My model is $y^.5= \beta_0 + \beta_1 * x^.5 + e$

After taking partial derivatives, I find that $\beta_1$ represents $(dy/y^.5)/(dx/x^.5)$

Now my question is how I can transform $\beta_1$ to be an elasticity. Should I multiply by $x^.5/y^.5$ to get the desired form? If so, what does that mean intuitively and how do I multiply by that factor with real data?


When $x$ has a causal effect on $y$, the elasticity is

the ratio of the percentage change in one variable to the percentage change in another variable


When $x$ is changed to $x + dx$, causing $y$ to change to $y+dy$, the percentage changes are $100 dx / x$ and $100 dy / y$, respectively. Their ratio is

$$\frac{100\, dy/y}{100\, dx/x} = \frac{dy}{dx} / \frac{y}{x}.$$

Letting $y$ stand for the fitted value at $x$, we may compute

$$y = (\beta_0 + \beta_1\sqrt{x} )^2$$


$$\frac{dy}{dx} = \frac{2 \sqrt{y} \beta_1 }{2\sqrt{x}} = \beta_1 \frac{\sqrt{y}}{\sqrt{x}}.$$

Consequently the elasticity is

$$\beta_1 \frac{\sqrt{y}}{\sqrt{x}} / \frac{y}{x} = \beta_1 \frac{\sqrt{x}}{\sqrt{y}} = \frac{\beta_1\sqrt{x}}{\beta_0 + \beta_1 \sqrt{x} }.$$

When analyzing data, you can only estimate the elasticity using estimates $b_0$ and $b_1$ instead of the true (but unknown) coefficients $\beta_0$ and $\beta_1$. Therefore the estimated elasticity at any point $x_0$ will be

$$\frac{b_1\sqrt{x_0}}{b_0 + b_1 \sqrt{x_0} }.$$

There are many ways to compute confidence intervals for those estimates. In the absence of the data--knowing only the variance-covariance matrix of the estimates--use the Delta method. When fitting using another procedure, such as Maximum Likelihood, you might be able to obtain confidence limits directly (such as by profiling the likelihood for a fixed $x_0$).

Note that the estimated elasticity is not constant unless $b_0=0$, or at least (approximately) when $|b_0| \ll |b_1 x_0|$.

As a double-check, consider the limiting case $\beta_0=0$. The original model for the fit

$$\sqrt{y} = \beta_1 \sqrt{x}, $$

can be rewritten

$$\log(y) = \log(\beta_1^2) + \log(x).$$

In a log-log model, the elasticity is the coefficient of $\log(x)$, here shown to be $1$. Indeed,

$$\frac{\beta_1 \sqrt{x}}{\beta_0 + \beta_1 \sqrt{x}} = \frac{\beta_1\sqrt{x}}{\beta_1 \sqrt{x}} = 1$$

agrees with that.

  • $\begingroup$ Wonderful response! One additional question: can you also direct me to how I can calculate the elasticity of y with respect to x (at the mean of x and y) when both x and y both contain large amounts of negative values? With a univariate regression, would it simply be $\beta_1 * xbar / ybar$ $\endgroup$ – Zslice Feb 18 '15 at 4:49
  • $\begingroup$ I cannot answer that one, because your model is inapplicable whenever $y \lt 0$ or $x \lt 0$: its square root is undefined. $\endgroup$ – whuber Feb 18 '15 at 16:01

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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