# If a MVUE exists it must be MLE

I'm trying to understand this theorem: Let $\theta=(\theta_1,...\theta_p)$ be parameters for a distribution. Then if $\theta$ has a minimum variance unbiased estimator $\hat{\theta}$, then $\hat{\theta}$ must be the maximum likelihood estimator obtained by solving $U(x;\hat{\theta})=0$.

It is not clear to me why this is true, although I understand that the MLE of $\theta$ maximizes the joint density function $f(x;\theta)$.

• Source? I don't think it could be true (but I could be thinking about it wrong). Think of something like an exponential distribution parametrized by $\lambda$ (not $\theta$). Shouldn't the sample mean still be be a complete, sufficient statistic there, so $1/\bar X$ scaled to be unbiased should be the UMVUE? However the MLE is just $1/\bar X$ and is biased. – spaceisdarkgreen May 15 '18 at 3:51

This is incorrect. There is no reason for MLE to be an unbiased estimator. E.g., in normal settings $N(\mu, \sigma^2)$, the MLE of $\sigma^2$ $$\frac{1}{n}\sum(X_i - \bar{X})^2,$$ while the UMVUE is $$\frac{1}{n-1}\sum(X_i - \bar{X})^2.$$