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I need to calculate the PDF of a random variable, which is quite similar to what was asked here. However, I have to deal with a shifted cosine function. Thus, my random variable is defined as $$Y:=cos(X+\omega)$$ where $X\sim \mathcal{U}[-\pi, \pi]$ and $\omega \in \mathbb{R}$. In this case, I cannot use the trick with the inversion of the cosine, which was used here to calculate the PDF of $Y$. How can I approach this problem?

Thanks in advance!

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It's actually the same PDF as $\cos X$, because $\mod_{[-\pi,\pi]} (X+w)$ and $\mod_{[-\pi,\pi]}(X)$ has the same uniform distribution. So, the cosine transformation doesn't feel the difference between the two.

Consider a simpler example, and let $X=\{0,1,2,3\}$ and $Y=\mod(X+10,4)$, and assume uniform distribution over $X$. $$P(Y=0)=P(X=2)=1/4\\P(Y=1)=P(X=3)=1/4\\P(Y=2)=P(X=0)=1/4\\P(Y=3)=P(X=1)=1/4$$ $Y$ is again uniform. The constant inside cosine transform has the same effect. it just shifts probabilities, but since the source distribution is uniform it has no effect.

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    $\begingroup$ Thanks, that makes sense! The minimal example helped and I now have a feeling about why cos(X) and cos(X+w) follow the same distribution! However, I'm unfamiliar with the $\mod_{[-\pi,\pi]}$ notation. Can you provide a small explanation of what it exactly does or post a referral link? Thank you! $\endgroup$
    – Thomas
    Commented Apr 13, 2020 at 14:18
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    $\begingroup$ It's actually an intuitive but abuse of notation I used for this specific case. It means whenever you have a number outside this range, you'll find the corresponding real number within the range by adding or subtracting $2\pi$ (since cosine has this period). For example, $$\mod_{[-\pi,\pi]}(3\pi)=\pi, \mod_{[-\pi,\pi]}(-6)=-6+2\pi$$ $\endgroup$
    – gunes
    Commented Apr 13, 2020 at 14:22
  • $\begingroup$ Cool, that clarifies a lot! But shouldn't it be "adding or subtracting a multiple of $2 \pi$". I mean, if $\omega$ is huge, it is certainly not enough to only subtract $2 \pi$. Sorry for being overly exact here, just want to get everything right ;-) $\endgroup$
    – Thomas
    Commented Apr 13, 2020 at 14:46
  • $\begingroup$ no problem, of course, I meant adding or subtracting $2\pi$ as much as it’s necessary. $\endgroup$
    – gunes
    Commented Apr 13, 2020 at 15:02

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