Likelihood function of linear model with variance equal to zero

Question

Let $Y_{j}=\beta$$x_{i}+\epsilon_{i}$ where $ \epsilon_{i} \sim N(0,T)$
IF we set variance T =0 then our model would be this $Y_{i}=\beta$$x_{i}$ which implies that it would fit data perfectly. I would like to know what is likelihood function of when T=0 , I struggle to find the likelihood function as if I were to use pdf of normal distribution its impossible as denominator would be zero

Answer 1

In general the likelihood for a normal distribution is given by $$ L(\beta; y_1, \dots, y_n) = \prod_{i=1}^n \mathcal{N}(y_i; \beta x_i, T) = \prod_{i=1}^N \frac{1}{\sqrt{2\pi T}} \exp \left[-\frac{(y_i - \beta x_i)^2}{T}\right]. $$ Now we might be temped to let $T \rightarrow \infty$ and observe that then $L(\beta; y_1, \dots, y_n) \rightarrow 0$, but before we do this we should rather ask ourself:

Does $T = 0$ really make sense?

No it doesn't and here's why: $T = 0$ implies that $\epsilon_i = 0$ for all $i$ which further means that $Y_i = \beta x_i$ for all $i$. In this case we may directly deduce $\beta$ as $$ \beta = (Y_2 - Y_1) / (x_2 - x_1), $$ so there really isn't any need for formulating the likelihood at all; The probability of observing $Y_2 = \beta x_2$ and $Y_1 = \beta x_1$ is 100%. There is zero probability of observing anything else.

And while you can set up the formula for the likelihood, I don't think you should. Likelihood is defined only in the context of something probabilistic having happened. But for $T = 0$, all the probabilistic contributions are zero $\epsilon_i = 0$ and thus our measurements were not the result of a probabilistic process. The likelihood isn't defined in this deterministic context.