Distribution of a sinusoid with random amplitude

Question

Let $X = A \sin(\phi)$ where $A \sim \mathcal{N}(\mu, \sigma^2)$ and $\phi$ is a uniform random variable defined over $[0,2\pi]$. Assume these two random variables are independent. What is the PDF of $X$?

According to [1], the PDF is given by:

$$f(x) = \int_x^\infty \frac{e^{-\frac{1}{2}\left(\frac{A-\mu}{\sigma}\right)^2}}{\sqrt{A^2-x^2}} dA$$

I had no luck solving for this integral.

[1] Random Data: Analysis and Measurement Procedures, 4th Edition Julius S. Bendat, Allan G. Piersol

Answer 1

The Gaussian density is

$$g(A)=\frac{e^{-\frac{(A-\mu )^2}{2 \sigma ^2}}}{\sqrt{2 \pi } \sigma }$$

and the density of $Z=\sin(\phi)$ where $\phi\sim U(0,2\pi)$ is

$$h(z)=\begin{equation} \begin{cases} \frac{1}{\pi \sqrt{(1-z) (z+1)}} & -1<z<1 \\ 0 & \textrm{otherwise} \end{cases} \\ \end{equation} $$

Using a formula for the product of two independent random variables (https://en.wikipedia.org/wiki/Distribution_of_the_product_of_two_random_variables), the density for $X=A \sin(\phi)=A Z$ is

$$f(x)=\int_{-\infty}^\infty g(A) h(x/A) /|A|dA$$

Because $Z$ ranges from -1 to 1 (and not $-\infty$ to $\infty$) we need to fiddle with the limits of integration:

$$f(x)=\int_{-\infty }^{-| x| } \frac{\sqrt{2} e^{-\frac{(A-\mu )^2}{2 \sigma ^2}}}{2 \pi ^{3/2} \sigma \sqrt{A^2-x^2}} \, dA +\int_{| x| }^{\infty } \frac{\sqrt{2} e^{-\frac{(A-\mu )^2}{2 \sigma ^2}}}{2 \pi ^{3/2} \sigma \sqrt{A^2-x^2}} \, dA$$

So that equation you post isn't correct or I've made a mistake in the algebraic manipulations or both are incorrect. (But in any event, the equation you post is certainly missing a multiplicative normalizing constant.) I don't think there's a nice closed-form for this integral so one would need to evaluated it numerically.

How to check? Simulations are handy. Let $\mu=5$ and $\sigma=1$. Using Mathematica we have the following:

f[x_, μ_, σ_] := 
  NIntegrate[(Sqrt[2] E^(-((A - μ)^2/(2 σ^2))))/(2 π^(3/2) σ Sqrt[A^2 - x^2]), {A, -∞, -Abs[x]}] +
  NIntegrate[(Sqrt[2] E^(-((A - μ)^2/(2 σ^2))))/(2 (π^(3/2)) σ Sqrt[A^2 - x^2] ), {A, Abs[x], ∞}]

n = 1000000;  (* Sample size *)
ϕ = RandomVariate[UniformDistribution[{0, 2 π}], n];
a = RandomVariate[NormalDistribution[5, 1], n];
x = a Sin[ϕ];
Show[Histogram[xx, "FreedmanDiaconis", "PDF"],
 Plot[f[x, 5, 1], {x, -10, 10}, PlotRange -> All]]

To do the equivalent in R:

# Integrand
  f <- function(a, x, mu, sigma) {
    exp(-(a-mu)^2/(2*sigma^2))*sqrt(2)/(2*pi^(3/2)*sigma*sqrt(a^2-x^2))
  }
  
# Probability density function of X = A*sin(phi)
  d <- function(x, mu, sigma) {
    integrate(f,  -Inf, -abs(x), x=-abs(x), mu=mu, sigma=sigma)$value +
    integrate(f, abs(x),    Inf, x= abs(x), mu=mu, sigma=sigma)$value
  }

# Random sample
  n = 1000000
  mu0 <- 5
  sigma0 <- 1
  a = rnorm(n, mu0, sigma0)
  phi = 2*pi*runif(n)
  hist(a*sin(phi), breaks=100, freq=FALSE, las=1, ylim=c(0,0.13), axes=FALSE)
  axis(1, c(-10:10), pos=0)
  axis(2, pos=-10, las=1)

# True density
  p <- NULL
  x <- NULL
  for (i in 1:499) {
      x[i] <- -10 + (10-(-10))*i/500
      p[i] <- d(x[i], mu0, sigma0)
  }
  lines(x, p, type="l", col="red", lwd=3)