I am having trouble finding the maximum likelihood estimate for $\sigma ^{2}$.
Find the maximum likelihood estimates of $N(\mu, \sigma ^{2})$
My work so far:
$f(x,\mu , \sigma^2)=\frac{1}{\sqrt{2\pi \sigma ^{2}}}e^{-\frac{(x-\mu )^{2}}{2\sigma ^{2}}}$
$l(\mu , \sigma^2) = -\frac{n}{2}ln(2\pi \sigma ^{2})-\frac{1}{2\sigma ^{2}}\sum(x_{i}-\mu)^{2}$
$\frac{dl(\mu, \sigma^2)}{d\sigma}=\frac{-n\sigma ^{3}+2\sigma ^{2}\sum(x_{i}-\mu)^{2}}{2\sigma ^{5}}=0$
$\sigma ^{2}=\frac{\sigma ^{3}n}{2\sum(x_{i}-\mu)^{2}}$
However, my professor wrote $\sigma ^{2}=\frac{\sum(x_{i}-\mu)^{2}}{n}$. Where did I go wrong!?
We can expand the log term a bit further. \begin{align*} l(\mu , \sigma ) &= -\frac{n}{2}ln(2\pi \sigma ^{2})-\frac{1}{2\sigma ^{2}}\sum(x_{i}-\mu)^{2} \\ &= -\frac{n}{2}ln(2\pi) -\frac{n}{2}ln(\sigma^2) - \frac{1}{2}\sum(x_{i}-\mu)^{2}\sigma^{-2} \end{align*} Taking derivatives and setting equal to zero, we can solve for the MLE. \begin{align*} \frac{dl(\mu, \sigma)}{d\sigma} = -n\sigma^{-1} + \sum(x_{i}-\mu)^{2}\sigma^{-3} &= 0 \end{align*} \begin{align*} -n\sigma^2 + \sum(x_{i}-\mu)^{2} &= 0 \end{align*} \begin{align*} \sigma ^{2}=\frac{\sum(x_{i}-\mu)^{2}}{n} \end{align*} It looks like you tried to combine the terms in $\frac{dl(\mu, \sigma)}{d\sigma}$ and perhaps messed up some algebra but it is not necessary to combine terms in the first place.