$ and 0 otherwise, we would get the same result irrespective of $k$. Indeed one can see that this would be the case since one can change the integration variable and remove $k$. So we can focus on

$$ F\left(X\right)=\frac{1}{2\pi}\int_0^{2\pi}d\theta\,I_{X}\left(\cos\left(\theta\right)\right) $$

Next we can use the fact that for this range of $\theta$ every value of $\cos$ is repeated twice, so we can reduce the range of integration:

$$ F\left(X\right)=\frac{1}{2\pi}\int_{0}^{2\pi}d\theta\dots=\frac{1}{\pi}\int_{0}^{\pi}d\theta\,I_{X}\left(\cos\left(\theta\right)\right) $$

Now we could use the fact that in the range $\theta=0\dots \pi$ $\cos$ is decreasing with $\theta$ so:

$$ F\left(X\right)=\frac{1}{\pi}\int_{\arccos\left(X\right)}^{\pi}d\theta\,1=1-\frac{\arccos\left(X\right)}{\pi} $$

And to get the pdf you need to differentiate, so:

$$ f\left(X\right)=\frac{dF}{dX}=\dots $$

Cryo
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