Why does adding any predictor to an OLS regression never lower its R-squared?

## The mechanism Write the statistic as `R2 = 1 - RSS/TSS`. The total sum of squares `TSS = sum (y_i - y_bar)^2` is built entirely from the outcome and its mean, so no change to the predictor set can touch it. The whole behaviour of R-squared therefore comes from the residual sum of squares `RSS = sum (y_i - y_hat_i)^2`. Ordinary least squares chooses coefficients to minimise RSS over all linear combinations of the columns it is given. Add a column, and every combination that was available before is still available - just take the same coefficients and put a zero on the new one. The estimator is minimising over a strictly larger set, so the minimum it finds is at most the previous minimum. Formally: `RSS_full <= RSS_reduced` for nested models on the same rows, hence `R2_full >= R2_reduced`. Equality happens only in the knife-edge case where the new column adds nothing at all to the fit. ## Why noise still moves the number That knife-edge case is essentially never reached with real data. A column of random numbers will, by chance, line up slightly with whatever residual structure remains, and least squares will happily exploit that chance alignment. This is why appending ten columns of pure noise to a model sitting at 0.62 can move it to 0.64: each column buys a small, entirely fictitious reduction in RSS. Push the idea to its limit and it becomes obvious - with `k + 1` estimated parameters and `n = k + 1` observations, the fit can pass through every point, `RSS = 0`, and `R2 = 1` while conveying zero information. ## What adjusted R-squared changes Adjusted R-squared replaces the two raw sums with unbiased variance estimates, dividing each by its degrees of freedom: `R2_adj = 1 - (RSS/(n - k - 1)) / (TSS/(n - 1))` which is algebraically `1 - (1 - R2) * (n - 1)/(n - k - 1)`. Here `k` counts the predictors excluding the intercept. Now adding a column pulls in two directions: RSS falls (helping) while `n - k - 1` falls (hurting). A column that buys almost nothing loses that tug-of-war and the adjusted value drops. There is a sharp characterisation of the tipping point for a single added predictor: adjusted R-squared rises exactly when that predictor's t-statistic exceeds 1 in absolute value. That is a much weaker bar than the usual significance threshold near 2. So adjusted R-squared is a mild anti-padding correction, not a model-selection rule - a variable nowhere near significant at the 5 percent level can still raise it. ## The negative case Unlike R-squared, the adjusted version has no floor at zero. If the model explains almost nothing and `k` is large relative to `n`, the ratio `(RSS/(n-k-1)) / (TSS/(n-1))` can exceed 1 and the adjusted value goes negative. Read that as a blunt verdict: once the parameters spent are charged for, this model is worse than the outcome's mean. It is also undefined at `n - k - 1 = 0` and unstable as you approach it. ## Interview framing The answer an interviewer is listening for is the nesting argument stated out loud: the smaller model is still reachable inside the larger one, so the optimiser's best score cannot get worse. Everything else follows from that. The second half of a strong answer is the consequence for practice - a rise in R-squared is not evidence that a new variable belongs in the model, because the statistic was guaranteed not to fall. Any claim that a feature helped needs evidence the statistic is structurally incapable of providing. ## Related traps Comparing R-squared between models fitted on different row sets, for instance because one model dropped rows with missing values on a new predictor, breaks the argument entirely: TSS is no longer the same total, and the larger model's R-squared can then move in either direction for reasons that have nothing to do with the predictor's value.