Why a Constant Odds Ratio Is Not a Constant Probability Change

Why the Same Odds Ratio Can Mean Different Probability Changes

A logistic regression coefficient gives a change on the log-odds scale. Exponentiating that coefficient gives an odds ratio for a specified predictor change. Neither quantity is itself a change in probability. To calculate the probability change, the starting probability must also be known.

This distinction matters when interpreting a fitted model. For example, a predictor increase might multiply the odds of an outcome by the same factor for every individual, while producing a relatively small probability change for one individual and a much larger change for another. The difference comes from their different starting probabilities, with all other predictors held fixed.

Definition: In a logistic regression model, the conditional probability $p(x)=P(Y=1\mid x)$ satisfies $$ \log\left(\frac{p(x)}{1-p(x)}\right)=\beta_0+\beta_1x, $$ when there is one predictor $x$. For an increase of $h$ units in $x$, with the model and other predictors held fixed, the odds ratio is $r=\exp(\beta_1h)$. The probability change is $p(x+h)-p(x)$, and depends on the starting probability $p(x)$ as well as on $r$.

Here $Y$ is a binary outcome, and the model describes its conditional probability. In a multiple-predictor model, the same reasoning applies when a specified predictor changes and the other predictors are held fixed: the change in the linear predictor determines the odds ratio. These are model-based conditional comparisons. An odds ratio alone does not establish that changing a predictor causes an outcome to change.

From Starting Probability to New Probability

Let the starting probability be $p$, where $0<p<1$, and let $r>0$ be the odds ratio for the stated predictor increase. The starting odds are $p/(1-p)$. Multiplying those odds by $r$ gives the new odds. Converting the new odds back to a probability gives the exact result.

\text{Starting odds}=\frac{p}{1-p},\qquad \text{new odds}=r\frac{p}{1-p},\qquad p_{\text{new}}=\frac{rp}{1-p+rp}.

The probability change is therefore

\Delta p=p_{\text{new}}-p =\frac{rp}{1-p+rp}-p =\frac{p(1-p)(r-1)}{1-p+rp}.

The sign of the change agrees with the direction of the odds change: if $r>1$, then $\Delta p>0$; if $0<r<1$, then $\Delta p<0$; and if $r=1$, then $\Delta p=0$. But the size of the probability change also involves $p$. Thus, fixing $r$ does not fix $\Delta p$.

Derivation: The logistic model says that increasing $x$ by $h$ changes the log odds by $\beta_1h$. Exponentiating this difference multiplies the odds by $r=\exp(\beta_1h)$. If the original odds are $p/(1-p)$, the new odds are $rp/(1-p)$. For odds $o$, the associated probability is $o/(1+o)$. Substitution gives $$ p_{\text{new}} =\frac{rp/(1-p)}{1+rp/(1-p)} =\frac{rp}{1-p+rp}. $$ Subtracting $p$ and combining terms over the common denominator gives $$ \Delta p =\frac{rp-p(1-p+rp)}{1-p+rp} =\frac{p(1-p)(r-1)}{1-p+rp}. $$

This is an exact formula for a finite predictor increase $h$. It is different from the local marginal effect, which describes the instantaneous rate of probability change as $x$ varies. Differentiating the logistic probability with respect to $x$ gives

\frac{dp}{dx}=\beta_1p(1-p).

The factor $p(1-p)$ varies with the probability. For a positive $\beta_1$, the local effect is greatest at $p=0.5$ and approaches zero as $p$ approaches either boundary. The derivative is a local rate, not the exact change over a nonzero predictor increment; for that, use the odds-ratio formula.

Worked Calculations

Worked Example: One Odds Ratio at Three Starting Risks

In a hypothetical model for a binary insurance-application outcome, a one-unit increase in a risk score has coefficient $\beta_1=\ln(3)$. Compare the predicted probability changes for applicants whose starting probabilities are $0.20$, $0.50$, and $0.80$. Assume the logistic model applies and all other predictors are held fixed.

Method. A one-unit increase has odds ratio $r=\exp(\beta_1)=\exp(\ln 3)=3$. For each starting probability, multiply the starting odds by 3, convert back to probability, then subtract the starting probability.

Starting probability $p=0.20$. The starting odds are $0.20/0.80=0.25$. The new odds are $3(0.25)=0.75$. Hence $$ p_{\text{new}}=\frac{0.75}{1+0.75}=\frac{3}{7}=0.428571\ldots, \qquad \Delta p=\frac{3}{7}-0.20=0.228571\ldots. $$

Starting probability $p=0.50$. The starting odds are $0.50/0.50=1$. The new odds are $3(1)=3$. Hence $$ p_{\text{new}}=\frac{3}{1+3}=0.75, \qquad \Delta p=0.75-0.50=0.25. $$

Starting probability $p=0.80$. The starting odds are $0.80/0.20=4$. The new odds are $3(4)=12$. Hence $$ p_{\text{new}}=\frac{12}{1+12}=\frac{12}{13}=0.923076\ldots, \qquad \Delta p=\frac{12}{13}-0.80=0.123076\ldots. $$

Rounded to four decimals, the probability changes are $0.2286$, $0.2500$, and $0.1231$, respectively. The odds ratio is 3 in every case, but the probability changes are not equal. Common mistake avoided: multiplying each starting probability by 3 would give invalid results, including a probability of $2.40$ from the starting value $0.80$. The odds, not the probability, are multiplied by the odds ratio.

