OLS estimator of a non-linear trend regression

Question

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?

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.