I was wondering if there are any distributions besides the normal where the mean and variance are independent of each other (or in other words, where the variance is not a function of the mean).
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Note: Please read answer by @G. Jay Kerns, and see Carlin and Lewis 1996 or your favorite probability reference for background on the calculation of mean and variance as the expectated value and second moment of a random variable.
A quick scan of Appendix A in Carlin and Lewis (1996) provides the following distributions which are similar in this regard to the normal, in that the same distribution parameters are not used in the calculations of the mean and variance. As pointed out by @robin, when calculating parameter estimates from a sample, the sample mean is required to calculate sigma.
$$E(X) = \mu$$ $$Var(X) = \Sigma$$
t and multivariate t:
$$E(X) = \mu$$ $$Var(X) = \nu\sigma^2/(\nu - 2)$$
Double exponential: $$E(X) = \mu$$ $$Var(X) = 2\sigma^2$$
Cauchy: With some qualification it could be argued that the mean and variance of the Cauchy are not dependent.
$E(X)$ and $Var(X)$ do not exist
In fact, the answer is "no". Independence of the sample mean and variance characterizes the normal distribution. This was shown by Eugene Lukacs in "A Characterization of the Normal Distribution", The Annals of Mathematical Statistics, Vol. 13, No. 1 (Mar., 1942), pp. 91-93.
I didn't know this, but Feller, "Introduction to Probability Theory and Its Applications, Volume II" (1966, pg 86) says that R.C. Geary proved this, too.