Assignment 12: Bootstrap Likelihood Ratio Test (BLRT)

Assignment Objectives

• Enhance understanding the procedure of Bootstrap hypothesis testing.

• Implement the procedures for detecting overfitting/underfitting issues in practical applications using bootstrap likelihood ratio test.

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Testing Overfitting/Underfitting

In Machine Learning and Statistics, overfitting occurs when a model is too complex and learns noise, leading to poor performance on new data, while underfitting happens when a model is too simple to capture important patterns, resulting in high errors overall; both issues are explained by the Bias–Variance Tradeoff and can cause unreliable predictions in real-world applications.

The probability density function (PDF) of the Weibull distribution is:

\[ f(t; \lambda, \beta) = \frac{\beta}{\lambda} \left( \frac{t}{\lambda} \right)^{\beta-1} \exp\left[ -\left( \frac{t}{\lambda} \right)^\beta \right], \quad t \ge 0 \] where \(\lambda > 0\) is the scale parameter (characteristic life) and \(\beta > 0\) is the shape parameter.

When \(\beta = 1\), the Weibull PDF simplifies to the exponential PDF:

\[ f(t; \lambda) = \frac{1}{\lambda} \exp\left( -\frac{t}{\lambda} \right) \] with constant hazard rate \(h(t) = 1/\lambda\).

This assignment focuses on performing a hypothesis test for the shape parameter (\(\beta\)) of the Weibull distribution within a reliability mode

\[\begin{align} H_0&: \beta = 1 \quad \text{(Exponential model, simpler)} \\ H_1&: \beta \neq 1 \quad \text{(Weibull model, more complex)} \end{align}\]

Steps of the BLRT

• Fit models under \(H_0\) and \(H_1\)} to the original data, compute \(\Lambda_{\text{obs}}\).

• Generate bootstrap samples under \(H_0\)}:

  • Estimate parameters under \(H_0\) from the original data.
  • Generate \(B\) datasets by sampling from the model under \(H_0\) (parametric bootstrap) or by resampling residuals/cases (nonparametric bootstrap; parametric is common for BLRT).

• For each bootstrap sample \(b = 1,\dots,B\):

  • Fit \(H_0\) and \(H_1\) models.
  • Compute \(\Lambda_b = -2[\ell_{0,b} - \ell_{1,b}]\).

• Approximate p-value:

\[ p = \frac{1}{B} \sum_{b=1}^B I(\Lambda_b \ge \Lambda_{\text{obs}}) \] (Often a small adjustment is made for stability: \((1 + \#\{\Lambda_b \ge \Lambda_{\text{obs}}\})/(B+1)\)).

Question: Reliability Application

A wind energy company monitors the reliability of gearboxes in 75 identical wind turbines located in a coastal wind farm. The gearbox is a critical component; its failure often leads to costly downtime and repairs. Previous studies suggest that the hazard rate (failure risk) may increase over time due to mechanical wear (fatigue, pitting, bearing degradation). Engineers want to test whether the failure time distribution follows an exponential model (constant hazard, random failures) or a Weibull model with shape parameter \(k>1\) (increasing hazard, indicative of aging/degradation). The failure times (in months) are:

   5.2,  7.8,  9.1, 11.3, 12.5, 13.0, 14.2, 15.1, 15.9, 16.7, 17.2, 17.8, 18.4, 18.9, 
  19.3, 19.7, 20.2, 20.6, 21.0, 21.5, 21.9, 22.3, 22.7, 23.1, 23.5, 23.9, 24.3, 24.7, 
  25.1, 25.5, 25.9, 26.3, 26.7, 27.1, 27.5, 27.9, 28.3, 28.7, 29.1, 29.5, 29.9, 30.3, 
  30.7, 31.1, 31.5, 31.9, 32.3, 32.7, 33.1, 33.5, 33.9, 34.3, 34.7, 35.1, 35.5, 35.9, 
  36.3, 36.7, 37.1, 37.5, 37.9, 38.3, 38.7, 39.1, 39.5, 39.9, 40.3, 40.7, 41.1, 41.5,
  41.9, 42.3, 42.7, 43.1, 43.5

This assignment focuses on hypothesis \(H_0: \beta = 1\) (exponential) against \(H_1: \beta \neq 1\) (Weibull). This framework detects overfitting (fitting a Weibull when exponential is true) and underfitting (fitting exponential when Weibull with \(\beta \neq 1\) is true).

a). Find the MLE of the Weibull parameters \(\lambda\) (scale) and \(\beta\) (shape), denoted by \(\hat{\lambda}\) and \(\hat{\beta}\), respectively, using the optim() procedure. [Hint: You should provide explicit expressions for the log-likelihood and gradient functions of the Weibull distribution parameters.]

