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This is a pretty basic question for all of you I'm sure, but I'm doing some intro classes on linear regression, where we are working some data which fits into a linear model as such:

$$W_i = 10.386 - .038*G_i$$

We're now asked to evaluate this same data using a nonlinear model:
$$ W_i = \gamma e^{\beta G_i} $$
where it is stated that we can rewrite this into linear form such that: $$ log(W_i) = \alpha + \beta G_i + \varepsilon_i $$ with with $G_i$ and $\alpha = log(\gamma)$

With this said, our previous model becomes:
$$ W_i = 2.341 - 0.0038*G_i$ $$ I've dropped the error terms for simplicity, but all of these are given as correct, and I'm having trouble figuring out why the intercept term was onverted from 10.386 to 2.341 (I know they just took the log, but why?) and why the slope term went from 0.038 to 0.0038 (no idea why).

I would be grateful for any insight/help. Thanks!

EDIT--and in the event that it is necessary to look at the data, here's what it looks like (not sure the best format for presenting this here):

Game <- c(1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15)
WinningTime <- c(10.3, 10.4, 10.5, 10.2, 10, 9.95, 10.14, 10.06, 10.25, 9.99 9.92, 9.96, 9.84, 9.87, 9.85)

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  • $\begingroup$ Which data are given? I think you cannot evaluate the exponential model from the linear model without knowing the data. $\endgroup$ – callculus Oct 24 '18 at 9:39
  • $\begingroup$ Thanks for the comment. I edited the post to add the data, which you can see is quite simple. I had the impression that the way the non-linear expression was written into linear form didn't rely so much on the data but, rather, basic algebra, but as I've mentioned, I don't understand how they did it, so clearly I'm missing something. Thanks for any help. $\endgroup$ – anguyen1210 Oct 24 '18 at 10:43
  • $\begingroup$ No, you really have to apply the linear regression with the use of the transformed data to obtain the value of the parameter of the exponential function. Feel free to ask if something is still unclear. $\endgroup$ – callculus Oct 24 '18 at 11:57
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You cannot directly convert the result of a linear regression into the result of a exponential regression. What you can do is to transform an exponential equation into an linear equation and then you can apply the linear regression.

$y_=\gamma\cdot e^{\beta \cdot G_i}$

$\ln(y_i)=\ln(\gamma)+\beta\cdot G_i$

The values for $\ln(y_i)$ are the following:


$$2,332143895\quad 2.341805806 \quad 2.351375257\quad 2.32238772\quad 2.302585093\quad 2.297572551\quad 2.316487998\quad 2.308567165\quad 2.327277706\quad 2.301584593\quad 2.294552921\quad 2.298577072\quad 2.286455711\quad 2.289499853\quad 2.287471455$$


The values for $G_i$ can be used without transformation. When I apply the RGP-function in Excel I get $\ln(\gamma)=2.340603915$ and $\beta=-0.003755949$. They can be rounded if you like.

All that is left is to transform back the ln-function.

$\ln(y_i)=2.340603915-0.003755949\cdot G_i$

$y_i=e^{2.340603915}\cdot e^{-0.003755949\cdot G_i}$

$$y_i=10.3875\cdot e^{-\large{0.003755949\cdot G_i}}\approx 10.388\cdot e^{-\large{0.0038\cdot G_i}}$$ Don´t care about differences at the positions behind the decimal point if they exist. It always depends how much decimal places you use in your calculation.

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    $\begingroup$ Thanks!!. I'm trying to do all the calculations by hand without running a simple function in Excel or R or whatever to get the coefficients, but if I understand this correctly, I need to take the log of all the dependent values I'm interested in first (in this case the WinningTime or $y_i$ values), and then use this newly transformed set to compute the $\beta$ value and then $\alpha$ coefficients as I normally would. That's a great help, thanks. $\endgroup$ – anguyen1210 Oct 24 '18 at 13:02
  • $\begingroup$ @anguyen1210 You´re welcome. You are perfectly correct. Indeed you don´t really need to use a software. But a calculator is helpful due the logarithm of the $y_i$-values. $\endgroup$ – callculus Oct 24 '18 at 19:46

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