I have two random variables, \alpha_i\sim \text{iid }U(0,1),\;\;i=1,2 where U(0,1) is the

uniform 0-1 distribution.Then, these yield a process, say:

P(x)=\alpha_1\sin(x)+\alpha_2\cos(x), \;\;\;x\in (0,2\pi)

Now, i was wondering whether there is a closed form expression for F^{-1}(P(x);0.75) the theoretical 75 percent quantile of P(x) for a given x\in(0,2\pi) –i suppose i can do it with with a computer and many realizations of P(x), but i’d prefer closed form–.

**Answer**

**This problem can quickly be reduced to one of finding the quantile of a trapezoidal distribution.**

Let us rewrite the process as

P(x) = U_1 \cdot \frac12 \sin x + U_2 \cdot \frac12 \cos x + \frac{1}{2} (\sin x + \cos x) \>,

where U_1 and U_2 are iid \mathcal U(-1,1) random variables; and, by symmetry, this has the same *marginal* distribution as the process

\overline P(x) = U_1 \cdot \left|\frac12 \sin x\right| + U_2 \cdot \left|\frac12 \cos x\right| + \frac{1}{2} (\sin x + \cos x) \>.

The first two terms determine a symmetric *trapezoidal density* since this is the sum of two mean-zero uniform random variables (with, in general, different half-widths). The last term just results in a translate of this density and the quantile is equivariant with respect to this translate (i.e., the quantile of the shifted distribution is the shifted quantile of the centered distribution).

**Quantiles of a trapezoidal distribution**

Let Y = X_1 + X_2 where X_1 and X_2 are independent \mathcal U(-a,a) and \mathcal U(-b,b) distributions. Assume without loss of generality that a \geq b. Then, the density of Y is formed by convolving the densities of X_1 and X_2. This is readily seen to be a trapezoid with vertices (-a-b,0), (-a+b,1/2a), (a-b,1/2a) and (a+b,0).

The quantile of the distribution of Y, for any p < 1/2 is, thus,

q(p) := q(p\,; a,b) =

\begin{cases}

\sqrt{8abp} - (a+b) \,, & p < b/2a \\

(2p - 1) a \,, & b/2a \leq p \leq 1/2 \>.

\end{cases}

By symmetry, for p > 1/2, we have q(p) = - q(1-p).

**Back to the case at hand**

The above already provides enough to give a closed-form expression. All we need is to break into two cases |\sin x| \geq |\cos x| and |\sin x| < |\cos x| to determine which plays the role of 2a and which plays the role of 2b above. (The factor of 2 here is only to compensate for the divisions by two in the definition of \overline P(x).)

For p < 1/2, on |\sin x| \geq |\cos x|, we set a = |\sin x|/2 and b=|\cos x|/2 and get

q_x(p) = q(p \,; a,b) + \frac{1}{2}(\sin x + \cos x) \>,

and on |\sin x| < |\cos x| the roles reverse. Similarly, for p \geq 1/2

q_x(p) = -q(1-p \,; a,b) + \frac{1}{2}(\sin x + \cos x) \>,

**The quantiles**

Below are two heatmaps. The first shows the *quantiles* of the distribution of P(x) for a grid of x running from 0 to 2\pi. The y-coordinate gives the probability p associated with each quantile. The colors indicate the value of the quantile with dark red indicating very large (positive) values and dark blue indicating large negative values. Thus each vertical strip is a (marginal) quantile plot associated with P(x).

The second heatmap below shows the quantiles themselves, colored by the corresponding probability. For example, dark red corresponds to p = 1/2 and dark blue corresponds to p=0 and p=1. Cyan is roughly p=1/4 and p=3/4. This more clearly shows the support of each distribution and the shape.

**Some sample R code**

The function `qproc`

below calculates the quantile function of P(x) for a given x. It uses the more general `qtrap`

to generate the quantiles.

```
# Pointwise quantiles of a random process:
# P(x) = a_1 sin(x) + a_2 cos(x)
# Trapezoidal distribution quantile
# Assumes X = U + V where U~Uni(-a,a), V~Uni(-b,b) and a >= b
qtrap <- function(p, a, b)
{
if( a < b) stop("I need a >= b.")
s <- 2*(p<=1/2) - 1
p <- ifelse(p<= 1/2, p, 1-p)
s * ifelse( p < b/2/a, sqrt(8*a*b*p)-a-b, (2*p-1)*a )
}
# Now, here is the process's quantile function.
qproc <- function(p, x)
{
s <- abs(sin(x))
c <- abs(cos(x))
a <- ifelse(s>c, s, c)
b <- ifelse(s<c, s, c)
qtrap(p,a/2, b/2) + 0.5*(sin(x)+cos(x))
}
```

Below is a test with the corresponding output.

```
# Test case
set.seed(17)
n <- 1e4
x <- -pi/8
r <- runif(n) * sin(x) + runif(n) * cos(x)
# Sample quantiles, then actual.
> round(quantile(r,(0:10)/10),3)
0% 10% 20% 30% 40% 50% 60% 70% 80% 90% 100%
-0.380 -0.111 -0.002 0.093 0.186 0.275 0.365 0.453 0.550 0.659 0.917
> round(qproc((0:10)/10, x),3)
[1] -0.383 -0.117 -0.007 0.086 0.178 0.271 0.363 0.455 0.548
[10] 0.658 0.924
```

**Attribution***Source : Link , Question Author : user603 , Answer Author : cardinal*