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Apply a regression model to produce a forecast

CFA® Level I · Quantitative Methods · Applications of Simple Linear Regression in Finance · QM.8.d

The idea

A prediction substitutes a value of the independent variable into the fitted equation: predicted y equals the intercept plus the slope times x. The point estimate is only half the answer; a prediction interval around it reflects both the scatter of the data and the uncertainty in the estimated coefficients, and it widens as the chosen x moves away from the mean of the sample. Predicting outside the range of the observed data assumes the fitted relationship continues to hold there, which the regression provides no evidence for at all.

Formulas

Prediction from a regression

predicted Y = b₀ + b₁ × X

b₀
the estimated intercept
b₁
the estimated slope
X
the value of the independent variable being predicted from

Sound inside the range of X the model was fitted on and a guess outside it. The prediction is a single number for something that has a distribution, so it comes with an interval whose width grows the further X sits from its mean.

A practice question

From the question bank, with the reason each wrong option is wrong.

A regression gives an intercept of 2.0 and a slope of 1.5. What is the predicted value when the independent variable is 4?

  1. 8.0 — correct
  2. 6.0Why not: 1.5 x 4 — the slope applied without adding the intercept. The line does not pass through the origin.
  3. 1.5Why not: The slope quoted back. It is the rate of change, not the predicted value at any particular point.

2.0 + 1.5 × 4 = 8.0. The intercept is added once and the slope multiplies the input; using the slope alone omits the anchor.

A harder one

Why does a prediction interval widen as the independent variable moves away from its sample mean?

  1. Because the residual variance grows with xWhy not: That would be heteroskedasticity, a separate problem. The interval widens even when residual variance is perfectly constant.
  2. Because uncertainty in the estimated slope has a larger effect further from the centre — correct
  3. Because R-squared falls at the extremesWhy not: R-squared is a single number for the whole regression. It does not vary from point to point along the line.

A small error in the slope pivots the line about its centre, so the vertical uncertainty grows with distance from the mean. The data anchor the fit most firmly in the middle.

The hardest tier

What is the central risk of using a regression to predict outside the range of the observed data?

  1. The arithmetic becomes unstableWhy not: The arithmetic is a multiplication and an addition, and works identically at any value. The model will produce a confident number; that is the danger.
  2. The intercept is no longer definedWhy not: The intercept is a fitted constant and does not change with the value being predicted.
  3. There is no evidence the fitted relationship holds in that region — correct

The regression describes the data it was given. A relationship that is linear across observed values may bend or break beyond them, and the model will extrapolate confidently regardless.

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