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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.

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    $\begingroup$ It's not true that $\sigma^2 = \frac{1}{n}\sum\limits_{i=1}^{n}(y_i-\mu)^2$. If $\mu$ is known, the RHS can be used to estimate $\sigma^2$. $\endgroup$ – Minus One-Twelfth May 22 at 23:53
  • $\begingroup$ Just a comment on notation: usually the lowercase "l" ($l$ or $\ell$) is reserved for the log likelihood. $\endgroup$ – K.defaoite May 23 at 0:58
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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.

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  • $\begingroup$ Thank you so much for your correction, I was too careless. I understand the maximum likelihood for $\sigma^2$ is $\frac{1}{n}\sum^n_{i=1}(y_i-\bar{y})^2$. I am curious if it is possible for $L(\sigma|\textbf{Y}, \mu) \propto \frac{1}{\sigma^n}e^{-\frac{n}{2}}$ under $\mu$ is known and we substitute $n\sigma^2$ for $\sum^n_{i=1}(y_i-\mu)^2$. $\endgroup$ – Electone May 23 at 3:41

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