Outline

• Bootstrapping
  • Bootstrapping variance, confidence interval
  • non-parametric Bootstrapping, parametric Bootstrapping
  • Bootstrapping p-values

International Prize in Statistics Awarded to Bradley Efron, for his contribution in Bootstrapping (annouced in 11/12/2018)

• Permutation test

Bootstrapping (Motivating example 1, variance of “mean estimator”)

• Given \(X_1, \ldots, X_n\) samples from \(N(\mu, \sigma^2)\). How to estimate the mean of this distribution, \(\hat{\mu}\)?
  • Use \(\frac{1}{n} \sum_{i=1}^n X_i\), by weak law of large number: \[\hat{\mu} = \frac{1}{n} \sum_{i=1}^n X_i \rightarrow E(X_i)\]
• How to estimate the variance of \(\hat{\mu}\)?
  • Similarly use weak law of large number. \[Var(\hat{\mu}) = Var(\frac{1}{n} \sum_{i=1}^n X_i) = \frac{1}{n^2} \sum_{i=1}^n Var(X_i) = \frac{1}{n} \sigma^2 \]

\[\hat{Var}(\hat{\mu}) =\frac{1}{n} \hat{\sigma}^2 = \frac{1}{n(n-1)} \sum_{i=1}^n (X_i - \bar{X})^2\]

Bootstrapping (Motivating example 2, variance of “median estimator”)

• Given \(X_1, \ldots, X_n\) samples from \(N(\mu, \sigma^2)\). How to estimate the median of this distribution, \(T_n\)?
  • Use \(T_n = median_{i=1}^n(X_i)\) be a statistic, where \(T_n\) is median of \(X_1, \ldots, X_n\).
• How to estimate \(V(T_n)\), the variance of \(T_n\)?
  • Seems difficult.

Aim of bootstrapping: how to estimate variance

• Suppose \(X_1, \ldots, X_n \sim F\) where \(F\) is a distribution.
• Let \(T_n = g(X_1, \ldots, X_n)\) be a statistic, where \(T_n\) is a function of the data.
• What is the variance of \(T_n\), \(V_F(T_n)\)?

Variance of “median estimator”

• Given \(X_1, \ldots, X_n\) samples from \(N(\mu_0, \sigma_0^2)\), where both \(\mu_0\) and \(\sigma_0^2\) are known. How to estimate the median of this distribution, \(T_n\)?
  • Let \(T_n = median_{i=1}^n(X_i)\) be a statistic, where \(T_n\) is median of \(X_1, \ldots, X_n\).
• How to estimate the variance of \(T_n\)?
  • We can use Monte Carlo simulation method to estimate the variance of \(T_n\).

Monte Carlo method

• For \(b = 1, \ldots, B\)
  • draw \(X^{b}_1, \ldots, X^{b}_n \sim N(\mu_0,\sigma_0^2)\)
  • Compute \(T^{(b)}_n = median_{i=1}^n(X^{b}_i)\)
• \(\bar{T}_n = \frac{1}{B} \sum_{b=1}^B T^{(b)}_n\)
• \(\hat{V}_F (T_n) = \frac{1}{B - 1} \sum_{b=1}^B \{T^{(b)}_n - \bar{T}_n \}^2\)
• By law of large number, \(\hat{V}_F (T_n) \rightarrow V_F(T_n)\)

Variance of “median estimator”, Monte Carlo method

mu <- 1
sigma <- 1
n <- 100
B <- 1000
Ts <- numeric(B)

for(b in 1:B){
  set.seed(b)
  ax <- rnorm(n, mu, sigma)
  Ts[b] <- median(ax)
}

varTest <- var(Ts)
print(varTest)

## [1] 0.014573

• What if we don’t know \(\mu\), and \(\sigma^2\)?
  • We only have observed data.

