Simulation-计算统计——Monte Carlo
Monte Carlo Integration
- 找到原函数,再计算
- 无法找到原函数,MC积分
Assume that we can generate \(U_1, . . . , U_n \sim Uniform (0, 1)\), and define \(\hat \theta_n = \frac{1}{n} \sum_{i=1}^{n} g(U_i)\)
By law of large numbers, if \(\int_{0}^{1}|g(u)|du < \infty\), we have \(\hat \theta_n \rightarrow \theta := \mathbb{E}g(U) = \int_0^1 g(u)du \quad a.s. \quad n \rightarrow \infty\)
a.s. means almost surely:the set of possible exceptions may be non-empty, but it has probability 0.
bounded interval
Eg. Calculate \(\theta = \int_0^1 x^2 dx\)
n <- 100
set.seed(1)
x <- runif(n)
theta.hat <- mean(x^2)
theta.hat
求:\(\theta = \int_a^{b} g(t) dt\)
利用\(\int_a^bg(t)dt = (b-a)\int_a^b g(t)\frac{1}{b-a}dt = (b-a)\mathbb{E}g(Y)\), where Y~ Uniform(a,b)
所以我们可以:
- 先生成\(Y_1,...,Y_n \sim Uniform(a,b)\)
- \(\hat \theta_n = (b-a) * \frac{1}{n} \sum_{i=1}^n g(Y_i)\)
Eg. Calculate \(\theta = \int_1^4 x^2dx\)
n <- 100
set.seed(1)
x <- runif(n, min = 1, max = 4)
theta.hat <- (4 - 1) * mean(x^2)
theta.hat
Unbounded Interval
Note that $$\Phi(x) = \mathbb{P}(Z \le x) = \mathbb{E}[I_{Z \le x}], \text{where } Z \sim N(0,1)$$
Therefore, we can estimate
事件发生的概率等于它的示性函数的期望。
Eg. Calculate \(\Phi(x) = \int_{-\infty}^x \frac{1}{\sqrt{2\pi}}e^\frac{t^2}{2}dt\) (1) when x = 0.3. (2) when \(x = (x_1, ..., x_T)\)
#(1)
x <- 0.3
n <- 100
set.seed(1)
z <- rnorm(n)
mean(z < x) # estimate
pnorm(x) # true value by pnorm()
#(2)
Phi.hat <- function(x, n = 1000){
z <- rnorm(n)
return(mean(z < x))
}
x <- seq(0, 2.5, len = 5)
Phi.hat(x) #但是这样有问题,输出只有一个值
##原因是当向量和数进行比较的时候,向量的每一个元素会分别跟数比较;而向量和向量比较时,是对应元素进行比较。Be Careful!
round(pnorm(x), digits = 3) # true value
#(2)解决方法
probab <- numeric(5)
for(i in 1:5){
probab[i] <- Phi.hat(x[i])
}
probab
For General Distributions
Calculate \(\theta = \int_A g(x)f(x) dx\)
- 先选一个满意的\(f(x)\)
- 生成一组\(X_1,...,X_n\) 服从\(f(x)\) (上节课的知识)
- 计算\(\hat \theta_n = \frac{1}{n} \sum{i=1}^n g(X_i)\)
Eg. Calculate \(\int_0^{\infty} \frac{e^{-x}}{1+x^2}dx\)
可以把\(e^{-x}\)视作均值为1的指数分布的密度函数,由此可以生成X。
Law of Large Numbers
Let\(\xi_1,\xi_2...\) be a sequence of i.i.d. random variables with finite expected value,by \(\mu\). Let
Then,
- Weak law of large numbers
- Strong law of large numbers)
Weak law of large numbers
Definition (Convergence in probability)
- Let \(\left(\bar{\xi}_n\right)_{n \geq 1}\) be a sequence of real-valued random variables defined on a probability space \((\Omega, \mathcal{F}, P)\).
- For any positive number \(\varepsilon\),
- Then, we say \(\bar{\xi}_n\) converges in probability to \(\mu\), denoted by
证明见老师ppt。
M <- 100000
n <- 100
eps <- 0.1
xi.bar <- numeric(M) # xi.bar[i] will record the sample mean of trial i
for (i in 1:M){
set.seed(i)
xi.vec <- sample(1:6, size = n, replace = TRUE)
xi.bar[i] <- mean(xi.vec)
}
mu <- 3.5
mean(abs(xi.bar - mu) > eps)
注意这里只跟n有关,m是试验数量。
如何从
Standard Error
估计这种积分形式的标准差。
如果能用数值解,就用数值解法。不知道分布的时候,可以用MC来估计。
where \(f(x)\)is a p.d.f.
, where \(X_i \sim f(x)\)
现求\(Var(\hat\theta_n)\)
standard error:标准误差
- The variance can be calculated by
where
- However, \(\sigma^2\) is not known in general.