Answer to Question 1 Part A

We are provided with the probability density function (PDF) of the Weibull distribution:

\[ f(t; \lambda, \beta) = \frac{\beta}{\lambda} \left( \frac{t}{\lambda} \right)^{\beta-1} \exp\left[ -\left( \frac{t}{\lambda} \right)^\beta \right], \quad t \ge 0 \]

We can use this to find the likelihood function:

\[ L(\lambda, \beta)=\prod_{i=1}^nf(t_i ; \lambda, \beta)=\prod_{i=1}^n \frac{\beta}{\lambda} \left( \frac{t_i}{\lambda} \right)^{\beta-1} \exp\left[ -\left( \frac{t_i}{\lambda} \right)^\beta \right] \]

Therefore the log-likelihood function is:

$$ \[\begin{aligned} \ell(\lambda, \beta) &= \log(L(\lambda, \beta)) \\ &= \sum_{i=1}^n\left[\log\beta-\log\lambda+(\beta-1)\log(t_i/\lambda)-(t_i/\lambda)^\beta\right] \\ &= n\log\beta-n\log\lambda+(\beta-1)\sum_{i=1}^n\log t_i-\sum_{i=1}^n(t_i/\lambda)^\beta \end{aligned}\]

$$

Using the log-likelihood function we can take the partial derivatives of \(\lambda\) and \(\beta\) to find the gradient functions:

$$ \[\begin{aligned} \frac{\partial\ell}{\partial\lambda} &=-\frac{n\beta}{\lambda}+\frac{\beta}{\lambda}\sum_{i=1}^n(t_i/\lambda)^\beta \\ \frac{\partial\ell}{\partial\beta} &=\frac{n}{\beta}-n\log\lambda+\sum_{i=1}^n\log t_i-\sum_{i=1}^n\log t_i-\sum_{i=1}^n(t_i/\lambda)^\beta\log(t_i/\lambda) \end{aligned}\]

$$

We can then use our log-likelihood function and gradient functions to find \(\hat{\lambda}\) and \(\hat{\beta}\):

turbine <- c(  5.2,  7.8,  9.1, 11.3, 12.5, 13.0, 14.2, 15.1, 15.9, 16.7, 17.2, 17.8, 18.4, 18.9, 
  19.3, 19.7, 20.2, 20.6, 21.0, 21.5, 21.9, 22.3, 22.7, 23.1, 23.5, 23.9, 24.3, 24.7, 
  25.1, 25.5, 25.9, 26.3, 26.7, 27.1, 27.5, 27.9, 28.3, 28.7, 29.1, 29.5, 29.9, 30.3, 
  30.7, 31.1, 31.5, 31.9, 32.3, 32.7, 33.1, 33.5, 33.9, 34.3, 34.7, 35.1, 35.5, 35.9, 
  36.3, 36.7, 37.1, 37.5, 37.9, 38.3, 38.7, 39.1, 39.5, 39.9, 40.3, 40.7, 41.1, 41.5,
  41.9, 42.3, 42.7, 43.1, 43.5)

loglik_weibull <- function(par, x){ #Used to impute log-likelihood
  lambda <- par[1]
  beta <- par[2]
  if (lambda <= 0 || beta <= 0) return (Inf)
  n <- length(x)
  ll <- n*log(beta)-n*beta*log(lambda)+(beta-1)*sum(log(x))-sum((x/lambda)^beta)
  
  return(-ll)
}

grad_weibull <- function(par, x){ #Used to impute gradient functions
  lambda <- par[1]
  beta <- par[2]
  if (lambda <= 0 || beta <= 0) return (c(Inf, Inf))
  n <- length(x)
  term <- (x/lambda)^beta
  d_lambda <- -n*beta/lambda+(beta/lambda)*sum(term)
  d_beta <- n/beta-n*log(lambda)+sum(log(x))-sum(term*log(x/lambda))
  
  return(-c(d_lambda, d_beta))
}

result <- optim( #Used to compute MLEs
  par=c(1,1),
  fn=loglik_weibull,
  gr=grad_weibull,
  x=turbine,
  method="L-BFGS-B",
  lower=c(1e-6, 1e-6)
)
result

$par
[1] 31.41820  3.37078

$value
[1] 275.103

$counts
function gradient 
      22       22 

$convergence
[1] 0

$message
[1] "CONVERGENCE: REL_REDUCTION_OF_F <= FACTR*EPSMCH"

lambda_hat <- result$par[1] #Gets MLEs from optim function
beta_hat <- result$par[2]

lambda_hat

[1] 31.4182

beta_hat

[1] 3.37078

The MLE of the Weibull parameters are \(\hat{\lambda}=31.41820\) and \(\hat{\beta}=3.37078\).

b). Find the MLE of the exponential parameter \(\lambda\) (scale), denoted by \(\hat{\lambda}\), using any procedure. [Hint: You should provide explicit expressions for the log-likelihood and gradient functions of the exponential distribution parameters.]