Emperical distribution

• \(F = P(X \le x)\) is a distribution function.
• We can estimate \(F\) with the empirical distribution function \(F_n\), the cdf that puts mass \(1/n\) at each data point \(X_i\). \[F_n (x) = \frac{1}{n} \sum_{i = 1}^n I(X_i \le x)\] where \[\begin{equation} I(X_i \le x) = \begin{cases} 1, & \text{if}\ X_i \le x \\ 0, & \text{if}\ X_i > x \end{cases} \end{equation}\]

Empirical process (visualization)

library(ggplot2)
n <- 1000
df <- data.frame(x = c(rnorm(n, 0, 1)))
base <- ggplot(df, aes(x)) + stat_ecdf()
base + stat_function(fun = pnorm, colour = "red") + xlim(c(-3,3))

## Warning: Removed 2 rows containing non-finite values (stat_ecdf).

Empirical process

Empirical distribution is close to the underlying distribution

• Glivenko-Cantelli Theorem \[\sup_x | F_n(x) - F(x)| \rightarrow 0\]

• Dvoretzky-Kiefer-Wolfowitz inequality, for any \(\varepsilon > 0\) \[P(\sup_x | F_n(x) - F(x)| > \varepsilon) \le 2\exp(-2n\varepsilon^2)\]

Variance of the “median estimator”, Bootstrapping method

Instead of drawing samples from the underlying distribution \(F \sim N(\mu, \sigma^2)\), we draw from the empirical distribution \(F_n\)

• For \(b = 1, \ldots, B\)
  • draw \(X^{b}_1, \ldots, X^{b}_n \sim\) N(mu,sigma^2) \(F_n\)
  • Compute \(T^{(b)}_n = median_{i=1}^n(X^{b}_i)\)
• \(\bar{T}_n = \frac{1}{B} \sum_{b=1}^B T^{(b)}_n\)
• \(\hat{V}_{F_n} (T_n) = \frac{1}{B-1} \sum_{b=1}^B \{T^{(b)}_n - \bar{T}_n \}^2\)

How to sample from the empirical distribution?

• Drawing \(X_1^*, \ldots, X_n^*\) from \(F_n\) is equivalent to draw \(n\) observations, with replacement from the original data \(\{X_1, \ldots, X_n\}\).
• Therefore, Bootstrapping sampling is also described as resampling data.
• For \(b = 1, \ldots, B\)
  • draw \(X^{b}_1, \ldots, X^{b}_n \sim \{ X_1, \ldots, X_n \}\) with replacement.
  • Compute \(T^{(b)}_n = median_{i=1}^n(X^{b}_i)\)
• \(\bar{T}_n = \frac{1}{B} \sum_{b=1}^B T^{(b)}_n\)
• \(\hat{V}_{F_n} (T_n) = \frac{1}{B-1} \sum_{b=1}^B \{T^{(b)}_n - \bar{T}_n \}^2\)

Variance of “median estimator”, Bootstrapping method

mu <- 1
sigma <- 1
n <- 100
set.seed(32611)
X <- rnorm(n, mu, sigma)
B <- 1000
Ts <- numeric(B)

for(b in 1:B){
  set.seed(b)
  ax <- sample(X, replace = T)
  Ts[b] <- median(ax)
}

varTest <- var(Ts)
print(varTest)

## [1] 0.02598333

Bootstrapping Variance Estimator

1. Draw a bootstrapping sample \(X_1^*, \ldots, X_n^* \sim F_n\), where \(F_n\) is the emperical CDF. Compute \({T^*}_n = g(X_1^*, \ldots, X_n^*)\).
2. Repeat the previous step \(B\) times, yielding estimators \({T^*}_n^{(1)}, \ldots, {T^*}_n^{(B)}\).
3. Compute \[\hat{Var}_{F_n}({T}_n) = \frac{1}{B-1}\sum_{b=1}^B ({T^*}_n^{(b)} - \bar{T}^*)^2,\] where \(\bar{T}^* = \frac{1}{B}\sum_{b=1}^B {T^*}_n^{(b)}\)
4. Output \(\hat{Var}_{F_n}({T}_n)\) as the bootstrapping variance of \({T}_n\).

Why Bootstrapping variance works?

• \(T_n = g(X_1, \ldots, X_n)\)
• \(mean_F(T_n) = \int g(X_1, \ldots, X_n) f(X) dX = \int g(X_1, \ldots, X_n) dF\)
• \(Var_F(T_n) = \int (g(X_1, \ldots, X_n) - mean_F(T_n))^2 dF\)

Since in general, we don’t know distribution \(F\), we will calculate using the empirical CDF \(F_n\).

