Why any normal distribution has an MLR?

Question

On Statistical Inference exercise 8.25, it reads "As $\bar{X}$ is sufficient and its distribution has an MLR (see Exercise 8.25)", I'm wondering if anyone could give a brief hint on this? Thanks!!!

Answer 1

We only need to prove
Suppose $X\textrm{~}\mathscr{N}(\mu,\sigma^2)$ for some $\mu$, $\sigma\in\mathbb{R}$. We have $f_X(x|\mu')/f_X(x|\mu)$ is monotonously increasing in $x$, where $\mu'>\mu$ and $f_X(\bullet|\mu)$ is the probability density function of $X$ given $\mu$.
In fact, since $f_X(x|\mu)=\frac{1}{\sqrt{2\pi\sigma^2}}\textrm{exp}(-\frac{(x-\mu)^2}{2\sigma^2})$, we have $$ \frac{f_X(x|\mu')}{f_X(x|\mu)}=\frac{\frac{1}{\sqrt{2\pi\sigma^2}}\textrm{exp}(-\frac{(x-\mu')^2}{2\sigma^2})}{\frac{1}{\sqrt{2\pi\sigma^2}}\textrm{exp}(-\frac{(x-\mu)^2}{2\sigma^2})} =\textrm{exp}(\frac{(x-\mu)^2-(x-\mu')^2}{2\sigma^2}) =\textrm{exp}(\frac{(\mu'-\mu)(2x-\mu-\mu')}{2\sigma^2}) $$, which is monotonously increasing because $\mu'>\mu$.