The previous lesson produced numbers — , for a hedge ratio — but numbers alone cannot answer the question a risk committee actually asks: is that slope reliably different from zero, or could five noisy days have produced this by chance? This note adds the one assumption needed to answer that — normality of the errors — and derives the exact sampling distributions of and . The payoff is threefold: a -test for , exact confidence intervals for the slope and intercept using the pivotal-quantity method already built, and a sharp distinction — routinely blurred in practice — between a confidence interval for the mean response and a prediction interval for a single new observation at .
Learning objectives. By the end of this note you should be able to:
- state the exact sampling distributions of and under normal errors, and prove them from the projection geometry of the previous lesson;
- construct and use the -statistic for testing , and the corresponding confidence interval for ;
- explain why OLS coincides with maximum likelihood under normal errors, and where the two estimators of diverge;
- distinguish a confidence interval for from a prediction interval for a new at , and compute both.
In this lesson
- Adding normality
- Sampling distributions of and
- Estimating and the -statistic
- Confidence interval vs. prediction interval
Before you start
- Simple Linear Regression: OLS Estimators and Geometry
- The Chi-Squared, t, and F Distributions
- Derivation — The Sampling Distribution of S² and Cochran's Theorem
- Pivotal Quantities and Exact Confidence Intervals
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