• \(Var_{F_n}(T_n) = \int (g(X_1, \ldots, X_n) - mean_{F_n}(T_n))^2 dF_n\)
• Finally, we used bootstrapping variance \(\hat{Var}_{F_n}(T_n)\) to estimate \(Var_{F_n}(T_n)\).

To summarize:

• Estimation error: \(Var_F(T_n) - Var_{F_n}(T_n) = O_p(1/\sqrt{n})\)
• Simulation error: \(Var_{F_n}(T_n) - \hat{Var}_{F_n}(T_n) = O_p(1/\sqrt{B})\)

The parametric Bootstrapping (Variance of “median estimator”)

• Calculate \(\hat{\mu} = \arg\max L(\mu;X_1^n)\), \(\sigma\) is known.
• For \(b = 1, \ldots, B\)
  • draw \(X^{b}_1, \ldots, X^{b}_n \sim N(\hat{\mu},\sigma^2)\) (\(\hat{F}\))
  • Compute \(\hat{T}^{(b)}_n = median_{i=1}^n(X^{b}_i)\)
• \(\bar{\hat{T}}_n = \frac{1}{B} \sum_{b=1}^B \hat{T}^{(b)}_n\)
• \(\hat{V}_F (T_n) = \frac{1}{B-1} \sum_{b=1}^B \{\hat{T}^{(b)}_n - \bar{\hat{T}}_n \}^2\)

The parametric Bootstrapping (Variance of “median estimator”)

mu <- 1
sigma <- 1
n <- 100
set.seed(32611)
X <- rnorm(n, mu, sigma)
muhat <- mean(X)
B <- 1000
Ts <- numeric(B)

for(b in 1:B){
  set.seed(b)
  ax <- rnorm(n,muhat,sigma)
  Ts[b] <- median(ax)
}

varTest <- var(Ts)
print(varTest)

## [1] 0.014573

Comparison study

• \(X_1, \ldots, X_n \sim POI(\lambda)\) (\(\lambda = 5\)), while we only observe the data \(X_1, \ldots, X_n\) and know data is from Poisson.
• \(n = 100\)
• \(T_n = (\frac{1}{n}\sum_{i=1}^nX_i)^2\)

• What is the variance of \(T_n\)?

• Methods:
  • delta method
  • non-parametric Bootstrapping
  • parametric Bootstrapping
  • simulation (requires knowning the underlying parameter \(\lambda\))

Delta method

• \(\hat{\lambda} = \bar{X}\)
• \(var(\bar{X}) = \frac{\lambda}{n}\)
• \(var(T_n) = var(\bar{X}^2) = (2 \lambda)^2 \times var(\bar{X}) = \frac{4\lambda^3}{n}\)
• \(\hat{var}_F(T_n) = \frac{4\hat{\lambda}^3}{n} = \frac{4\bar{X}^3}{n}\)

lambda <- 5
n <- 100
set.seed(32611)
X <- rpois(n, lambda)
var_hat1 <- 4*mean(X)^3/n
print(var_hat1)

## [1] 4.97006

Bootstrapping (non-parametric)

lambda <- 5
n <- 100
set.seed(32611)
X <- rpois(n, lambda)
B <- 1000
TB <- numeric(B)
for(b in 1:B){
  set.seed(b)
  aX <- sample(X,n,replace = T)
  TB[b] <- (mean(aX))^2
}
var_hat2 <- var(TB)
print(var_hat2)

## [1] 4.44549

Parametric Bootstrapping

lambda <- 5
n <- 100
set.seed(32611)
X <- rpois(n, lambda)
lambdaHat <- mean(X)
B <- 1000
TB <- numeric(B)
for(b in 1:B){
  set.seed(b)
  aX <- rpois(n, lambdaHat)
  TB[b] <- (mean(aX))^2
}
var_hat3 <- var(TB)
print(var_hat3)

## [1] 5.261807

Bootstrapping using R package

library(boot)
myMean <- function(data, indices){
  d <- data[indices]
  mean(d)^2
}

## non parametric bootstrap
set.seed(32611)
boot_nonpara <- boot(data=X, statistic = myMean, R = B)
var(boot_nonpara$t)

##          [,1]
## [1,] 4.495394

## parametric bootstrap
genPois <- function(data, lambda){
  rpois(length(data), lambda)
}
boot_para <- boot(data=X, statistic = myMean, R = B, sim="parametric", ran.gen = genPois, mle = mean(X))
var(boot_para$t)

##          [,1]
## [1,] 5.199379

Simulation

• We know the underlying parameters in simulations

lambda <- 5
n <- 100
B <- 1000
Ts <- numeric(B)
for(b in 1:B){
  set.seed(b)
  aX <- rpois(n, lambda)
  Ts[b] <- (mean(aX))^2
}
print(var(Ts))

## [1] 5.28793

Summary

• Assume \(\sigma\) is known, \(\mu\) is the only unknown parameter.