Worked Example: A Two-Unit Increase With a Per-Unit Coefficient

A hypothetical logistic model for a binary claim outcome has coefficient $\beta_1=\ln(1.5)$ for a continuous exposure measure. Compare a two-unit increase for two cases with starting probabilities $0.40$ and $0.10$. Assume a linear predictor with no interaction involving this measure, and hold other predictors fixed.

Method. The two-unit change in log odds is $2\beta_1=2\ln(1.5)=\ln(2.25)$, so the total odds ratio is $\exp(2\beta_1)=2.25$. Using the per-unit odds ratio $1.5$ only once would understate the effect of a two-unit increase.

Starting probability $p=0.40$. The starting odds are $0.40/0.60=2/3$. The new odds are $2.25(2/3)=1.5$. Therefore $$ p_{\text{new}}=\frac{1.5}{1+1.5}=0.60, \qquad \Delta p=0.60-0.40=0.20. $$

Starting probability $p=0.10$. The starting odds are $0.10/0.90=1/9$. The new odds are $2.25(1/9)=0.25$. Therefore $$ p_{\text{new}}=\frac{0.25}{1+0.25}=0.20, \qquad \Delta p=0.20-0.10=0.10. $$

The same two-unit increase multiplies odds by $2.25$ in both cases, yet the probability increases are $0.2000$ and $0.1000$, respectively. Check: the odds at the new probabilities are $0.60/0.40=1.5$ and $0.20/0.80=0.25$, exactly the computed new odds.

Worked Example: A Negative Coefficient and the Direction of Change

For a hypothetical binary outcome, a one-unit increase in a predictor has coefficient $\beta_1=\ln(0.5)$. An individual’s starting probability is $0.60$. Calculate the new probability and the probability change, assuming other predictors are fixed.

Method. The odds ratio for the one-unit increase is $\exp(\ln(0.5))=0.5$. The odds are halved; the probability is not halved.

The starting odds are $0.60/0.40=1.5$. The new odds are $0.5(1.5)=0.75$. Converting back to probability, $$ p_{\text{new}}=\frac{0.75}{1+0.75} =\frac{3}{7} =0.428571\ldots. $$ Thus $$ \Delta p=\frac{3}{7}-0.60 =\frac{15}{35}-\frac{21}{35} =-\frac{6}{35} =-0.171428\ldots. $$

Rounded to four decimals, the new probability is $0.4286$ and the probability change is $-0.1714$. The negative sign is consistent with an odds ratio below 1. Common mistake avoided: halving $0.60$ gives $0.30$, but that is not the logistic-model probability. The correct operation halves the odds, then converts those odds to probability.

Interpreting the Result Carefully

An odds ratio is constant for a specified predictor change under the stated logistic model. For a one-predictor model, that means a fixed $h$ and coefficient $\beta_1$ produce the same $\exp(\beta_1h)$ at every value of $x$. In a multiple-predictor model, the comparison must also specify what is held fixed. If an interaction changes the coefficient associated with the predictor, the odds ratio itself may depend on another predictor; that is a different issue from the probability change varying with the starting risk.

A probability difference should therefore be reported with its starting conditions. Saying “the odds are multiplied by 3 for a one-unit increase” is a statement about relative odds. Saying “the probability rises by 0.20” is a statement about an absolute probability change at a particular starting probability. Those claims are not interchangeable. In particular, the same odds ratio can correspond to different absolute changes, as the worked calculations show.

For a finite increase, use this sequence:

Specify the predictor change.
Identify the increment $h$ and, for an interpretation in a multiple-predictor model, state which other predictors are held fixed.

Find the odds ratio for that increment.
For coefficient $\beta_1$, calculate $r=\exp(\beta_1h)$, not merely $\exp(\beta_1)$ unless $h=1$.

Use the starting probability.
Calculate $p_{\text{new}}=rp/(1-p+rp)$, then subtract $p$ to obtain the probability change.

Check the interpretation.
The new probability must lie between 0 and 1, and its direction must agree with whether $r$ is above or below 1.

This procedure is useful when communicating fitted logistic models in risk analysis. Reporting only a coefficient or odds ratio may be appropriate for a log-odds comparison, but a question about absolute probability impact cannot be answered without a starting probability. The conversion also helps detect a common exam error: confusing an odds multiplier with a probability multiplier.

Key takeaway: A fixed odds ratio does not imply a fixed probability change. Given starting probability $p$ and odds ratio $r$, calculate the new probability as $rp/(1-p+rp)$, and calculate the absolute change by subtracting $p$.

Check Your Understanding

For each problem, use the stated logistic-model comparison and round probabilities and probability changes to four decimals where needed.

A one-unit predictor increase has odds ratio $1.5$. If the starting probability is $0.20$, calculate the new probability and the probability change. (Answer: $p_{\text{new}}=0.2727$; change $=0.0727$.)
A coefficient is $\ln(2)$, and the predictor increases by three units. Starting from $p=0.20$, calculate the total odds ratio, new probability, and probability change. (Answer: odds ratio $=8$; $p_{\text{new}}=0.6667$; change $=0.4667$.)
An odds ratio of $0.5$ applies to a specified predictor increase. If the starting probability is $0.60$, calculate the new probability and signed probability change. (Answer: $p_{\text{new}}=0.4286$; change $=-0.1714$.)
The same predictor increase has odds ratio $3$. Compare probability changes from starting probabilities $0.10$ and $0.50$. (Answer: from $0.10$, $p_{\text{new}}=0.2500$ and change $=0.1500$; from $0.50$, $p_{\text{new}}=0.7500$ and change $=0.2500$.)
A specified predictor increase has odds ratio $2$, and the resulting probability is $0.60$. Find the starting probability and the probability change. (Answer: starting probability $=0.4286$; change $=0.1714$.)

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