Maximum likelihood estimator for model without the predictor variable

We have the following simple linear model without the predictor variable: $$y_t = \beta_t + \epsilon_t \quad \text{ where } t = 1, 2, \ldots,n; \ \epsilon_t \sim N(0, \sigma^2)$$

What is the maximum likelihood estimator for $$\beta_t$$?

Proceeding with the log-likelihood function and removing the constant term and the term with $$\sigma^2$$, we find that the problem reduces to the following minimization problem:

$$\min_{\beta_t}\sum_{t=1}^{n}(y_t - \beta_t)^2$$

The maximum likelihood estimator for $$\beta$$ is therefore the corresponding $$y$$. Is the interpretation correct?

• Is $\sigma^2$ known? – StubbornAtom Apr 10 '20 at 14:42
• I think I have followed, and described, an MLE method, but agree that the final result for a single $\beta_{MLE}$ in our model, if single $\beta$ were the case, would be the same as the least square solution of single $\beta$. But it is the MLE interpretation about different $\beta$s that I am not fully certain yet about. – Mathophile Apr 10 '20 at 14:46
• Yes, $\sigma^2$ is know. Say, it equals 1. – Mathophile Apr 10 '20 at 14:48

If you're estimating a separate $$\beta_t$$ parameter for each data point, the maximum likelihood solution is to set $$\epsilon = 0$$, and so $$\sigma = 0$$. This model can perfectly fit your data, but is useless, since it has one parameter per data point.
A more useful model is $$y_t = \beta + \epsilon_t$$, with only one $$\beta$$ parameter. The MLE of $$\beta$$ is just the mean of the $$y$$ values. The MLE of $$\sigma^2$$ isn't exactly the same as the sample variance, but it's close.
• Thanks. I agree it's more interesting to have only one $\beta$ in the model for practical purpose. However, on a pure theoretical pursuit, if we have $\sigma \ne 0$ and hence $\epsilon = 0$ does not hold true, what can we say about the MLE of $\beta$? – Mathophile Apr 10 '20 at 14:18
• $\sigma = 0, \beta_t = y_t$ is the MLE, and has a likelihood of 1, since all values of $y$ are exactly where the model predicts them to be. If you restrict your analysis to, for instance, $\sigma \geq .1$, then $\sigma = .1, \beta_t = y_t$ is the MLE. – Eoin Apr 14 '20 at 10:12