- We can use the estimated variance:
- The estimated variance of \(\hat{\theta}_n\) is
f(x) 是自己取的,{\(X_i\)}是根据f(x)生成的。n是MC取的样本数。
由前面的内容,
Eg. Calculate \(\int_0^1 xdx\)
Choose f(x)=1, then g(x) = x, 然后可以代入公式计算。
Central limit theorems
重复生成如10000次\(\hat \theta_n\)(每一次都要投100次骰子,总的会投很多很多次骰子), 当n足够大时,$$P(\frac{\theta_n-\theta}{\sigma/\sqrt{n}}) \rightarrow \Phi(x)$$
Eg. Variance of Dice Rolling
sigma <- sqrt(35/12)
xi.bar.standardized <- (xi.bar - mu)/(sigma/sqrt(n))
hist(xi.bar.standardized, probability = TRUE)
By CLT(Central Limit Theorem) $$ P(-1.96 \le \frac{\hat \theta_n - \theta}{\sigma/\sqrt{n}}) \le 1.96 \sim 0.95 $$, So,$$ \hat{\theta}_n \in\left[\theta-1.96 \times \frac{\sigma^{\prime}}{\sqrt{n}}, \theta+1.96 \times \frac{\sigma}{\sqrt{n}}\right] . $$
但是\(\sigma\)不一定知道,可以用估计值代替,于是就有了t-分布。
- As \(\sigma^2\) is unknown in general, people usually use \(\hat{\sigma}^2\) to replace it in practice. In this case,
So, by limit distribution,
Eg. Dice Rolling
# student's t-distribution
M <- 100
xi.student <- numeric(M)
for (i in 1:M){
set.seed(i)
xi.vec <- sample(1:6, size = n, replace = TRUE)
xi.student[i] <- (mean(xi.vec) - 3.5) / (sd(xi.vec) / sqrt(n))
}
Note在n越来越大的时候,t分布会和normal越来越像。
M<-100
n<-200
u<-runif(n)
set.seed(1)
g <- numeric(M)
for (i in 1:M){
g[i]<-sin(u[i])^4*exp(-u[i])
theta<-g[i]
}
theta<-theta/M
theta
Variance and Efficiency
Definition (Efficiency)
Let \(\hat{\theta}_1\) and \(\hat{\theta}_2\) be two estimators of \(\theta\). We say \(\hat{\theta}_1\) is more efficient (in a statistical sense) than \(\hat{\theta}_2\) if
Remark.
If the variances of estimators \(\hat{\theta}_1\) and \(\hat{\theta}_2\) are unknown, we can estimate efficiency by substituting a sample estimate of the variance for each estimator.
Antithetic variables (对照变量)
如果自己生成两个负相关的变量,然后就可以生成一个variance更小的变量。
- We can also use
to estimate \(\theta\).
- Note that
- If \(\hat{\theta}_1\) and \(\hat{\theta}_2\) are negatively correlated, then
and hence,
想要构造\(\hat\theta_1, \hat\theta_2\)
使得他们负相关。下面构造的\(X_i\)和\(\tilde X_i\)就是负相关的。
- If $$ \hat{\theta}1=\frac{1}{n} \sum^n g\left(X_i\right) $$
where
- Assume that \(X_i\) is generated from the inverse transform method, then
- If the function \(g\) is a monotone function, then
- Therefore,
is also an estimator of \(\theta\)
Eg.\(\int_0^1x^2dx\)
u<-runif(100)
theta1 <-mean(u^2)
theta2<-mean((1-u)^2)
theta<-mean(theta1,theta2)
theta
Eg. Normal distribution function
x <- 1
pnorm(1)
n <- 100
V <- runif(100, 0, 1)
V.tilde <- 1 - V
theta1 <- mean(exp(-V^2/2)*(1/sqrt(2*pi)))
theta2 <- mean(exp(-V.tilde^2/2)*(1/sqrt(2*pi)))
theta <- mean(theta1, theta2)
theta
控制方差的程度和g(x)有关,也就是和f(x)的选择有关,好的时候能控制到90%以下
Control variates
- Choose $$ c^*=-\frac{\operatorname{Cov}(g(X), f(X))}{\operatorname{Var}(f(X))} $$
- Therefore, the improved estimator is defined as
where
Eg. \(\theta = \int_0^1 e^u du\)
u <- runif(100)
gu <- exp(u)
theta.hat <- mean(gu)
fu <- u
mu <- 1/2
sample.cov <- cov(gu,fu)
sample.var <- var(fu)
c.star <- -sample.cov / sample.var
theta.star <- mean(gu) + c.star*mean(fu-1/2)
print(theta.hat)
print(theta.star)
这里取f(u)=u