Question 1. Single-Factor (Market) Model [25 points]

Given Data

alpha      <- 0.0017
beta       <- 0.98
se_alpha   <- 0.0020
se_beta    <- 0.17
R2_q1      <- 0.50
E_mkt_prem <- 0.0070
n_q1       <- 96
critical_t <- 1.98

(a) t-statistic for β and test H₀: β = 0

t_beta <- beta / se_beta
cat("t-statistic for beta:", round(t_beta, 4), "\n")

## t-statistic for beta: 5.7647

cat("Critical value: ±", critical_t, "\n")

## Critical value: ± 1.98

cat("Reject H0 (beta = 0)?", abs(t_beta) > critical_t, "\n")

## Reject H0 (beta = 0)? TRUE

Interpretation: Since |t| = 5.7647 > 1.98, we reject H₀: β = 0. β = 0.98 means the fund moves ~0.98% for every 1% market move — nearly identical systematic risk to the market.

(b) Test H₀: β = 1

t_beta1 <- (beta - 1) / se_beta
cat("t-statistic for H0: beta = 1:", round(t_beta1, 4), "\n")

## t-statistic for H0: beta = 1: -0.1176

cat("Reject H0 (beta = 1)?", abs(t_beta1) > critical_t, "\n")

## Reject H0 (beta = 1)? FALSE

Interpretation: |t| = 0.1176 < 1.98 → fail to reject H₀: β = 1. The fund’s systematic risk is not statistically different from the market.

(c) t-statistic for α (Jensen’s Alpha)

t_alpha <- alpha / se_alpha
cat("t-statistic for alpha:", round(t_alpha, 4), "\n")

## t-statistic for alpha: 0.85

cat("Reject H0 (alpha = 0)?", abs(t_alpha) > critical_t, "\n")

## Reject H0 (alpha = 0)? FALSE

Interpretation: |t| = 0.85 < 1.98 → α is not statistically significant. The marketing claim of “positive risk-adjusted performance” is not statistically justified.

(d) Interpret R² = 0.50

systematic_pct    <- R2_q1 * 100
diversifiable_pct <- (1 - R2_q1) * 100
cat("Systematic variation:", systematic_pct, "%\n")

## Systematic variation: 50 %

cat("Diversifiable variation:", diversifiable_pct, "%\n")

## Diversifiable variation: 50 %

Interpretation: 50% of return variation is systematic (market risk); 50% is diversifiable (idiosyncratic) risk.

(e) CAPM-implied Expected Monthly Excess Return

expected_excess_return <- beta * E_mkt_prem
cat("Expected monthly excess return:", round(expected_excess_return * 100, 4), "%\n")

## Expected monthly excess return: 0.686 %

Result: β × E[R_m − R_f] = 0.98 × 0.70% = 0.686% per month

Question 2. Fama–French Three-Factor Model [25 points]

Given Data

a <- 0.0029;  se_a <- 0.0018
b <- 0.97;    se_b <- 0.08
s <- 0.75;    se_s <- 0.11
h <- -0.13;   se_h <- 0.13
R2_q2     <- 0.92
adj_R2_q2 <- 0.918
n_q2      <- 144

(f) t-statistics for all four coefficients

t_a <- a / se_a
t_b <- b / se_b
t_s <- s / se_s
t_h <- h / se_h

results <- data.frame(
  Term        = c("Intercept (alpha)", "MKT (b)", "SMB (s)", "HML (h)"),
  Estimate    = c(a, b, s, h),
  Std.Error   = c(se_a, se_b, se_s, se_h),
  t_stat      = round(c(t_a, t_b, t_s, t_h), 4),
  Significant = abs(c(t_a, t_b, t_s, t_h)) > critical_t
)
print(results)

##                Term Estimate Std.Error  t_stat Significant
## 1 Intercept (alpha)   0.0029    0.0018  1.6111       FALSE
## 2           MKT (b)   0.9700    0.0800 12.1250        TRUE
## 3           SMB (s)   0.7500    0.1100  6.8182        TRUE
## 4           HML (h)  -0.1300    0.1300 -1.0000       FALSE

Significant at 5%: MKT (b) and SMB (s). Intercept and HML are not significant.