• For \(b = 1, \ldots, B\)
  • draw \(X^{b}_1, \ldots, X^{b}_n\) from
    • Simulation: \(N(\mu,\sigma^2)\)
    • Non-parametric bootstrapping: \(X^{b}_1, \ldots, X^{b}_n \sim \{ X_1, \ldots, X_n \}\) with replacement.
    • Parametric bootstrapping: \(N(\hat{\mu},\sigma^2)\)
  • Compute \(T^{(b)}_n = median_{i=1}^n(X^{b}_i)\)
• \(\bar{T}_n = \frac{1}{B} \sum_{b=1}^B T^{(b)}_n\)

• \(\hat{V}_{F_n} (T_n) = \frac{1}{B-1} \sum_{b=1}^B \{T^{(b)}_n - \bar{T}_n \}^2\)

Summary

• Goal: to estimate the variance of an estimator

Comparison between Simulation and Non-parametric bootstrapping

• Simulation: Monte Carlo Simulation from underlying distribution.
• Non-parametric bootstrapping: Monte Carlo Simulation from the empirical distribution.

The rest of the percedures are the same for these two methods

Bootstrapping confidence interval

• Percentiles
• normal approximation
• Pivotal Intervals

Bootstrapping confidence interval via Percentiles

1. Draw a bootstrapping sample \(X_1^*, \ldots, X_n^* \sim F_n\). Compute \({T^*}_n = g(X_1^*, \ldots, X_n^*)\).
2. Repeat the previous step \(B\) times, yielding estimators \({T^*}_n^{(1)}, \ldots, {T^*}_n^{(B)}\).
3. Rank \({T^*}_n^{(1)}, \ldots, {T^*}_n^{(B)}\) such that \({T^r}_n^{(1)} \le {T^r}_n^{(2)} \le \ldots \le {T^r}_n^{(B)}\)

We can define 95% confidence interval using (if B = 10,000) \[[{T^r}_n^{(250)}, {T^r}_n^{(9750)}]\]

Calculate Bootstrapping confidence interval via Percentiles (1)

lambda <- 5
n <- 100
set.seed(32611)
X <- rpois(n, lambda)
B <- 1000
TB <- numeric(B)
for(b in 1:B){
  set.seed(b)
  aX <- sample(X,n,replace = T)
  TB[b] <- (mean(aX))^2
}
quantile(TB, c(0.025, 0.975))

##     2.5%    97.5% 
## 20.97411 29.16000

Performance of Bootstrapping confidence interval via Percentiles (2)

• Underlying truth:

\[E(\bar{X}^2) = E(\bar{X})^2 + var(\bar{X}) = \lambda^2 + \lambda/n\]

lambda <- 5
n <- 100
truth <- lambda^2 + lambda/n
B <- 1000
Repeats <- 100

counts <- 0

plot(c(0,100),c(0,Repeats), type="n", xlab="boot CI", ylab="repeats index")
abline(v = truth, col=2)

for(r in 1:Repeats){
  set.seed(r)
  X <- rpois(n, lambda)
  TB <- numeric(B)
  for(b in 1:B){
    set.seed(b)
    aX <- sample(X,n,replace = T)
    TB[b] <- (mean(aX))^2
  }
  segments(quantile(TB, c(0.025)), r, quantile(TB, c(0.975)), r)
  if(quantile(TB, c(0.025)) < truth & truth < quantile(TB, c(0.975))){
    counts <- counts + 1
  }
}

counts/Repeats

## [1] 0.93

Calculation of Bootstrapping confidence interval via Percentiles (3)

• We can also obtain this Percentiles CI by boot package

library(boot)
myMean <- function(data, indices){
  d <- data[indices] ## in this example, data is a vector
  mean(d)^2
}
boot_nonpara <- boot(data=X, statistic = myMean, R = B)
boot.ci(boot_nonpara, type="perc")

