# R^2 and r^2 in multivariate regression to exponential function

#### NonSensei

##### New Member
R^2 (model fit, correlation of Y-hat with Y) is r^2 (correlation of X with Y) in "simple linear regression."

I have two explanatory variables and an OLS model fitted to an exponential curve, y = ae^(b1x1 + b2x2), so I gather the two are different. Is it still possible to derive one from the other? Or -- if that is a relationship I really want to explore -- is it better to log-transform everything to have a linear situation?

#### Dason

##### Ambassador to the humans
It's not clear to me what things you're referring to in your questions.

#### NonSensei

##### New Member
Although R^2 is not trivially equal to r^2, is there a way to derive R^2 from, or relate R^2 to, the combination of the separate correlations of the two independent variables with the dependent?

#### Dason

##### Ambassador to the humans
How is that related to your exponential stuff you had going on in the previous post

#### NonSensei

##### New Member
Same model. But essentially an unrelated question.

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#### NonSensei

##### New Member
“Researchers and reviewers should be aware that R2 is inappropriate when used for demonstrating the performance or validity of a certain nonlinear model. It should ideally be removed from scientific literature dealing with nonlinear model fitting or at least be supplemented with other methods such as AIC or BIC or used in context to other models in question.”

#### Buckeye

##### Member
I am not sure what you are asking. But r^2 lower case or capitalize is the same measure in a linear model. It is the proportion of the variance in the response that is explained by the linear relationship with the predictor(s). Thus, r^2 would not be an appropriate measure if the relationship is not linear.

#### NonSensei

##### New Member
r^2 and R^2 are not equal if there are more than one dependent variable. For example, in the linear case, r1^2 + r2^2 = R^2. I was hoping there was an analogous formula for an exponential function, something like R^2 = e^(r1^2 + r2^2), but it looks like (1) the answer is no and (2) you shouldn't be using R^2 to evaluate non-linear models anyway.

#### Buckeye

##### Member
In this case, I have no idea how to help. And I have not heard of what you're talking about.

#### GretaGarbo

##### Human
For example, in the linear case, r1^2 + r2^2 = R^2
I dont think this is true.

Imagine r1^2 = 0.8 and r2^2 =0.8. What would R^2 be then? The equation is not possible.

#### Dason

##### Ambassador to the humans
Their previous post which is now deleted did mention the caveat that it only holds when the predictors are uncorrelated. In which case I believe it's a true statement. But that certainly wouldn't hold for the specific nonlinear case mentioned.