📘 Section 11.8 – Correlation & Bivariate Normal Regression

Goal: Understand regression when both variables \( X \) and \( Y \) are random, assuming bivariate normality.

🔍 Core Concepts

In this setting, both \( X \) and \( Y \) come from a joint distribution: \( f(x, y) \).
We assume this distribution is bivariate normal with correlation \( \rho \), means \( \mu_X, \mu_Y \), and variances \( \sigma_X^2, \sigma_Y^2 \).
The conditional mean of \( Y \mid X = x \) is linear: \( E(Y \mid x) = \beta_0 + \beta_1 x \)
The slope is related to correlation: \( \beta_1 = \rho (\sigma_Y / \sigma_X) \)

📌 Key Formulas

Correlation: \( \rho = \frac{\sigma_{XY}}{\sigma_X \sigma_Y} \)
Conditional Mean: \( E(Y\mid x) = \mu_Y + \rho \frac{\sigma_Y}{\sigma_X}(x - \mu_X) \)
Conditional Variance: \( \sigma^2_{Y|x} = \sigma_Y^2(1 - \rho^2) \)
Sample Correlation: \( \hat{\rho} = \frac{S_{XY}}{\sqrt{S_{XX} S_{YY}}} \)
\( R^2 = \hat{\rho}^2 \) in this special case.

🧪 Example: Wire Bond Pull Strength

A dataset contains n = 25 paired values of wire length (X) and bond strength (Y). Regression gives:

Slope: \( \hat{\beta}_1 = 2.9027 \)
Intercept: \( \hat{\beta}_0 = 5.115 \)
Sample correlation: \( \hat{\rho} = 0.9818 \), so \( R^2 = 0.964 \)

Interpretation: 96.4% of the variation in bond strength is explained by wire length. The correlation is very strong.

📉 Visualization: Simulated Correlation Patterns

📊 Hypothesis Test for Zero Correlation

H₀: \( \rho = 0 \)
Test statistic: \( T_0 = \frac{\hat{\rho}\sqrt{n - 2}}{\sqrt{1 - \hat{\rho}^2}} \sim t_{n-2} \)
Reject H₀ if \( |T_0| > t_{\alpha/2, n−2} \)

📐 Confidence Interval for ρ

Use Fisher’s z-transformation: \( z = \text{arctanh}(\hat{\rho}) = \frac{1}{2}\ln\frac{1+\hat{\rho}}{1-\hat{\rho}} \)
\( z \sim N(\mu = \text{arctanh}(\rho), \sigma^2 = 1 / (n - 3)) \)
CI: \( \rho \in \left[ \tanh(z - z_{\alpha/2}\sqrt{1/(n-3)}), \tanh(z + z_{\alpha/2}\sqrt{1/(n-3)}) \right] \)