Wyrażenie w postaci zamkniętej dla kwantyli

16

Mam dwie zmienne losowe, αiiid U(0,1),i=1,2 , gdzieU(0,1) jest jednolity rozkład 0-1.

Następnie dają proces, powiedzmy:

P(x)=α1sin(x)+α2cos(x),x(0,2π)

Zastanawiałem się teraz, czy istnieje wyrażenie w postaci zamkniętej dla F1(P(x);0.75) teoretycznego 75-procentowego kwantylu P(x) dla danego x(0,2π) przypuszczam, że i mogę to zrobić za pomocą komputera i wielu realizacji P(x) , ale wolę formę zamkniętą--.

użytkownik603
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1
Myślę, że chcesz założyć, że α 1 i α 2 są statystycznie niezależne. 12
Michael R. Chernick,
@Procrastinator: czy możesz napisać to jako odpowiedź?
user603,
4
(+1) Punkt widzenia „procesu” wydaje się tutaj trochę czerwonym śledziem. Napisz W którym β I = α i - 1 / 2 ~ U ( - 1 / 2 , 1 / 2 ) . Następnie, dla każdego ustalonego x , pierwsze dwa terminy określają funkcjęgęstościtrapezoidalnej,a ostatni składnik jest tylko średnim przesunięciem. Do wyznaczenia gęstości trapezoidalnej wystarczy wziąć pod uwagę x [ 0 , π / 2 ) .
P(x)=β1sinx+β2cosx+12(sinx+cosx),
βi=αi1/2U(1/2,1/2)xx[0,π/2)
kardynał
2
Liczbowo można to zrobić po prostu za pomocą quant = function(n,p,x) return( quantile(runif(n)*sin(x)+runif(n)*cos(x),p) )i quant(100000,0.75,1).

Odpowiedzi:

19

Problem ten można szybko sprowadzić do znalezienia kwantyla rozkładu trapezoidalnego .

Przeredagujmy proces jako

P(x)=U112sinx+U212cosx+12(sinx+cosx),
where U1 and U2 are iid U(1,1) random variables; and, by symmetry, this has the same marginal distribution as the process
P¯(x)=U1|12sinx|+U2|12cosx|+12(sinx+cosx).
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

Y=X1+X2X1X2U(a,a)U(b,b)ab. Then, the density of Y is formed by convolving the densities of X1 and X2. This is readily seen to be a trapezoid with vertices (ab,0), (a+b,1/2a)(ab,1/2a) and (a+b,0).

Y, for any p<1/2 is, thus,

q(p):=q(p;a,b)={8abp(a+b),p<b/2a(2p1)a,b/2ap1/2.
By symmetry, for p>1/2, we have q(p)=q(1p).

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 |sinx||cosx| and |sinx|<|cosx| 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 P¯(x).)

For p<1/2, on |sinx||cosx|, we set a=|sinx|/2 and b=|cosx|/2 and get

qx(p)=q(p;a,b)+12(sinx+cosx),
and on |sinx|<|cosx| the roles reverse. Similarly, for p1/2
qx(p)=q(1p;a,b)+12(sinx+cosx),

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 2π. 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

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.

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
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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