Why do we put this term to the power of n/2 instead of n in this MLE?

Question

Let say we want the estimator of something within the normal distribution. Let's take $\sigma^2$ as an example.

We would therefore write $$L(\sigma^2) = \prod_{i=1}^n \dfrac{1}{\sqrt{2\pi \sigma^2}} e^{\frac{- (x - \mu)^2}{2\sigma^2}}$$

When trying to extract terms outside the product symbol we get this: $$\left(\frac{1}{\sqrt{2\pi}}\right)^n \left(\frac{1}{\sigma^2} \right)^{n/2} \prod_{i=1}^n e^{ \frac{-1}{2} \cdot \frac{(x - \mu)^2}{\sigma^2}}$$

My question is why when this $\dfrac{1}{\sigma^2}$ is put out of the product symbol do we make it to the power of $n/2$ instead of $n$ in contrast to what we did to $\dfrac{1}{\sqrt{2\pi}}$?

Answer 1

\begin{align} \text{wrong: } & L(\sigma^2) = \prod_{i=1}^n \dfrac{1}{\sqrt{2\pi \sigma^2}} e^{-(x - \mu)^2/(2\sigma^2)} \\[10pt] \text{right: } & L(\sigma^2) = \prod_{i=1}^n \dfrac{1}{\sqrt{2\pi \sigma^2}} e^{-(x_i - \mu)^2/(2\sigma^2)} \end{align} If nothing that changes as $i$ goes from $1$ to $n$ appears to the right of $\text{“}\prod_{i=1}^n \text{''},$ then what is that symbol there for? And $e^{-(x-\mu)/(2\sigma^2)}$ does not change as $i$ goes from $1$ to $n,$ then why not make it $\Big( e^{-(x-\mu)/(2\sigma^2)}\Big)^n\text{ ?} \phantom{\frac{\displaystyle\sum}{}}$

To address the $n/2:$

$$\frac 1 {\sqrt{\sigma^2}} = \left( \frac 1 {\sigma^2} \right)^{1/2}. $$

Answer 2

You could have written $\left(\frac{1}{\sqrt{2\pi \sigma^2}}\right)^n \prod_{i=1}^n e^{ \frac{-1}{2} \cdot \frac{(x_i - \mu)^2}{\sigma^2}}$

or $\left(\frac{1}{\sqrt{2\pi}}\right)^n \left(\frac{1}{\sqrt{\sigma^2}} \right)^{n}\prod_{i=1}^n e^{ \frac{-1}{2} \cdot \frac{(x_i - \mu)^2}{\sigma^2}}$

or $\left(\frac{1}{\sqrt{2\pi}}\right)^n \left(\frac{1}{\sigma} \right)^{n}\prod_{i=1}^n e^{ \frac{-1}{2} \cdot \frac{(x_i - \mu)^2}{\sigma^2}}$

or $\left(\frac{1}{\sqrt{2\pi }}\right)^n\left(\frac{1}{\sigma^2} \right)^{n/2}\prod_{i=1}^n e^{ \frac{-1}{2} \cdot \frac{(x_i - \mu)^2}{\sigma^2}}$

as they are all equivalent,

and in any case multiplying the likelihood by a positive constant does not affect the maximum likelihood estimate