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I was wondering about the statistical variance of $A\sin(F x)+ Bx$, and the covariance (or correlation) between $A_{1} \sin(F_{1} x) + B_{1} x$ and $A_{2} \sin(F_{2} x) + B_{2} x$, assuming that $x$ is uniformly distributed in $[0, U]$, and $A, A_1, A_2, F, F_1, F_2, B, U$ are constants.

How can I proceed to compute these quantities in the simplest way?

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  • $\begingroup$ $X$ is uniform where? We need an interval to start. $\endgroup$ Commented May 9, 2023 at 15:03
  • $\begingroup$ uniform on [0, U] $\endgroup$
    – Jada
    Commented May 9, 2023 at 15:32
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    $\begingroup$ Have you tried writing down the definition of the variance/covariance? That should give you an integral, right? Have you tried to evaluate the integral? $\endgroup$
    – Kenny Wong
    Commented May 9, 2023 at 15:48
  • $\begingroup$ I have not succeeded, as the definition seemed difficult to apply: hoping in some simpler way. Note that there is probably no need to refer to a distribution, it could be simply a Lebesgue measure. $\endgroup$
    – Jada
    Commented May 9, 2023 at 15:51
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    $\begingroup$ At an intuitive level, I would expect the correlation to be close to the geometric mean of the B's. $\endgroup$
    – Jada
    Commented May 9, 2023 at 16:01

1 Answer 1

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I'm going to simplify your notation some. I will have capital letters refer to random variables and lowercase letters refer to the parameters. To that end, suppose $X \sim \mathcal{U}(0, u)$ (that is, $X$ follows a uniform distribution on $[0, u]$). Consider the random variable $Y = a\sin(\lambda X) + bX.$ I'll leave it to you to compute the variance, knowing that $$\sigma_X ^2 = E(X^2) - (E(X))^2$$

(hint: integrate by parts). Once you have the variance, now use the correlation formula $$\rho_{X, Y} = \frac{\mathrm{Cov}(X, Y)}{\sigma_X \sigma_Y}$$

where the covariance is a double integral in this case, i.e.

$$ \mathrm{Cov}(X, Y) =E\left[ (X - E(X))(Y - E(Y) \right].$$

This, again, may require parts.

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  • $\begingroup$ Yes, thank you. With your formula above, the variance seems easier to compute (with respect to the classic definition). I need to see how to do the correlation which seems a bit more complicated :-) Perhaps a similar simplification can be done for that too by using mixed moments ... $\endgroup$
    – Jada
    Commented May 9, 2023 at 20:37

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