(g) Investment Style Classification

cat("SMB loading (s):", s, "→ Large positive → Small-cap tilt\n")

## SMB loading (s): 0.75 → Large positive → Small-cap tilt

cat("HML loading (h):", h, "→ Small negative (insignificant) → Slight growth tilt\n")

## HML loading (h): -0.13 → Small negative (insignificant) → Slight growth tilt

cat("Style: Small-cap Growth fund\n")

## Style: Small-cap Growth fund

(h) Interpret the Intercept (Alpha)

cat("Alpha:", a, " | t-stat:", round(t_a, 4), "\n")

## Alpha: 0.0029  | t-stat: 1.6111

cat("Statistically significant?", abs(t_a) > critical_t, "\n")

## Statistically significant? FALSE

Interpretation: α is not significant (t = 1.6111) → we cannot conclude the manager adds value beyond the three factor exposures.

(i) R² Rise from 0.75 to 0.92 and Why Adjusted R²

R2_single  <- 0.75
cat("Single-factor R²:", R2_single, "\n")

## Single-factor R²: 0.75

cat("Three-factor R²:", R2_q2, "\n")

## Three-factor R²: 0.92

cat("Adjusted R²:", adj_R2_q2, "\n")

## Adjusted R²: 0.918

cat("Improvement:", R2_q2 - R2_single, "\n")

## Improvement: 0.17

Interpretation: SMB and HML explain substantial additional variation. Adjusted R² penalizes for extra predictors — it only rises if new factors genuinely improve fit, making it the correct metric for comparing models with different numbers of predictors.

Question 3. Logistic Regression for Market Direction [25 points]

Given Data

beta0     <- -0.02
beta1     <-  5.4
beta2     <- -0.38
r_lag     <-  0.010
dVIX      <-  1.5
threshold <-  0.5

(j) Predicted Probability of “Up”

logit_val <- beta0 + beta1 * r_lag + beta2 * dVIX
prob_up   <- 1 / (1 + exp(-logit_val))

cat("Logit value:", round(logit_val, 4), "\n")

## Logit value: -0.536

cat("P(Up):", round(prob_up, 4), "\n")

## P(Up): 0.3691

cat("Predicted class:", ifelse(prob_up >= threshold, "Up", "Down"), "\n")

## Predicted class: Down

Result: P(Up) = 0.3691 < 0.5 → predicted class is “Down”.

(k) Economic Interpretation of β₁ and β₂

cat("beta1 = +5.4 (lagged return): Momentum effect\n")

## beta1 = +5.4 (lagged return): Momentum effect

cat("  Positive yesterday return → higher P(Up) tomorrow\n\n")

##   Positive yesterday return → higher P(Up) tomorrow

cat("beta2 = -0.38 (delta VIX): Fear effect\n")

## beta2 = -0.38 (delta VIX): Fear effect

cat("  Rising VIX (fear) → lower P(Up) tomorrow\n")

##   Rising VIX (fear) → lower P(Up) tomorrow

(l) Confusion Matrix Metrics

TP <- 67; FP <- 44; FN <- 33; TN <- 56; Total <- 200

accuracy    <- (TP + TN) / Total
sensitivity <- TP / (TP + FN)
specificity <- TN / (TN + FP)
precision   <- TP / (TP + FP)

metrics <- data.frame(
  Metric = c("Accuracy", "Sensitivity (TPR for Up)", "Specificity", "Precision for Up"),
  Value  = round(c(accuracy, sensitivity, specificity, precision), 4),
  Pct    = paste0(round(c(accuracy, sensitivity, specificity, precision)*100, 2), "%")
)
print(metrics)

##                     Metric  Value    Pct
## 1                 Accuracy 0.6150  61.5%
## 2 Sensitivity (TPR for Up) 0.6700    67%
## 3              Specificity 0.5600    56%
## 4         Precision for Up 0.6036 60.36%

(m) Naïve Majority-Class Rule

naive_accuracy <- 100 / 200   # balanced classes
cat("Naïve (majority class) accuracy:", naive_accuracy * 100, "%\n")

## Naïve (majority class) accuracy: 50 %

cat("Model accuracy:", round(accuracy * 100, 2), "%\n")

## Model accuracy: 61.5 %

cat("Model beats naïve rule?", accuracy > naive_accuracy, "\n\n")

## Model beats naïve rule? TRUE

cat("Accuracy is inadequate for trading: a wrong prediction on a large-move day\n")

## Accuracy is inadequate for trading: a wrong prediction on a large-move day

cat("costs far more than a correct prediction on a flat day.\n")

## costs far more than a correct prediction on a flat day.

cat("Better criterion: Sharpe ratio or P&L of the resulting trading strategy.\n")

## Better criterion: Sharpe ratio or P&L of the resulting trading strategy.

Question 4. Resampling and Regularization in a Backtest [25 points]

