Is MLE of the mean of a distribution always the sample average? From the connotation of "Maximum likelihood estimator" I am inclined to think that the maximum likelihood estimator of the mean of a distribution should equal the mean of the sample values drawn from that distribution. What else could the "maximum likelihood estimate" of the mean be? 
Also, by calculus, the least squares estimator of the mean is again equal to the mean of the sample values of a sample drawn from the distribution. So is the MLE of the mean of a distribution always equal to the least squares estimator of the mean? If it is not, can someone give a counter-example?
 A: A generic contradiction to your intuition is that the MLE is invariant by transformations, while the mean is not. In particular, in exponential families, the MLE is the empirical mean of the natural statistics, but not of other transforms of the sample. For instance, in a Normal $X\sim \mathcal N(\theta,1)$ sample, the MLE of $\theta$, mean of $X$, is $X$, but the MLE of the mean of $\exp(X)$, $\exp\{\theta+1/2\}$, is $\exp\{X+1/2\}$ and not $\exp\{X\}$.
See also the connected discussion on when is the MLE a biased estimator of the mean.
A: Following @ZhanXiong's Comment. Suppose we look at $n = 10^5$ samples of
size $n = 5$ from a Laplace (double exponential) population centered
at $10.$ That is, population mean and median are both 10.
The following simulation in R, illustrates that the sample means $\bar X = A$ and $\tilde X = H$ have $E(A) = E(H) = 10,$ so that both the sample mean and median are
unbiased estimators of the center. However, the sample means have
a larger standard deviation than the sample medians.
Thus, according to one frequently-used criterion, the sample median is a "better" estimator of the center than the sample mean.
set.seed(1112)
m = 10^5;  n = 5
x = rexp(m*n)-rexp(m*n)+10
DTA = matrix(x, nrow=m)
a = rowMeans(DTA)
mean(a);  sd(a)
[1] 9.997945          # aprx E(A) = 10 
[1] 0.6317852         # aprx SD(A) = sqrt(2/5) =  0.6325

h = apply(DTA,1,median)
mean(h);  sd(h)
[1] 9.997512          # aprx E(H) = 10
[1] 0.5910876         # SD(H) < SD(A)


par(mfrow=c(2,1))
 hist(a, prob=T, br=40, col="skyblue2", xlim=c(6,15), 
      main="Aprx Dist'n of Sample Meane")
 hist(h, prob=T, br=40, col="skyblue2", xlim=c(6,15), 
      main="Aprx Dist'n of Sample Medians")
par(mfrow=c(1,1))

A: 
maximum likelihood estimator of the mean of a distribution

I don't think I've ever seen the mean be computed via MLE.  Remember, MLE is about parameters, not moments of the distribution.
For a lot of distributions, the parameters just happen to be best estimated by the sample mean (see $\mu$ for the normal, $\lambda$ for the poisson), but this isn't always the case (see $\lambda$ for the exponential, but this depends on the parameterization).
