w*******1
发帖数: 112 | 1 random number generator generates standard normal variable x. how to use
monte carlo simulation and importance sampling to find the prob(x>10)? thank
you very much |
l******n
发帖数: 9344 | 2 N = 10000
alpha = 10
set.seed(12345)
samples = alpha-log(1-runif(N))
z = 1/(sqrt(2*pi))*exp(-alpha)*exp(samples-samples^2/2)
mean(z)
sd(z)
thank
【在 w*******1 的大作中提到】
: random number generator generates standard normal variable x. how to use
: monte carlo simulation and importance sampling to find the prob(x>10)? thank
: you very much
|
w*******1
发帖数: 112 | 3 能稍微解释下思路么？多谢，多谢
【在 l******n 的大作中提到】
: N = 10000
: alpha = 10
: set.seed(12345)
: samples = alpha-log(1-runif(N))
: z = 1/(sqrt(2*pi))*exp(-alpha)*exp(samples-samples^2/2)
: mean(z)
: sd(z)
:
: thank
|
K*****Y
发帖数: 629 | 4 题目是已经给定可以生成standard normal random variable x, 所以不清楚为什么楼
上的解法还要用到 1/(sqrt(2*pi))和对数变换，感觉用不着这么麻烦。
用R测试并写的程序：
N=10000 # number of random samples, which can be specified
x=rnorm(N) # generate N standard normal random variables
alpha=10 # calculate P(x>alpha)
est=mean(as.numeric(x>0)*exp(-alpha*x-0.5*alpha^2)) #est gives the required
value
思路很简单。要求P(x>alpha)，E[1_{x>alpha}],做一个变换x'=x-alpha，也就是要求
期望值E{1_{x'>0}*w(x')}，其中w(x')=exp(-alpha*x'-0.5*alpha^2) is the weight
function, 也就是新旧变量的likelihood ratio。自己写一下积分推导一下很容易得到
。 |
m****r
发帖数: 141 | 5 est=mean(as.numeric(x>0)*exp(-alpha*x-0.5*alpha^2))
should be
est=mean(as.numeric(x>0)*exp(-x^2/2-alpha*x-0.5*alpha^2))
right?
required
weight
【在 K*****Y 的大作中提到】
: 题目是已经给定可以生成standard normal random variable x, 所以不清楚为什么楼
: 上的解法还要用到 1/(sqrt(2*pi))和对数变换，感觉用不着这么麻烦。
: 用R测试并写的程序：
: N=10000 # number of random samples, which can be specified
: x=rnorm(N) # generate N standard normal random variables
: alpha=10 # calculate P(x>alpha)
: est=mean(as.numeric(x>0)*exp(-alpha*x-0.5*alpha^2)) #est gives the required
: value
: 思路很简单。要求P(x>alpha)，E[1_{x>alpha}],做一个变换x'=x-alpha，也就是要求
: 期望值E{1_{x'>0}*w(x')}，其中w(x')=exp(-alpha*x'-0.5*alpha^2) is the weight
|
K*****Y
发帖数: 629 | 6 No. The quadratic term should not be there.
Suppose alpha=0, then in principal the answer should be 0.5 (est will just
be mean(as.numeric(x>0)). But your solution will give some value around 0.
35.
【在 m****r 的大作中提到】
: est=mean(as.numeric(x>0)*exp(-alpha*x-0.5*alpha^2))
: should be
: est=mean(as.numeric(x>0)*exp(-x^2/2-alpha*x-0.5*alpha^2))
: right?
:
: required
: weight
|
