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.]

############################################################
# (a) Weibull MLE
############################################################

t <- c(
  12.4, 18.7, 25.3, 30.1, 33.5, 35.2, 38.9, 40.3, 42.7, 45.1,
  47.6, 49.8, 52.4, 55.0, 57.3, 60.2, 62.8, 65.1, 67.9, 70.5,
  72.3, 75.6, 78.2, 80.9, 83.4, 85.7, 88.1, 90.6, 93.2, 95.8,
  98.4, 101.0, 104.5, 107.3, 110.6, 113.2, 116.8, 120.1, 123.7, 127.4,
  130.9, 134.5, 138.2, 142.0, 146.3, 150.7, 155.2, 160.8, 168.4, 175.9
)

weibull_nll <- function(par, t) {
  lambda <- par[1]
  beta   <- par[2]
  
  if (lambda <= 0 || beta <= 0) return(Inf)
  
  ll <- sum(log(beta/lambda) +
            (beta - 1)*log(t/lambda) -
            (t/lambda)^beta)
  
  return(-ll)
}

fit <- optim(c(mean(t), 1), weibull_nll, t = t, hessian = TRUE)

lambda_hat <- fit$par[1]
beta_hat   <- fit$par[2]

lambda_hat

[1] 99.02759

beta_hat 

[1] 2.205876

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.]

############################################################
# (b) Exponential MLE
############################################################

lambda_exp_hat <- mean(t)
lambda_exp_hat

[1] 87.61

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

############################################################
# (c) Likelihood Ratio Test
############################################################

ll_weibull <- -fit$value
ll_exp <- sum(-log(lambda_exp_hat) - t/lambda_exp_hat)

LRT_stat <- 2 * (ll_weibull - ll_exp)
pval_LRT <- 1 - pchisq(LRT_stat, df = 1)

LRT_stat

[1] 34.44995

pval_LRT

[1] 4.373535e-09

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.]

############################################################
# (d) Wald Test
############################################################

vcov_mat <- solve(fit$hessian)
var_beta <- vcov_mat[2,2]

W_stat <- (beta_hat - 1)^2 / var_beta
pval_Wald <- 1 - pchisq(W_stat, df = 1)

W_stat

[1] 23.10901

pval_Wald

[1] 1.530721e-06

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.

############################################################
# (e) Density Plot
############################################################

x_vals <- seq(min(t), max(t), length.out = 200)

# Weibull density
weibull_density <- (beta_hat / lambda_hat) *
                   (x_vals / lambda_hat)^(beta_hat - 1) *
                   exp(-(x_vals / lambda_hat)^beta_hat)

# Exponential density
exp_density <- (1 / lambda_exp_hat) * exp(-x_vals / lambda_exp_hat)

# Plot
hist(t, probability = TRUE,
     main = "Weibull vs Exponential Fit",
     xlab = "Time to Failure")

lines(x_vals, weibull_density, lwd = 2)
lines(x_vals, exp_density, lwd = 2, lty = 2)

legend("topright",
       legend = c("Weibull", "Exponential"),
       lwd = 2,
       lty = c(1,2))

Both statistical tests and the density fit confirm that the Weibull model significantly outperforms the exponential model and the system exhibits increasing failure risk over time.