Answer to Question 1 Part B

We are provided with the probability density function (PDF) of the Weibull distribution when \(\beta=1\):

\[ f(t; \lambda) = \frac{1}{\lambda} \exp\left( -\frac{t}{\lambda} \right) \]

We can use this to find the likelihood function:

\[ L(\lambda)=\prod_{i=1}^nf(t; \lambda)=\prod_{i=1}^n\frac{1}{\lambda}e^{-t_i/\lambda} \]

Then we can find the log-likelihood function:

\[ \ell(\lambda)=\log L(\lambda)=\sum_{i=1}^n\left[-\log\lambda-\frac{t_i}{\lambda}\right] = -n\log\lambda-\frac{1}{\lambda}\sum_{i=1}^n t_i \]

We can then find the gradient function and solve for \(\hat{\lambda}\):

$$ \[\begin{aligned} &\frac{\partial\ell}{\partial\lambda}=-\frac{n}{\lambda}+\frac{1}{\lambda^2}\sum_{i=1}^n t_i \\ &\Rightarrow -\frac{n}{\lambda}+\frac{1}{\lambda^2}\sum_{i=1}^n t_i=0 \\ &\Rightarrow -n\lambda+\sum_{i=1}^n t_i =0 \\ &\Rightarrow \lambda=\frac{1}{n}\sum_{i=1}^n t_i \end{aligned}\]

$$

Therefore,

\[ \hat{\lambda}=\frac{1}{n}\sum_{i=1}^n t_i \]

We can use this to find \(\hat{\lambda}\):

lambda_hat2 = mean(turbine) #Gets lambda MLE

lambda_hat2

[1] 28.18533

The MLE of the Weibull parameter when \(\beta=1\) is \(\hat{\lambda}=28.18533\).

c). Use a) and b) to perform the regular likelihood ratio \(\chi^2\) test for \(\beta = 1\) and report the p-value.

Answer to Question 1 Part C

ll_weibull <- function(x, lambda, beta){ #Computes Weibull log-likelihood
  n <- length(x)
  n*log(beta)-n*beta*log(lambda)+(beta-1)*sum(log(x))-sum((x/lambda)^beta)
}

ll_exp <- function(x, lambda){ #Computes exponential log-likelihood
  n <- length(x)
  -n*log(lambda)-sum(x)/lambda
}

lambda_hat_weibull <- 31.41820 #Plugs in MLEs
beta_hat_weibull <- 3.37078
lambda_hat_exp <- 28.18533

ll1 <- ll_weibull(turbine, lambda_hat_weibull, beta_hat_weibull) #Gets maximized log-likelihoods
ll0 <- ll_exp(turbine, lambda_hat_exp)

LR <- 2*(ll1 - ll0) #Computes likelihood ratio
LR

[1] 100.6144

p_value <- 1 - pchisq(LR, df=1) #Gets p-value
p_value

[1] 0

The p-value is approximately zero.

d). Use the BLRT algorithm to perform a bootstrap likelihood ratio test and report the bootstrap p-value. Note that you are expected to translate the BLRT algorithm into R code to perform the BLRT. [Hint: The chi-square distribution should not be used in this part of the analysis.]

Answer to Question 1 Part D

set.seed(123)
B <- 1000
n <- length(turbine)
LR_boot <- numeric(B)

for (b in 1:B){
  x_boot <- rexp(n, rate = 1/lambda_hat2)
  lambda_exp_star <- mean(x_boot)
  fit <- optim(
    par = c(1,1),
    fn = loglik_weibull,
    gr = grad_weibull,
    x = x_boot,
    method = "L-BFGS-B",
    lower = c(1e-6, 1e-6)
  )
  lambda_weib_star <- fit$par[1]
  beta_weib_star <- fit$par[2]
  
  ll0_star <- ll_exp(x_boot, lambda_exp_star)
  ll1_star <- ll_weibull(x_boot, lambda_weib_star, beta_weib_star)
  LR_boot[b] <- 2*(ll1_star-ll0_star)
}

p_value2 <- (sum(LR_boot >= LR) + 1) / (B + 1)

p_value2

[1] 0.000999001

The p-value is 0.000999001.

e). Write a summary of the above analyses to address the following:

• Whether the two tests generated the same results.

• Which model is recommended for the data.

Answer to Question 1 Part E:

Both tests indicated that \(\beta \ne 1\) (at the 0.05 significance level) given that the p-values for the tests were \(\text{p-value}= 0 < 0.05\) and \(\text{p-value}= 0.000999001 < 0.05\). Therefore, the Weibull model would be recommended for this data given that we have evidence that \(\beta \ne 1\).