解析解为的位数

我有两个随机变量， $\alpha_i\sim \text{iid }U(0,1),\;\;i=1,2$ ，其中是均匀的分布0-1。 $U(0,1)$

然后，这些产生一个过程，说：

P (x) = α_{1} \sin (x) + α_{2} \cos (x), x \in (0, 2 π)

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

现在，我想知道对于给定的，的理论上75％的分位数是否存在的闭式表达式。 -我想我可以用计算机和许多实现来做到这一点，但我更喜欢封闭形式-。 $F^{-1}(P(x);0.75)$ $P(x)$ $x\in(0,2\pi)$ $P(x)$

distributions probability stochastic-processes

— 用户603
source

我想你想承担α

和α

是统计独立。

_{1}

$_1$

_{2}

$_2$

— Michael R. Chernick

@Procrastinator：你能写这个作为答案吗？

— user603 2012年

（+1）在这里，“过程”观点似乎有点像鲱鱼。写其中。然后，对于每个固定的，前两项确定梯形密度函数，而后一项只是平均偏移。为了确定梯形密度，我们只需要考虑。

P (x) = β_{1} \sin x + β_{2} \cos x + \frac{1}{2} (\sin x + \cos x),

$P(x) = \beta_1 \sin x + \beta_2 \cos x + \frac{1}{2} (\sin x + \cos x) \>,$

β_{i} = α_{i} - 1 / 2 \sim U (- 1 / 2, 1 / 2)

$\beta_i = \alpha_i - 1/2 \sim \mathcal U(-1/2,1/2)$

x

$x$

x \in [0, π / 2)

$x \in [0,\pi/2)$

— 主教

在数字上，可以简单地通过使用quant = function(n,p,x) return( quantile(runif(n)*sin(x)+runif(n)*cos(x),p) )和来完成quant(100000,0.75,1)。

这个问题可以很快地解决成为寻找梯形分布的分位数之一。

让我们重写过程其中和是独立同分布的随机变量; 并且，通过对称性，这具有相同的边缘分布作为处理

P (x) = U_{1} \cdot \frac{1}{2} \sin x + U_{2} \cdot \frac{1}{2} \cos x + \frac{1}{2} (\sin x + \cos x),

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

U_{1}

$U_1$

U_{2}

$U_2$

U (- 1, 1)

$\mathcal U(-1,1)$

前两个项确定对称梯形密度，因为这是两个均值零均匀随机变量（通常具有不同的半角宽度）的总和。最后一项仅导致该密度的平移，并且分位数相对于该平移是等变的（即，位移分布的分位数是居中分布的位移分位数）。

\bar{P} (x) = U_{1} \cdot | \frac{1}{2} \sin x | + U_{2} \cdot | \frac{1}{2} \cos x | + \frac{1}{2} (\sin x + \cos x) .

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

梯形分布的分位数

让，其中和是独立和分布。假设不失一般性的是损失。然后，通过对和的密度进行卷积来形成的密度。这可容易地看出，与顶点的梯形 $Y = X_1 + X_2$ $X_1$ $X_2$ $\mathcal U(-a,a)$ $\mathcal U(-b,b)$ $a \geq b$ $Y$ $X_1$ $X_2$ ，，和。 $(-a-b,0)$ $(-a+b,1/2a)$ $(a-b,1/2a)$ $(a+b,0)$

的分布的四分，对于任何是，因此， $Y$ $p < 1/2$ 通过对称，为，我们有。

q (p) := q (p; a, b) = {\begin{cases} \sqrt{8 a b p} - (a + b), & p < b / 2 a \\ (2 p - 1) a, & b / 2 a \leq p \leq 1 / 2 . \end{cases}

$q(p) := q(p\,; a,b) = \begin{cases} \sqrt{8abp} - (a+b) \,, & p < b/2a \\ (2p - 1) a \,, & b/2a \leq p \leq 1/2 \>. \end{cases}$

p > 1 / 2

$p > 1/2$

q (p) = - q (1 - p)

$q(p) = - q(1-p)$

回到手头的情况

为，在，我们设置和并得到 $p < 1/2$ $|\sin x| \geq |\cos x|$ $a = |\sin x|/2$ $b=|\cos x|/2$

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

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

| \sin x | < | \cos x |

$|\sin x| < |\cos x|$ the roles reverse. Similarly, for

p \geq 1 / 2

$p \geq 1/2$

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

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

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 $P(x)$ .

Quantiles as a function of x

$p = 1/2$ $p=0$ $p=1$ . Cyan is roughly $p=1/4$ and $p=3/4$ . This more clearly shows the support of each distribution and the shape.

Quantile plot

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

— cardinal
source

I wish i could upvote more. This is the reason i love this website: the power of specialization. i didn't know of the trapezoid distribution. It would have taken me some time to figure this out. Or I would have had to settle for using Gaussians instead of Uniforms. Anyhow, it's awesome.

— user603