Maximum Likelihood, normal distribution

Question

Let $Z$~$N(\mu,\sigma^2+1)$, find the maximum likelihood estimator for $\mu$ and $\sigma^2$.

I did but I want to check that this right actually

$f_z(z)=\frac{1}{\sqrt{2\pi}\sqrt{\sigma^2+1}}e^{-\frac{1}{2\sigma^2+1}(z-\mu)^2}=(2\pi)^{-\frac{1}{2}}(\sigma^2+1)^{-\frac{1}{2}}e^{-\frac{1}{2\sigma^2+1}(z-\mu)^2}$

$L=(2\pi)^{-\frac{n}{2}}(\sigma^2+1)^{-\frac{n}{2}}e^{-\frac{1}{2\sigma^2+1}\sum(z-\mu)^2}$

$logL=-\frac{n}{2}log(2\pi)-\frac{n}{2}log(\sigma^2+1)-\frac{1}{2(\sigma^2+1)}\sum (z-\mu)^2$

$\frac{dlogL}{d\mu}=\sum (z-\mu)=0\Leftrightarrow \sum z-n\mu=0\Leftrightarrow \hat{\mu}=\overline{Z}$

Similarly

$$\frac{dLogL}{d\sigma^2}=-\frac{n}{2(\sigma^2+1)}+\frac{1}{2(\sigma^2+1)^2}\sum (z-\mu)^2=-n+\frac{1}{(\sigma^2+1)}\sum (z-\mu)^2=0$$ $$\Leftrightarrow \sum(z-\mu)^2=n\sigma^2+n\Leftrightarrow \hat{\sigma}^2=\frac{\sum(z-\mu)^2}{n}-1$$

I want to take and ask for help in this matter on conditional distribution Conditional

Answer 1

Your estimator for $\sigma^2$ should use the estimator for $\mu$, not the unknown parameter itself. You may also wish to use subindices in things over which you are summing.

Other than that, yes. You could have also simplified things by using that $\sum_i(z_i-\bar z)^2/n$ is well-known to estimate the variance of the normal distribution, here $\sigma^2+1$. By the invariance properties of MLEs, $\sum_i(z_i-\bar z)^2/n-1$ therefore is the MLE of $\sigma^2$.