# getting elasticity after sqrt transformation

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

(Wikipedia).

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

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

and

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

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.

• 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 $\beta_1 * xbar / ybar$ – Zslice Feb 18 '15 at 4:49
• I cannot answer that one, because your model is inapplicable whenever $y \lt 0$ or $x \lt 0$: its square root is undefined. – whuber Feb 18 '15 at 16:01