## BOOTSTRAP CONFIDENCE INTERVAL CALCULATIONS
## Based on 1000 bootstrap replicates
## 
## CALL : 
## boot.ci(boot.out = boot_nonpara, type = "perc")
## 
## Intervals : 
## Level     Percentile     
## 95%   (22.56, 30.80 )  
## Calculations and Intervals on Original Scale

Normal approximation

\[[\hat{T}_n - Z_{1 - \alpha/2}\hat{\sigma}_B, \hat{T}_n - Z_{\alpha/2}\hat{\sigma}_B],\] Where \(Z_\alpha = \Phi^{-1}(1-\alpha)\), \(\Phi\) is the cdf of standard Normal distribution.

• \(Z_{0.025} = -1.96\)
• \(Z_{0.975} = 1.96\)

Where \(\hat{T}_n\) is the estimator from the original sample and \(\hat{\sigma}_B\) is bootstrapping se.

Implementation for Normal approximation

lambda <- 5
n <- 100
B <- 1000

set.seed(32611)
X <- rpois(n, lambda)
lambdaHat <- mean(X)
That <- lambdaHat^2 
TB <- numeric(B)
for(b in 1:B){
  set.seed(b)
  aX <- sample(X,n,replace = T)
  TB[b] <- (mean(aX))^2
}
ci_l <- That - 1.96*sd(TB)
ci_u <- That + 1.96*sd(TB)

c(ci_l, ci_u)

## [1] 20.76757 29.03263

Calculation of Bootstrapping confidence interval via Normal approximation (4)

• We can also obtain this Percentiles CI by boot package

library(boot)
myMean <- function(data, indices){
  d <- data[indices] ## in this example, data is a vector
  mean(d)^2
}
boot_nonpara <- boot(data=X, statistic = myMean, R = B)
boot.ci(boot_nonpara, type="norm")

## BOOTSTRAP CONFIDENCE INTERVAL CALCULATIONS
## Based on 1000 bootstrap replicates
## 
## CALL : 
## boot.ci(boot.out = boot_nonpara, type = "norm")
## 
## Intervals : 
## Level      Normal        
## 95%   (20.91, 28.95 )  
## Calculations and Intervals on Original Scale

Evaluation for Normal approximation

lambda <- 5
n <- 100
truth <- lambda^2 
B <- 1000
Repeats <- 100

counts <- 0

plot(c(0,100),c(0,Repeats), type="n", xlab="boot CI", ylab="repeats index")
abline(v = truth, col=2)

for(r in 1:Repeats){
  set.seed(r)
  X <- rpois(n, lambda)
  lambdaHat <- mean(X)
  That <- lambdaHat^2 
  TB <- numeric(B)
  for(b in 1:B){
    set.seed(b)
    aX <- sample(X,n,replace = T)
    TB[b] <- (mean(aX))^2
  }
  ci_l <- That - 1.96*sd(TB)
  ci_u <- That + 1.96*sd(TB)
  segments(ci_l, r, ci_u, r)
  if(ci_l < truth & truth < ci_u){
    counts <- counts + 1
  }
}

counts/Repeats

## [1] 0.93

Bootstrapping confidence interval via Pivotal Intervals

• Won’t be covered this year.
• If you are interested in this, checkout previous lecture notes https://caleb-huo.github.io/teaching/2017FALL/lectures/week11_bootstrapAndPermutation/bootstrap/bootstrap.html

Summary Bootstrapping confidence interval

Bootstrapping p-value

• Using linear regression an example.
  • We can obtain the pvalue for lcavol using lm pacakge
  • How to get thie pvalue from bootstrapping procedure?

library(ElemStatLearn)
prostate$train <- NULL
alm <- lm(lpsa ~ ., data = prostate)
summary(alm)

