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Transform this non linear model $y=\beta_0+\beta_1x_1^{\beta_2}+\epsilon$ to a linear model $y^*=X\beta^*+\epsilon^*$

I am attending an introductory course to linear regression and this is one of the problems that was posed to me. I can do log transforms fairly easily, but applying the log transform on this model gives me a lot of problems.

$\log{y}=\log{(\beta_0+\beta_1x_1^{\beta_2}+\epsilon)}$

and I am simply unable to continue. I need to get a linear model where the parameters are linear.

(eg) $y^*=\beta_0^*+\beta_1^*x_1+\epsilon^*$

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    $\begingroup$ If $\beta_2$ is a known constant $c$, then this problem is trivial: define $z_1 = x_1 ^c$ and regress $y$ on $z_1$ and a constant. If $\beta_2$ is a parameter you have to estimate, I don't see how this can possibly be written as a linear model. $\beta_0 + \beta_1 x_1^{\beta_2}$ is not linear in $\mathbf{\beta}$. $\endgroup$
    – Matthew Gunn
    Sep 19, 2016 at 12:53
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    $\begingroup$ If your interest is focused on "solving this particular question" that was posed to you as a result of your class, then it's clearly a self-study question. Please add the self-study tag, read the tag wiki and edit your question to show what you've tried and what steps you need help with. (Simply taking logs and giving up doesn't seem to be quite sufficient.) In response to the present answer you've clearly eliminated answers that would convey understanding of the concepts as an option. A pity; that's actually likely to be most useful. $\endgroup$
    – Glen_b
    Sep 19, 2016 at 12:58
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    $\begingroup$ One hacky trick that can sometimes be acceptable is to take the first order Taylor expansion to linearize the model. Of course, this may work poorly if the linearized model is sufficiently different from the true, non-linear one. $\endgroup$
    – Matthew Gunn
    Sep 19, 2016 at 13:01
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    $\begingroup$ No such transformation exists. $\endgroup$
    – whuber ♦
    Sep 19, 2016 at 18:43
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    $\begingroup$ If all three parameters are being estimated, you have an intrinsically nonlinear model. You might find it useful to search around on that term. $\endgroup$
    – Glen_b
    Sep 20, 2016 at 2:21

1 Answer 1

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Think about how to define the vector $X$ . For example you can choose the first coordinate to be $x_1$ or some polynomial on $x_1$ . Then think about what the second coordinate should be and how to define a vector $\beta$ that when multiplied equals the original formula.

I am not sure by what is allowed in your transformation and what is not. So there are a few options.

One thing you could do is define a vector $X = (x^c, 1) $ and a vector $\beta = (\beta_1 , \beta_0)$ and keep epsilon the same. From here it should be obvious what you should set c to be so as to make this equal to your original problem.

If you want to use dimension preserving transformations only then you could do a transformation $\hat{y} = y - \beta_0 -\epsilon $ and now you have $ \hat{y} = \beta_1 {x}_1 ^ {\beta_2} $ . From here it should be obvious what log transformation you should apply to both $\hat{y}$ and $x_1$ to get your final $y^\ast$ and $X$.

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    $\begingroup$ That doesn't help at all. You are giving me a general way to think of a transformation while I need to solve this specific question. $\endgroup$
    – A New Guy
    Sep 19, 2016 at 10:42
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    $\begingroup$ I edited my response. Hope that helps more $\endgroup$
    – user3494047
    Sep 19, 2016 at 12:01
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    $\begingroup$ In practice I assume you observe only $Y$ and $X$ -- $\beta_0$ and $\epsilon$ are unknown. In that case, your $\hat{y} \equiv y - \beta_0 - \epsilon$ would not be very helpful... $\endgroup$
    – Adrian
    Sep 19, 2016 at 12:15
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    $\begingroup$ Yes, you are correct. I assumed all of the parameters were given, since I am not used to an exponent factor being a parameter that you try to learn, but I guess I misunderstood. @ANewGuy is that correct? And if this is true, then my other suggestion is also not helpful (where c = $\beta_2 $ $\endgroup$
    – user3494047
    Sep 19, 2016 at 12:20
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    $\begingroup$ $\beta_0$ is known in my opinion. $\endgroup$
    – Metariat
    Sep 19, 2016 at 12:26

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