I'm trying to find the OLS estimator $\hat{\beta}$ of the following model:

$$Y_t = \beta t^{(3/2)}+\epsilon_t~~~~~~~~, ~ \epsilon \sim NID(0, \sigma^2)$$

So I started in this way. Firts the sum of squares function is:

$$S(\hat{B}) = \sum_{t=1}^n (Y_t - \beta t^{(3/2)})^2 \\ = \sum_{t=1}^n (Y_t^2-2Y_t\beta t^{3/2}+\beta^2t^3)$$

Then I took the first derivative of it:

$$\dfrac{ \partial S(\hat{\beta})}{\partial \hat{\beta}}=-2\sum_{t=1}^n (Y_tt^{3/2}-\beta t^3)$$

Then I equal it to zero to find the condition of the first order:

$$-2\sum_{t=1}^n (Y_tt^{3/2}-\beta t^3) = 0$$

Reached this point I am not able to continue. Have I missing something?


1 Answer 1


When we speak of a linear model, what we are assuming is that it is linear in the parameters. This is the case for your model, and the usual OLS estimator is correct.

If you want to continue with your derivation, you can divide out by -2 and split the sum like this:

$$\sum_{t=1}^n Y_t t^{3/2} = \beta \sum_{t=1}^n t^3$$

Then divide to find that:

$$\hat{\beta} = \frac{\sum_{t=1}^n Y_t t^{3/2}}{\sum_{t=1}^n t^3}$$

If you define $X_t = t^{3/2}$, then this is just:

$$\hat{\beta} = \frac{\sum_{t=1}^n Y_t X_t}{\sum_{t=1}^n X_t^2}$$

This is the usual OLS solution when there is no intercept.

  • $\begingroup$ Many thanks... from this point how is it possible to derive the distribution of this estimator? In particular the expected value and the variance? $\endgroup$
    – Archimede
    Commented Feb 11, 2017 at 15:37
  • 1
    $\begingroup$ Yes, simply refer to the usual properties of the OLS estimator, and plug in $X_t = t^{3/2}$ as needed. $\endgroup$
    – Chris Haug
    Commented Feb 11, 2017 at 15:40
  • $\begingroup$ I need to see the passages, moreover I would like the solution without the trick of $X_t = t^{3/2}$. I'm going to open a new question answer there if you want. $\endgroup$
    – Archimede
    Commented Feb 11, 2017 at 15:44

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