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A model $Y=(\beta_0+\beta_1x)^{-1}+\epsilon$, where $\epsilon \sim N(0,\sigma^2)$ is to be fitted to the data $(x_1,Y_1), (x_2,Y_2), \dots (x_n,Y_n)$.

Is this a linear or non linear model, and why?

Show how it can be made into a GLM and state the link function?

Any help would be great!

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  • $\begingroup$ It is linear in the parameters, So it is a linear model. However it is undefined at x=0. $\endgroup$ Commented Aug 7, 2017 at 0:57
  • $\begingroup$ Please add the actual question to the body of post. $\endgroup$
    – Firebug
    Commented Aug 7, 2017 at 11:45
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    $\begingroup$ @Michael Please explain what meaning you ascribe to "linear." As far as I can tell, this model is not linear in any of the parameters or its variable $x$ in the standard mathematical or statistical meaning of the term. And why, exactly, would it be undefined at $x=0$? Except when $\beta_0$ also is zero, one can readily make sense of and compute $(\beta_0+\beta_1(0))^{-1}$ Luvspi, have you seen stats.stackexchange.com/questions/148638? It explains the meaning of "linear" regression model. $\endgroup$
    – whuber
    Commented Aug 7, 2017 at 13:47
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    $\begingroup$ I thought the parentheses were missing in the original post. So only x was raised to the -1 power. $\endgroup$ Commented Aug 7, 2017 at 16:44
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    $\begingroup$ @Digio No need for that: whatever you might think of the outcome, your analysis was interesting. $\endgroup$
    – whuber
    Commented Aug 7, 2017 at 22:41

1 Answer 1

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This is just a somewhat unusual generalised linear model with a Gaussian response $Y\sim N(\mu,\sigma^2)$ where the mean of the response $\mu$ is linked to the linear predictor $\eta$ via $$ g(\mu)=\underbrace{\beta_0 + \beta_1 x}_\eta. $$ where $g$ is the inverse link function $g(\mu)=1/\mu$,

The linear predictor $\eta$ is clearly linear in $x$ and, as implied by its name, in the regression coefficients $\beta_0$ and $\beta_1$. However, because of the non-linear link function, $\mu=EY$ is non-linear in the model parameters and in $x$.

To fit this in R do glm(y ~ x, gaussian(link = "inverse")).

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  • $\begingroup$ Much appreciated. So the model is in fact non-linear that can be put into the GLM form by using the inverse link function? $\endgroup$
    – Luvpsi
    Commented Aug 9, 2017 at 11:43
  • $\begingroup$ @Luvpsi Yes, absolutely. $\endgroup$ Commented Aug 9, 2017 at 11:47
  • $\begingroup$ Thanks again. Would it require much work to get it into GLM? $\endgroup$
    – Luvpsi
    Commented Aug 9, 2017 at 13:05
  • $\begingroup$ No, you just have to define to vectors x and y containing the data and then run the above command. $\endgroup$ Commented Aug 9, 2017 at 13:32
  • $\begingroup$ in the question given I am not given the data to use with R, but rather asked what I would need to do to get it into GLM? $\endgroup$
    – Luvpsi
    Commented Aug 10, 2017 at 22:34

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