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Say how forecasting a quantity can draw on a linear regression and on a logistic regression, and what each does there.

FRM® Part I · Quantitative Analysis · Machine Learning and Prediction · FRM1.QA.15.a

The idea

Linear and logistic regression answer different forecasting questions. A linear regression forecasts a continuous quantity, such as next quarter's credit loss or a loss given default. You fit Y = b0 + b1 × X1 + b2 × X2 + … by least squares, then plug the new predictor values into the fitted equation to get a point forecast. A logistic regression forecasts a binary event, such as default or no default. It models the log-odds, ln(p ÷ (1 − p)), as a linear function of the predictors. The forecast is a probability that always lies between 0 and 1. p = 1 ÷ (1 + e^(−z)). Here z is the fitted linear score. A coefficient acts on the odds, multiplying them by e raised to the coefficient per unit of the predictor. The usual mistake is reading the log-odds or the odds as a probability. The two are often combined, with PD from the logistic model and LGD from the linear model. EL = PD × LGD × exposure.

Formulas

Probability from a logistic regression

p = 1 ÷ (1 + e^(−z))

z
the fitted linear score, the log-odds b₀ + b₁ × X₁ + …

Always between 0 and 1. z is the log-odds, not the probability, and e^z is the odds.

Odds multiplier in a logistic regression

odds multiplier = e^(b × Δx)

b
the coefficient on the predictor
Δx
the change in the predictor, in its own units

The coefficient acts on the log-odds, so the odds are multiplied, not the probability shifted. A dummy coefficient compares with the reference category.

Expected loss from a PD model and an LGD model

EL = PD × LGD × exposure

PD
the probability of default, from the logistic model
LGD
loss given default as a share of exposure, from the linear model
exposure
the amount at risk at default

Convert the logistic log-odds into a probability before multiplying.

A practice question

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

A bank wants to forecast the probability that a corporate borrower defaults within one year. What does a logistic regression contribute to this forecast?

  1. It models the log-odds of default as a linear function of the predictors, so the fitted output is a probability between 0 and 1. - correct
  2. It predicts the loss in currency units as a linear function of the predictors, so fitted values can exceed 1.Why not: Confuses logistic regression with linear regression, which forecasts a continuous quantity rather than a probability.
  3. It finds the maximum-margin boundary between defaulters and non-defaulters and returns only a class label.Why not: Describes a support vector machine, which gives a class rather than a probability from a logit link.
  4. It forecasts the default rate as a weighted average of past default rates with exponentially declining weights.Why not: Describes an exponentially weighted moving average, which uses no predictors and no link function.

Logistic regression sets ln(p/(1-p)) = b0 + b1 x X1 + ..., so p = 1/(1+e^-z) is bounded between 0 and 1. Linear regression is the tool for a continuous quantity.

A harder one

A linear regression of quarterly credit loss (USD million) gives: loss = 2.0 + 0.6 x U - 0.04 x H, where U is the unemployment rate in percent and H is a house price index change in percent. For U = 7 and H = 20, the forecast loss is closest to:

  1. USD 3.4 millionWhy not: Omits the intercept of 2.0 and uses only the slope terms: 4.2 - 0.8.
  2. USD 7.0 millionWhy not: Flips the sign of the house price term, adding 0.8 instead of subtracting it.
  3. USD 5.4 million - correct
  4. USD 6.2 millionWhy not: Drops the house price term altogether and uses only 2.0 + 0.6 x 7.

Loss = 2.0 + 0.6 x 7 - 0.04 x 20 = 2.0 + 4.2 - 0.8 = 5.4, so USD 5.4 million.

The hardest tier

In a logistic default model the coefficient on the debt-to-income ratio (in percentage points) is 0.08. Holding other variables fixed, a 10-point increase in the ratio multiplies the odds of default by a factor closest to:

  1. 1.08Why not: Applies e^0.08, the effect of a single percentage point, rather than the effect of ten points.
  2. 1.80Why not: Treats 10 x 0.08 = 0.8 as a proportional rise in the odds, adding it to 1 instead of exponentiating.
  3. 0.45Why not: Uses e^-0.8, the wrong sign on the exponent, which would describe a decrease in the odds.
  4. 2.23 - correct

The log-odds rise by 10 x 0.08 = 0.8, so the odds are multiplied by e^0.8 = 2.23. The coefficient acts on the log-odds, not directly on the probability or the odds.

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