d******3
发帖数: 209 | 1 想向各位请教一个问题，关于随机分布,一直挣扎，希望各位不吝赐教，非常感谢！
问题的表述是:
现在有10，000个随机分布的变量（假设A），每个都独立并且在一个时间段都独立地按
照期望为0、方差为sigma的正态随机分布变化。 如果想用1000个新的变量（假设B）来
代表原来的10，000个A，B的分布应该是什么样的呢？ | g******2
发帖数: 234 | 2 what do you mean by representing?
Does 10000 mean dimension or sample? | d******3
发帖数: 209 | 3 谢谢回复。
这10000个是sample,从一维的先考虑吧。
另一种表述：
10，000个A微粒和10，000个Z微粒在一个一维系统中，每个都独立，大小重量都一样，
都按
照布朗运动，假设这些布朗运动步幅都是期望为0、方差为sigma的（sigma和dt 成正比
）正态随机变量。在一个小时间段，dt, 每对任意A和Z微粒相碰撞的概率记为P1，当P1
>0.5时，A消失。
现在想用 1000个新的微粒（假设B）来代表原来的10，000个A微粒，同理，在一个小时
间段，dt, 每对任意B和Z微粒相碰撞的概率记为P2，当P2>0.5时，B消失.
B如何量化才能使得 B和A消失的比例一样？ B(loss)/B0 = A(loss)/A0.
非常感谢！
[在 dongd333 (“受侮不辩，闻谤不答”) 的大作中提到：]
：想向各位请教一个问题，关于随机分布,一直挣扎，希望各位不吝赐教，非常感谢！
：
：........... | g******2
发帖数: 234 | 4 I'm still not very clear on the question.
Suppose A has start point d_a_t0, and Z has start point d_z_t0, where d_z_t0
> d_a_t0. If d_a_{t0+dt} > d_z_{t0+dt}, then you consider this as collision
? If that's the case, the you can look at d_z_{t0+dt}-d_a_{t0+dt}, which
follows N(d_z_t0-d_a_t0, 2*sigma^2). P(A,Z collide)=pnorm(0, d_z_t0-d_a_t0,
2*sigma^2). If I'm correct, your P1 is
1- prod_{i=1}^10000(1 - pnorm(0, |d_zi_t0-d_a_t0|, 2*sigma^2)).
I think you can take one of the following two approaches:
1. subsample 1000 point from A.
2. use 1000 empirical quantiles of A.
you can calculate which approach gives you better accuracy. | d******3
发帖数: 209 | 5 I am trying to describe the problem more in detail.
In a system (1-dimension for example), there are two types of particles, A
and B, with the number of particles as NA and NB.
A and B particles are independent and randomly distributed in a system.
During any short time interval, Δt, each particle moves a random distance (
Brownian motion), the distance has zero mean and variance of DA* Δt and DB*
Δt.
if A and B particles are close to each other, they may disappear at the
ratio of 1:1. The probability that a pair of A and B particles to be
annihilated depends on their distance. Assuming the distance between any
pair (one A and one B) particle is s. The probability density function is
written as P=C1/sqrt(2π(D_A+D_B)Δt) exp[-s^2/(2(D_A+D_B)Δt)].Where C1 is
a constant.
If P>0.5, the pair of particles would be annihilated. Otherwise, they would
stay alive.
Now, I would like to using C particles (total NC) to represent the mass of
mA, e.g., NC=NA/m. At the same time, the movement of C particles resembles
the movements of (m) A particles. In other words, C particles move like a
cluster of A particles under Brownian motion.
Similar to A and B pair, when a pair of C and B particles are close, they
would be annihilated at the ratio of 1/m:1 during a short time interval Δt.
If the total annihilated C particles is noted as Cr, A particles is Ar.
The question is: how to quantify the movement of C particles, and how does
the probability function changes to achieve the goal that
Cr/NC = Ar/NA or Cr/NC ≈ Ar/NA at any time step?
Thank you very much, | g******2
发帖数: 234 | 6 1. do you know the initial position of each particle of A?
2. The probability formula you provided is not probability, but a density.
If you calculate probability, the probability for any given one pair to be
annihilated is always less than 0.5. The probability for an A particle to be
annihilated with any B particle is probably the right probability you want
to consider, in which case you should use the formula I wrote above.
3. I think my suggestion above should be valid, either use a random
subsample of A or empirical quantiles of A (if you know the initial position
of A particles). You should have Cr/NC ≈ Ar/NA.
Some R code for justification:
A <- runif(10000, -1000, 1000)
B <- runif(10000, -1000, 1000)
sigma.a <- 0.1
sigma.b <- 0.1
pAnnihilate <- function(a) {
1-exp(sum(pnorm(0, abs(a-B), sqrt(sigma.a^2+sigma.b^2), lower.tail=FALSE,
log.p=TRUE)))
}
p.a <- sapply(A, pAnnihilate)
mean(p.a>0.5)
C <- sample(A, 1000, replace=TRUE)
p.c <- sapply(C, pAnnihilate)
mean(p.c>0.5)
C <- quantile(A, probs=seq(0,1,length.out=1000))
p.c <- sapply(C, pAnnihilate)
mean(p.c>0.5) | d******3
发帖数: 209 | 7 Thank you very much, I will search random subsample and empirical quantiles.
The locations of all particles are randomly distributed in the system.
be
want
position
【在 g******2 的大作中提到】
: 1. do you know the initial position of each particle of A?
: 2. The probability formula you provided is not probability, but a density.
: If you calculate probability, the probability for any given one pair to be
: annihilated is always less than 0.5. The probability for an A particle to be
: annihilated with any B particle is probably the right probability you want
: to consider, in which case you should use the formula I wrote above.
: 3. I think my suggestion above should be valid, either use a random
: subsample of A or empirical quantiles of A (if you know the initial position
: of A particles). You should have Cr/NC ≈ Ar/NA.
: Some R code for justification:
| o*******w
发帖数: 349 | 8 你的问题不是sampling问题，查”quantile", "empiric distribution" 没有用。请参
见在数学版不才的进一步表述。
quantiles.
【在 d******3 的大作中提到】
: Thank you very much, I will search random subsample and empirical quantiles.
: The locations of all particles are randomly distributed in the system.
:
: be
: want
: position
|
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