Likelihood Function of Normal Variables

Question

We know that for $i.i.d.$ random variable $Y \sim N(\mu, \sigma^2)$, $Var(Y) = \frac{1}{n}\sum_{i = 1}^n (y_i - \mu)^2$, and the likelihood function for $Y_1, Y_2, ..., Y_n$ is $$l(\mu, \sigma|Y) \propto e^\frac{\sum_{i = 1}^n(y_i - \mu)^2}{2\sigma^2}$$ I am curious if I substitute $\sum_{i = 1}^n(y_i - \mu)^2$ to $n\sigma^2$, then $l(\mu, \sigma|Y)$ will be $$l(\mu, \sigma|Y) \propto e^\frac{n}{2}$$ I know it is wried, but I am curious whether it is possible to happen.

Answer 1

You have some mistakes in your likelihood. If both $\mu$ and $\sigma$ are unknown, then $$\mathcal L(\mu, \sigma \mid \boldsymbol y) \propto \color{red}{\sigma^{-n}} e^{\color{red}{-} (2\sigma^2)^{-1} \sum_{i=1}^n (y_i - \mu)^2},$$ where the red highlighted symbols are missing from your likelihood and cannot be ignored. The factor $\sigma^{-n}$ may only be ignored if the likelihood is for $\mu$; i.e., if $\sigma$ is known.

Moreover, while one can compute the joint maximum likelihood estimates for $\mu$ and $\sigma$ as $$\hat \mu = \bar y, \quad \hat \sigma = \sqrt{\frac{1}{n} \sum_{i=1}^n (y_i - \bar y)^2},$$ you can see that substituting this back into the likelihood doesn't give you the result you expect. Note that your choice $$n \hat \sigma^2 = \sum_{i=1}^n (y_i - \mu)^2$$ is invalid as a maximum likelihood estimate of $\sigma^2$, because the parameter $\mu$ is contained on the right-hand side, whereas the correct choice as I have illustrated above is not a function of any unknown parameters.