MLE for mean of symmetric but otherwise unknown distribution Given i.i.d. draws $x_1,...,x_n$ from $X$, where:


*

*$X$ has a finite mean $E[X]=\mu < \infty$,

*$X$ is symmetric about its mean, meaning $f_X(\mu+c)=f_X(\mu-c)$ for all $c$,

*The probability density function $f_X$ is not otherwise known. 


Is it possible to prove the following?
Proposition. The MLE for the mean of $X$ is the sample mean, $\hat \mu_{MLE}=\bar x = \sum_{i=1}^n x_i$.
A proof or a counterexample would be great. I am willing to additionally assume that $X$ has a finite variance $Var[X]=\sigma^2 < \infty$, or other common basic assumptions, if that becomes necessary for the proposition to hold, or if it greatly simplifies the proof.
I suspect that it may be possible to use the invariance of the MLE to transformations of the data to prove this, but it might follow from simpler facts about the sample mean.
 A: Consider the single-parameter Exponential Family of distributions, i.e. distributions whose probability density (or mass) function can be written as
$$f(x) = h(x)\cdot \exp{\big\{\eta(\theta)T(x)-A(\theta)\big\}}$$
The log-likelihood from an i.i.d sample of size $n$ is then
$$\tilde L = \sum_{i=1}^n\ln h(x_i) + \eta(\theta)\sum_{i=1}^nT(x_i) - nA(\theta)$$
and the derivative with respect to $\theta$ is 
$$\frac {\partial \tilde L}{\partial \theta}=\eta'(\theta)\sum_{i=1}^nT(x_i)-nA'(\theta) = 0$$
$$\Rightarrow  \frac 1n \sum_{i=1}^nT(x_i) = \frac {A'(\hat \theta_{MLE})}{\eta'(\hat \theta_{MLE})}$$
it is obvious from the above that, to arrive at "the sample mean is the MLE for the mean", the involved functions must have suitable forms.  
Examples where the result holds
1) For the Normal distribution (with known variance $\sigma^2$) : $T(x_i) = x_i/\sigma$, $A(\theta)=\mu^2 / 2\sigma^2 \Rightarrow A'(\theta) = \mu / \sigma^2$, $\eta(\theta) = \mu/\sigma\Rightarrow \eta'(\theta) = 1/\sigma$ 
2) For the Bernoulli(p) distribution,  $T(x_i) = x_i$, $A(\theta) -\ln (1-p)\Rightarrow  A'(\theta) = 1/(1-p) $, $\eta (\theta) = \ln(p/(1-p)\Rightarrow \eta'(\theta) = 1/p(1-p)$
In these cases, indeed the MLE for the mean is the sample mean.  It is perhaps easier to find counter-examples, as Whuber hinted.
