Standard Deviation Likelihood

Question

Suppose $Y_i$ i.i.d $N(0,\sigma^2)$, i = 1, ..., n. Derive the likelihood for $\sigma$.

$L(\mu, \sigma^2) = \Pi_{i=1}^n f(X_i | \mu, \sigma^2) $
$= \Pi_{i=1}^n \frac{1}{\sqrt{2\pi\sigma^2}}e^{-(X_i-\mu)^2/2\sigma^2}$
However, I am lost how to get rid of the $\mu$ term and just derive $\sigma$. Any help?

Answer 1

In your setting, $\mu=0$ since the mean of each $Y_i = 0$

$$L(\sigma^2) = \prod_{i=1}^n \frac{1}{\sqrt{2\pi\sigma^2}}e^{-(X_i)^2/2\sigma^2}$$

Edit:

$$L(\sigma^2) = \frac{1}{(\sqrt{2\pi\sigma^2})^n}e^{-\sum_{i=1}^n(X_i)^2/2\sigma^2}$$