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Suppose that $X$ is uniformly distributed on $[-2\pi,2\pi]$. Find the probability density function of $Y=\tan(X).$

What I have so far:

First I note that $$f_{X}(x)= \begin{cases} \displaystyle\frac{1}{4\pi}&\text{if }x\in[-2\pi,2\pi]\\0&\text{otherwise. }\end{cases}$$ Now we observe that to find the PDF of $Y$, we can first find its CDF and then differentiate. By definition, we have for $y\in \mathbb{R},$ $$F_{Y}(y)=\mathsf P(Y\leq y)=\mathsf P(\tan(X)\leq y)=\mathsf P(X\leq \arctan(y)).$$ Therefore we find that $$F_{Y}(y)=\frac{\arctan(y)}{4\pi}$$ if $y\in[-2\pi,2\pi]$ and $F_{Y}(y)=0$ otherwise. Now we can differentiate $F_{Y}$ to obtain $$f_{Y}(y)= \begin{cases}\displaystyle\frac{1}{4\pi(1+y^2)}&\text{if }y\in[-2\pi,2\pi]\\0&\text{otherwise. }\end{cases}$$


Is the approach above the correct one? If not, could you please provide a hint to point me in the right direction, please, no solutions, only hints.

Thank you for your time, and appreciate any feedback.

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    $\begingroup$ Is $\tan x \le y$ equivalent to $x \le \arctan(y)$ for $x\in[-2\pi,2\pi]$? Also remember that the range for $Y$ should be all of $\mathbb{R}$, not just $[-2\pi,2\pi]$ (that is for $X$). $\endgroup$ Commented Mar 5, 2019 at 11:33
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    $\begingroup$ $\arctan (\tan(X )$ is not the same as $X$. Your $f_Y$ is not a density function. $\endgroup$ Commented Mar 5, 2019 at 11:44
  • $\begingroup$ @KaviRamaMurthy Should I break it into cases depending on the value of $y$? For instance, apply $\arctan$ if $y\in[-\pi/2,\pi/2]$, and then handle the other cases separately. $\endgroup$
    – Stackman
    Commented Mar 5, 2019 at 12:01
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    $\begingroup$ You have to split $x\in [-2\pi,2\pi]$ into $x\in [-\pi /2,\pi /2]$, etc. $\endgroup$ Commented Mar 5, 2019 at 12:04

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If we instead had $X\sim U\left[-\frac{\pi}{2},\,\frac{\pi}{2}\right]$ so $Y$ would be monotonic in $X$, we'd have $$P(Y\le y)=P(X\le\arctan x)=\frac{1}{\pi}\arctan x+\frac{1}{2}\implies f_Y(y)=\frac{1}{\pi(1+y^2)}.$$This respects unitarity, i.e. $\int_{\Bbb R}f(y) dy=1$, so we can't e.g. change $\pi$ to $4\pi$. The effect of switching back to $X\sim U\left[-2\pi,\,2\pi\right]$ is to run over four periods of $Y$'s dependence on $X$. What's more, these periods inject, albeit not in an order-preserving way because of asymptotes at half-odd multiples of $\pi$. Therefore, the above pdf is actually correct in this example too.

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  • $\begingroup$ I find pdf of $Y$ by dividing in 4 part. and got same pdf. But how you conclude that we take o $X\sim U\left[-2\pi,\,2\pi\right]$ or o $X\sim U\left[\frac{-\pi}{2},\,\frac{\pi}{2}\right]$. we get same pdf .please explain logic behind Therefore, the above pdf is actually correct in this example too.? $\endgroup$
    – Meet Patel
    Commented Oct 14, 2023 at 12:21
  • $\begingroup$ @MeetPatel The conditional PDF on each of four monotonic cases is the same. Now average, as each has probability $\frac14$. $\endgroup$
    – J.G.
    Commented Oct 14, 2023 at 12:31

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