Given Data

mean_return <- 0.0070
sd_return   <- 0.0550
n_months    <- 48

(n) Monthly and Annualized Sharpe Ratio

SR_monthly <- mean_return / sd_return
scaling    <- sqrt(12)
SR_annual  <- SR_monthly * scaling

cat("Monthly Sharpe Ratio:", round(SR_monthly, 4), "\n")

## Monthly Sharpe Ratio: 0.1273

cat("Scaling factor: sqrt(12) =", round(scaling, 4), "\n")

## Scaling factor: sqrt(12) = 3.4641

cat("Annualized Sharpe Ratio:", round(SR_annual, 4), "\n")

## Annualized Sharpe Ratio: 0.4409

(o) Bootstrap Procedure for Sharpe SE

set.seed(42)
monthly_returns <- rnorm(n_months, mean = mean_return, sd = sd_return)

B       <- 10000
boot_SR <- numeric(B)

for (i in 1:B) {
  samp      <- sample(monthly_returns, size = n_months, replace = TRUE)
  boot_SR[i] <- mean(samp) / sd(samp) * sqrt(12)
}

SE_boot <- sd(boot_SR)
cat("Bootstrap SE of Annualized Sharpe Ratio:", round(SE_boot, 4), "\n")

## Bootstrap SE of Annualized Sharpe Ratio: 0.516

Why i.i.d. bootstrap is inappropriate: Monthly returns often show serial correlation. The i.i.d. bootstrap destroys time structure → underestimates true uncertainty.

Fix: Use a block bootstrap to preserve temporal dependence.

(p) Lasso: Which λ to Deploy?

lambda_min_cv <- 0.030
lambda_1se    <- 0.065

cat("Lambda min-CV:", lambda_min_cv, "→ 14 factors\n")

## Lambda min-CV: 0.03 → 14 factors

cat("Lambda 1-SE: ", lambda_1se,    "→  7 factors\n\n")

## Lambda 1-SE:  0.065 →  7 factors

cat("Deploy: lambda =", lambda_1se, "(1-SE rule)\n")

## Deploy: lambda = 0.065 (1-SE rule)

cat("Reason: Simpler model generalizes better OOS and reduces overfitting risk.\n")

## Reason: Simpler model generalizes better OOS and reduces overfitting risk.

(q) Walk-Forward Validation Scheme

cat("Walk-Forward (Time-Respecting) Scheme:\n\n")

## Walk-Forward (Time-Respecting) Scheme:

cat("Fold 1: Train months  1-36  | Test months 37-48\n")

## Fold 1: Train months  1-36  | Test months 37-48

cat("Fold 2: Train months  1-48  | Test months 49-60\n")

## Fold 2: Train months  1-48  | Test months 49-60

cat("Fold 3: Train months  1-60  | Test months 61-72\n")

## Fold 3: Train months  1-60  | Test months 61-72

cat("...always expand training window forward...\n\n")

## ...always expand training window forward...

cat("Rule: NEVER use future data in the training set.\n\n")

## Rule: NEVER use future data in the training set.

cat("Why standard k-fold CV is UNSAFE:\n")

## Why standard k-fold CV is UNSAFE:

cat("  - Randomly mixes past and future observations\n")

##   - Randomly mixes past and future observations

cat("  - Creates look-ahead bias → inflated performance estimates\n")

##   - Creates look-ahead bias → inflated performance estimates

cat("  - Walk-forward respects the arrow of time → honest OOS evaluation\n")

##   - Walk-forward respects the arrow of time → honest OOS evaluation

Final Examination 20260608

Student

2026-06-08

Question 1. Single-Factor (Market) Model [25 points]

Given Data

(a) t-statistic for β and test H₀: β = 0

(b) Test H₀: β = 1

(c) t-statistic for α (Jensen’s Alpha)

(d) Interpret R² = 0.50

(e) CAPM-implied Expected Monthly Excess Return

Question 2. Fama–French Three-Factor Model [25 points]

Given Data

(f) t-statistics for all four coefficients

(g) Investment Style Classification

(h) Interpret the Intercept (Alpha)

(i) R² Rise from 0.75 to 0.92 and Why Adjusted R²

Question 3. Logistic Regression for Market Direction [25 points]

Given Data

(j) Predicted Probability of “Up”

(k) Economic Interpretation of β₁ and β₂

(l) Confusion Matrix Metrics

(m) Naïve Majority-Class Rule

Question 4. Resampling and Regularization in a Backtest [25 points]

Given Data

(n) Monthly and Annualized Sharpe Ratio

(o) Bootstrap Procedure for Sharpe SE

(p) Lasso: Which λ to Deploy?

(q) Walk-Forward Validation Scheme