Regression estimator where exponents are freely varying?

Is there a regression estimation methodology that can estimate the following:

$$Y_t = \alpha + \beta X_t^x + \gamma Z_t^z + \epsilon_t$$

where $x,z\in \mathbb{R}$, are freely varying and are chosen by the estimator according to some statistical criteria.

If such an estimator exists why have I never seen it in the financial econometrics literature?

• I think power-law functions may be of interest. But also realize that if you define an objective function to optimize (some measure of error), then to get estimates you can use a numerical optimizer. – assumednormal Dec 1 '12 at 14:36
• You are probably looking for fractional polynomials. – tchakravarty Dec 1 '12 at 14:47
• @fg ... or Box-Cox transformations, which I am confident have been mentioned in the financial literature, because they are so fundamental and useful. (These "freely varying" exponents are mathematically equivalent to separate Box-Cox transformations of the independent variables.) – whuber Dec 3 '12 at 22:54
• Taking powers usually makes little sense and rarely corresponds with anything meaningful or interpretable unless all values are non-negative. Of course values can be forced to be non-negative by subtracting some constant smaller than the minimum, and on occasion this approach will work. – whuber Dec 4 '12 at 6:32
• In my experience, Jase, squaring a variable is rare (although it happens). Far more often I see square and cube roots, logarithms, reciprocals, and reciprocal roots, because these are interpretable. For instance, the cube root of a volume represents a length. In general, physical law--which includes laws derived for the physical and dynamical systems studied in chemistry and biology--tends to involve relationships which become linear when one of these simpler transformations is applied. It is the rare physical law indeed that posits a 1.95 power relationship! – whuber Dec 5 '12 at 15:06

You will need to use non-linear optimization to solve this problem. Excel's solver should be able to find the parameters easily. Simply set the objective to minimize the sum of the squared residuals between the actual values you observed and the estimated values from your model. He is an example.

http://www.csupomona.edu/~seskandari/documents/Curve_Fitting_William_Lee.pdf

If you are looking for a programmatic way to do this, consider using NLOPT.

http://ab-initio.mit.edu/wiki/index.php/NLopt

On a side note, taking the natural log of the your dependent variable and doing least squares regression via the normal equations will create an exponential model. If we ignore the error term, then...

$$\ln(y) = b_0 + b_1x_1 + b_2x_2$$

will yield

$$y = e^{(b_0 + b_1x_1 + b_2x_2)}$$

which yields

$$y = e^{(b_0)}e^{(b_1x_1)}e^{(b_2x_2)}$$

This information might be helpful in future endeavors. Let me know if need anymore help.

• If we're only minimizing SSR, what's stopping the exponent estimates from assuming absurdly large/small values (and the slope coefficient estimates getting smaller/larger to account for this) that aren't sensible economically? – Jase Dec 4 '12 at 2:33
• More to the point, it is not the case that the logarithm of the model $y = \exp(b_0+b_1 x_1+b_2 x_2) +\varepsilon$ can be written as $\log(y) = b_0+b_1 x_1 + b_2 x_2 + \eta$. The problem, when we are explicit about the error structure (as reflected by the random variables $\varepsilon$ and $\eta$), is apparent: the logarithm of a sum is not the sum of the logarithms. The two models are different. – whuber Dec 4 '12 at 6:36