Double Exponential

[Art Vandelay]

I take it you meant "Plot it out and pick the B minimizing r^2 as your estimate of b, and the corresponding estimate of a as your estimate of a"?

There certainly is a least squares solution, it just isn't linear.

 

[chipotle]

I take it you meant "Plot it out and pick the B minimizing r^2 as your estimate of b, and the corresponding estimate of a as your estimate of a"?

There certainly is a least squares solution, it just isn't linear.

Error minimizing will be r^2 maximizing. Once the b is set as a constant the equation is linearizable and you can use linear methods.

 

[Art Vandelay]

Hmm, there seems to be some ambiguity, as "r" can stand for "regression" or "residual". If you mean "regression", I think that you should use the term "SSR". So, now, how is the equation linear with a constant b?

 

[kmortis]

Well sure, in reality. But as we all know reality has an annoying habit of not conforming to theory.

"In theory, reality and theory are the same. In reality, they ain't." - Yogi Berra.

 

[chipotle]

Hmm, there seems to be some ambiguity, as "r" can stand for "regression" or "residual". If you mean "regression", I think that you should use the term "SSR". So, now, how is the equation linear with a constant b?

In regression, r^2 is 1-SSR/SSE. The equation isn't linear, it's linearizable. I am going to use some simpler examples since I'm not 100% sure of the notation. (V/m is, what, the units?)

Consider Y = k * exp(aX). That's not linear. But set Y2 = log(Y) = log(k) + aX. Now that can be solved with regression.

For something a little harder, try
Y = k * exp(aX) + b

What you need to do is set b = B. Now set Y1 = Y - B. Then you have Y1 = k * exp(aX) which can be solved like above to get the best a for b = B. search through the space of candidate values for B to get the error minimizing one.