Let \(R\) be a random Rayleigh-distributed variable. Let \(\phi\) be a random uniformly-distributed variable.

The time series defined by:

\[ Y_t = R\cos{(2\pi(ft + \phi))} \]

with \(t\in\mathbb{Z}\), is equivalent to the time series

\[ Y_t' = U + V \]

if \(U\) and \(V\) are independently distributed standard normal variables:

\[ U, V \sim N(0,1) \]

Equivalence in this context means that \(Y_t\) and \(Y'_t\) have the same PMFs.

Proof

Distribution of Cosine

The PDF of \(\cos(2\pi(ft + \phi))\) can be derived from the PDF of \(\phi\) using the rule for functions of a random variable: For a random variable \(X\) with PDF \(f\), and an invertible function \(Y(X)\), the PDF \(g\) of \(Y\) is given by:

\[ g(y) = f(x(y))\left|\frac{d x}{d y}\right| \]

In the case that \(Y = \cos(2\pi(ft + \phi))\) and \(f\) is the uniform distribution, we get:

\[ g(y) = \frac{1}{\sqrt{1-y^2}} \]

With this, we can calculate the PMF of \(Y\) in terms of \(\phi\):

\[ \text{Pr}(g(Y)) = \int g(y) dy \] \[ = \int\frac{1}{\sqrt{1-y^2}} dy \]

\[ =\int\frac{\sin(2\pi(ft+\phi))}{2\pi\sqrt{1-\cos^2(2\pi(ft + \phi))}}d\phi \]

\[ =\int \frac{d\phi}{2\pi} \]

In other words

\[ \phi \sim U(0,1) \implies \cos(2\pi(ft + \phi)) \sim U(0,2\pi) \]

Equivalence of Series

The PMF of \(Y_t\) can be expressed as a product, since \(R\) and \(\phi\) are independent:

\[ \text{Pr}(R\cos(2\pi(ft + \phi))) = \text{Pr}( R)\text{Pr}(\cos(2\pi(ft + \phi))) \]

\[ = \int r e^{-r^2/2} dr \int \frac{1}{2\pi} d\phi \]

\[ = \frac{1}{2\pi}\iint r e^{-r^2/2} dr d\phi \]

Change our variables of integration:

\[ u = r\cos(\phi), \qquad v = r\sin(\phi) \]

Then we can write: \[ \text{Pr}(R\cos(2\pi(ft + \phi))) = \frac{1}{2\pi}\iint e^{-(u^2+v^2)/2} du dv \]

Which is the PMF of the time series defined by \(Y_t' = U + V\). \(\square\)