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What will be the $Var(\sin(X))$ when $X\sim \mathcal{N}(\mu,\,\sigma^2)$.

Muhammad Waqas
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    Does this help?https://math.stackexchange.com/questions/3121914/variance-of-sine-and-cosine-of-a-random-variable – UnsinkableSam Jun 26 '20 at 09:58
  • For $\mu \neq 0$ this won't work. I wonder if there is some nice closed form solution. – Muhammad Waqas Jun 26 '20 at 10:29
  • Have you encountered Characteristic function before? The above link should be quite useful since the only thing you need to do is find $E(sin X)$, which can be done using cf – UnsinkableSam Jun 26 '20 at 10:33

1 Answers1

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It's: $$\frac12(1-e^{-\sigma^2})(1+e^{-\sigma^2}\cos 2\mu )\qquad (*)$$

Note that in the special case $\mu=0$, this simplifies to the answer given here.

You can derive this as follows. The characteristic function of a standard normal is: $$E e^{itZ} = e^{-t^2/2}$$ However, from the identity $e^{ix} = i \sin x + \cos x$ and linearity of the expectation, this means: $$i E\sin (tZ) + E\cos (tZ) = e^{-t^2/2}$$ implying $E \sin tZ = 0$ (which was obvious by symmetry of $Z$) and: $$E \cos (tZ) = e^{-t^2/2}$$ which was less obvious.

Now, it's just trigonometric identities. First, by the usual identity for the sine of sums, we have: $$\sin(\mu + \sigma Z) = \sin \mu \cos(\sigma Z) + \sin (\sigma Z) \cos \mu$$ and therefore, for $X = \mu + \sigma Z$, we have: $$E \sin X = \sin \mu\, E \cos(\sigma Z) + \cos \mu\, E \sin(\sigma Z) = \sin \mu \times e^{-\sigma^2/2} + \cos \mu \times 0$$ Second, we have \begin{align} \sin^2(\mu + \sigma Z) &= (1 - \cos [2(\mu + \sigma Z)])/2 & \hbox{as $\cos 2\theta = 1-2\sin^2\theta$}\\ &= (1 - \cos (2\mu) \cos(2\sigma Z) + \sin(2\mu)\sin(2\sigma Z))/2 & \hbox{identity for $\cos (\alpha+\beta)$}\\ \end{align} and therefore, we have $$E (\sin X)^2 = \left(1 - \cos(2\mu)e^{-2\sigma^2} + 0\right)/2 $$

This gives: \begin{align} \mbox{Var}(\sin X) &= \left(1-e^{-2\sigma^2}\cos(2\mu)\right)/2 - [e^{-\sigma^2/2}\sin \mu]^2 \\ &= \left(1-e^{-2\sigma^2}\cos(2\mu)\right)/2 - e^{-\sigma^2}(1-\cos(2\mu))/2 \end{align} with the second equality following from $\cos 2\theta = 1 - 2\sin^2 \theta$.

Factoring this gives the expression (*) above.

K. A. Buhr
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