Question: How to construct standard errors and test statistics for a ratio of regression coefficients?

Background: I am running the same regression on two related outcomes $$ Y_1 = \alpha_1 + \theta_1 X_1 + \beta 'X +\epsilon_1 $$

$$ Y_2 = \alpha_1 + \theta_2 X_1 + \beta 'X +\epsilon_2 $$ (where $\epsilon_1, \epsilon_2$ are independent).

I am interested in the ratio of the coefficients from these two regressions. In particular, I want to test the following $$ H_0: \frac{\theta_1}{\theta_2}=1, H_A: \frac{\theta_1}{\theta_2}\neq1 $$ and also the one-sided version of this test for $ \frac{\theta_1}{\theta_2}>1$.

  • 1
    $\begingroup$ If, say, you know $\theta_2$ is positive, then your test can be re-expressed in terms of the difference $\theta_1-\theta_2,$ thereby reducing the question to an ordinary regression of $Y_1-Y_2$ against $(X_1,X)$ (presumably without intercept, since you posit the same intercept in both models). Do we need to be concerned, then, with the possibility that $\theta_2\le 0$? $\endgroup$
    – whuber
    Mar 27, 2022 at 14:05
  • 1
    $\begingroup$ Theoretically either $\theta_1 < 0$ and $\theta_2 < 0$ or $\theta_1> 0$ and $\theta_2 > 0$, so, yes the reformulation in terms of differences would work in my application. $\endgroup$
    – Papayapap
    Mar 28, 2022 at 9:35

1 Answer 1


Assuming (as you do in a comment) that the $\theta_i$ have the same sign and $\theta_2$ is nonzero, the null hypothesis is algebraically equivalent to

$$H_0: \theta_1 = \theta_2$$

while the two alternative hypotheses are equivalent to

$$H_A:\theta_1\ne\theta_2\quad\text{and}\quad H_A^\prime:\theta_1\lt\theta_2\lt 0\text{ or } 0\lt \theta_2\lt\theta_1.$$

Since all the $\epsilon_1$ are independent of the $\epsilon_2,$ all the $Y_1$ responses are independent of the $Y_2$ responses and so (presuming the estimates are separately computed, one set for the $Y_1$ data and another set for the $Y_2$ data) the parameter estimates $\hat\theta_i$ are independent.

How you proceed depends on circumstances. To sketch the general approach, let's suppose you would use a $t$ test or $Z$ test in either regression alone. This means the combined information of assumptions and data is strong enough to suggest the sampling distributions of the $\hat\theta_i$ are approximately Normal with estimated variances $\sigma_i^2$ respectively. Consequently, the test statistic

$$\hat\theta = \hat\theta_1 - \hat\theta_2$$

is approximately Normal with sampling variance

$$\operatorname{Var}(\hat\theta) = \operatorname{Var}(\hat\theta_1) + \operatorname{Var}(\hat\theta_2) = \sigma_1^2 + \sigma_2^2.$$

You would therefore refer the test statistic

$$Z = \frac{\hat\theta_2 - \hat\theta_2}{\sqrt{\sigma_1^2 + \sigma_2^2}}$$

to the Standard Normal distribution with distribution function as $\Phi.$ The critical region for testing $H_0$ against $H_A$ for a test with confidence $1-\alpha$ therefore is

$$\mathcal{C}(\alpha) = (-\infty, \Phi^{-1}(\alpha/2)]\ \cup\ [\Phi^{-1}(1-\alpha/2),\infty).$$


Unless you're pretty sure what the common sign of the $\theta_i$ is, $\mathcal{C}(\alpha)$ would be the critical region for $H_A^\prime,$ too.

When both the coefficient estimates $\hat\theta_i$ indicate they are significantly different from zero with the same sign, for testing $H_A^\prime$ you could instead replace $\mathcal C(\alpha)$ by the one-sided region

$$\mathcal C^\prime(\alpha) = (-\infty, \Phi^{-1}(\alpha)]$$

(for a common negative estimate) or its negative (for a common positive estimate), thereby achieving greater power. Although the confidence of this test would be slightly less than $1-\alpha,$ the difference shouldn't matter. A quick simulation study adapted to data like yours would remove any doubts.

When one (or both) of the datasets is "small" (around $20$ or less, approximately), consider a Welch t-test instead of a Z test.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.