## 
## Call:
## lm(formula = lpsa ~ ., data = prostate)
## 
## Residuals:
##      Min       1Q   Median       3Q      Max 
## -1.76644 -0.35510 -0.00328  0.38087  1.55770 
## 
## Coefficients:
##              Estimate Std. Error t value Pr(>|t|)    
## (Intercept)  0.181561   1.320568   0.137  0.89096    
## lcavol       0.564341   0.087833   6.425 6.55e-09 ***
## lweight      0.622020   0.200897   3.096  0.00263 ** 
## age         -0.021248   0.011084  -1.917  0.05848 .  
## lbph         0.096713   0.057913   1.670  0.09848 .  
## svi          0.761673   0.241176   3.158  0.00218 ** 
## lcp         -0.106051   0.089868  -1.180  0.24115    
## gleason      0.049228   0.155341   0.317  0.75207    
## pgg45        0.004458   0.004365   1.021  0.31000    
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 0.6995 on 88 degrees of freedom
## Multiple R-squared:  0.6634, Adjusted R-squared:  0.6328 
## F-statistic: 21.68 on 8 and 88 DF,  p-value: < 2.2e-16

Bootstrapping p-value

• We are interested in calculating the bootstrapping for lcavol

1. Estimate the bootstrapping se
2. Then calculate the Z statistics \[Z = \frac{\hat{\beta}}{\hat{se(\hat{\beta})}}\]
3. Under the NULL, \(Z\sim N(0,1)\)

B <- 1000
coef_lcavol <- rep(NA,B)
for(b in 1:B){
  set.seed(b)
  bindex <- sample(nrow(prostate), replace = TRUE)
  bprostate <- prostate[bindex,]
  blm <- lm(lpsa ~ ., data = bprostate)
  coef_lcavol[b] <- coef(blm)["lcavol"]
}
se_lcavol <- sd(coef_lcavol)
Z_lcavol <- coef(alm)["lcavol"] / se_lcavol
pnorm(Z_lcavol, lower.tail = FALSE) * 2 ## two-sided bootstrapping p-value

##      lcavol 
## 6.96673e-13

Bootstrapping p-value

• lasso estimator (when \(\lambda = 10\))
• this is hard to directly obtain a p-value
• But we can use bootstrapping method to get a p-value

library(lars)

## Loaded lars 1.2

x <- as.matrix(prostate[,1:8])
y <- prostate[,9]
lassoFit <- lars(x, y, normalize = FALSE) ## lar for least angle regression
coef(lassoFit, s=10, mode="lambda") ## get beta estimate when lambda = 2. 

##       lcavol      lweight          age         lbph          svi          lcp 
##  0.576840683  0.023404575 -0.005103704  0.076617068  0.000000000  0.000000000 
##      gleason        pgg45 
##  0.000000000  0.006724607

B <- 1000
coef_lcavol <- rep(NA,B)
for(b in 1:B){
  set.seed(b)
  bindex <- sample(nrow(prostate), replace = TRUE)
  bprostate <- prostate[bindex,]
  bx <- as.matrix(bprostate[,1:8])
  by <- bprostate[,9]
  blassoFit <- lars(bx, by, normalize = FALSE) ## lar for least angle regression
  bcoef <- coef(blassoFit, s=10, mode="lambda") ## get beta estimate when lambda = 2. 
  coef_lcavol[b] <- bcoef["lcavol"]
}
se_lcavol <- sd(coef_lcavol)
Z_lcavol <- coef(lassoFit, s=10, mode="lambda")["lcavol"] / se_lcavol
pnorm(Z_lcavol, lower.tail = FALSE) * 2 ## two-sided bootstrapping p-value

##       lcavol 
## 2.213797e-11

HW, Large scale Bootstrapping exercise

• For the HAPMAP data on hiperGator.
• Perform SVD and obtain the largest singular value.
• what is the Bootstrapping variance of the largest singular value.
• what is the Bootstrapping confidence interval of the largest singular value.

Permutation test

• Distribution free.
• Doesn’t involve any asymptotic approximations.

Problem setting:

• Suppose we have data
  • \(X_1, \ldots, X_n \sim F\)
  • \(Y_1, \ldots, Y_m \sim G\)
• We want to test: \(H_0: F=G\) versus \(H_1: F \ne G\)

Permutation test procedure

• If we want to test if the mean parameter of two distributions are the same.
• Let \(Z = (X_1, \ldots, X_n, Y_1, \ldots, Y_m)\).
• Create labels \(L = (1,1,\ldots, 1,2,2,\ldots, 2)\) such that there are \(n\) copies of 1 and \(m\) copies of 2.
• A test statistics can be written as a function of \(Z\) and \(L\). For example, \[T = |\bar{X}_n - \bar{Y}_m|\] or

\[T = | \frac {\sum_{i = 1}^NZ_i I(L_i=1)} {\sum_{i = 1}^N I(L_i=1)} - \frac {\sum_{i = 1}^NZ_i I(L_i=2)} {\sum_{i = 1}^N I(L_i=2)} |,\] where \(N = n + m\). Therefore the test statistic \(T = g(L,Z)\) is a function of \(L\) and \(Z\).

