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If $k$ is a positive integer and $U \sim \text{Unif}(0,1)$ is the standard uniform distributed, why does it hold that

$$ \sin(2\pi kU) \sim \sin(2\pi U) $$

Is there is an intuition behind this or a proof?

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    $\begingroup$ Show that $m\{ x \in [0,1] | \sin (2 \pi k x) \le \alpha \} = m\{ x \in [0,1] | \sin (2 \pi x) \le \alpha \} $. The left is just $n$ copies of a ${1 \over n}$ scaled version of the right. $\endgroup$
    – copper.hat
    Jul 30 at 18:25

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We know that $$\varphi_U(u) = \frac{e^{iu} - 1}{iu}.$$ So $$\varphi(2l\pi) = \begin{cases}1 & \text{if $l=0$}\\ 0 & \text{if $l\neq 0$}\end{cases}$$ and if $p\in \mathbb N$, $p \neq 0$, $$\varphi(2\pi pl) = \varphi(2\pi l)$$ Let $Y_k = \sin\left(2\pi k U\right)$

\begin{align} \mathbb E\left[Y_k^n\right] &= \mathbb E\left[\left(\frac{e^{i(2\pi kU)} - e^{-i(2\pi kU)}}{2i}\right)^n\right]\\ &= \frac{1}{(2i)^n}\sum_{m=0}^{n}\binom{n}{m} \mathbb E\left[e^{i(2\pi kU)m}e^{-i(2\pi kU)(n-m)}\right]\\ &= \frac{1}{(2i)^n}\sum_{m=0}^{n}\binom{n}{m} \varphi_{U}(2\pi k(2m-n))\\ &= \frac{1}{(2i)^n}\sum_{m=0}^{n}\binom{n}{m} \varphi_{U}(2\pi (2m-n))\\ &= \frac{1}{(2i)^n}\sum_{m=0}^{n}\binom{n}{m} \mathbb E\left[e^{i(2\pi U)m}e^{-i(2\pi U)(n-m)}\right]\\ &= \mathbb E\left[\left(\frac{e^{i(2\pi U)} - e^{-i(2\pi U)}}{2i}\right)^n\right]\\ &= \mathbb E\left[Y_1^n\right] \end{align}

This proves that $Y_k$ and $Y_1$ have the same moments and so the same distribution.


Another simple approach of proving it: Let $X$ be uniform in the set of integers $\{0,\ldots,k-1\}$ and $V\sim Uni(0,1)$, So $\frac{X+V}{k}\sim U$.

And $$\sin(2\pi kU)\sim\sin \left(2\pi (X+V)\right) = \sin (2\pi U)$$

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  • $\begingroup$ This is a pretty complicated approach to a simple question! $\endgroup$
    – copper.hat
    Jul 30 at 19:56
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    $\begingroup$ @copper.hat I add a simpler way of doing it. $\endgroup$
    – Youem
    Jul 30 at 23:03

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