Permutation test procedure (2)

• Define \[p = \frac{1}{N!}\sum_\pi I(g(L_\pi,Z) \ge g(L,Z)),\]

where \(L_\pi\) is a permutation of the labels and the sum is over all permutations.

• Under \(H_0\), permuting the labels does not change the distribution.
  • In other word, \(g(L,Z)\) has equal chances of having any rank among all permuted values.
  • Under \(H_0\), \(p \sim Unif(0,1)\)
  • If we reject when \(p<\alpha\), then we have a level \(\alpha\) test

Permutation test procedure (3)

Sometimes summing over all possible permutations is infeasible. But it suffices to use a random sample of permutation.

Permutation test procedure:

1. Calculate the observed test statistic \(T = g(L,Z)\)
2. Compute a random permutation of the labels \(L_k\) and compute \(T^{(k)} = g(L_k,Z)\). Do this \(K\) times, which will genreate values of \(T^{(1)}, \ldots, T^{(K)}\).
3. Compute the p-value as: \[\frac{1}{K} \sum_{k=1}^K I(T^{(k)} \ge T )\]

An illustrating example (no difference in mean)

• \(n = 100\), \(m = 100\)
• \(X_i \sim N(0,1^2)\)
• \(Y_i \sim N(0,2^2)\)

If we observe \(X_1, \ldots, X_n\) and \(Y_1, \ldots, Y_m\), how can we test if the mean of \(X\) and mean of \(Y\) have a difference. Consider the follow one-sided test:

• \(H_0: \mu_x = \mu_y\)
• \(H_A: \mu_x < \mu_y\)

n <- 100
m <- 100

set.seed(32611)
x <- rnorm(n,0,1)
y <- rnorm(n,0,2)

adataFrame <- data.frame(data=c(x,y),label=gl(2,n))

T <- with(adataFrame, mean(data[label==2] - data[label==1]))

B <- 1000

TB <- numeric(B)

for(b in 1:B){
  set.seed(b)
  bdataFrame <- adataFrame
  bdataFrame$label <- sample(bdataFrame$labe)
  TB[b] <- with(bdataFrame, mean(data[label==2] - data[label==1]))
}

pvalue = mean(TB >= T)
pvalue

## [1] 0.647

An illustrating example (with a difference in mean)

• \(n = 100\), \(m = 100\)
• \(X_i \sim N(0,1^2)\)
• \(Y_i \sim N(1,2^2)\)

If we observe \(X_1, \ldots, X_n\) and \(Y_1, \ldots, Y_m\), how can we test if the mean of \(X\) and mean of \(Y\) have a difference.

n <- 100
m <- 100

set.seed(32611)
x <- rnorm(n,0,1)
y <- rnorm(n,0.5,2)

adataFrame <- data.frame(data=c(x,y),label=gl(2,n))

T <- with(adataFrame, mean(data[label==2] - data[label==1]))

B <- 1000

TB <- numeric(B)

for(b in 1:B){
  set.seed(b)
  bdataFrame <- adataFrame
  bdataFrame$label <- sample(bdataFrame$labe)
  TB[b] <- with(bdataFrame, mean(data[label==2] - data[label==1]))
}

mean(TB >= T) 

## [1] 0.031

Reference

• http://www.stat.cmu.edu/~larry/=stat705/Lecture13.pdf
• http://users.stat.umn.edu/~helwig/notes/bootci-Notes.pdf
• proof of p-value follows \(UNIF(0,1)\) under the null
  • Under the null, a test statistic \(S\) has distribution \(F(s)\)
  • \(s\) is the observed test statistics
  • p-value \(p = Pr(S \le s) = F(s)\)
  • \(Pr(P \le p) = Pr(F^{-1}(P) \le F^{-1}(p)) = Pr(S \le s) = p\)
  • QED (completion of proof)