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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1$\begingroup$ I am not sure if I understand the question correctly. Are you asking if there are any distributions apart from the normal that are completely specified by the mean and the variance? In some sense, variance is a function of the mean as it is a measure of the dispersion around the mean but I guess this is not what you have in mind. $\endgroup$– user28Commented Nov 9, 2010 at 18:38
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$\begingroup$ you mean the sample mean $\bar{X}=\frac{1}{n}\sum_{i=1}^nX_i$ and sample variance $\frac{1}{n}\sum_{i=1}^n(X_i-\bar{X})^2$ are independant. Good question ! maybe projecting a gaussian random variable will keep independance ? $\endgroup$– robin girardCommented Nov 9, 2010 at 18:58
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4$\begingroup$ Srikant is right. If the question is asking about "sample mean and variance" then the answer is "no". If the question is about population mean and variance, then the answer is yes; David gives good examples below. $\endgroup$– user1108Commented Nov 9, 2010 at 19:15
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1$\begingroup$ Just to clarify, what I meant is this. For the normal distribution, the mean $\mu$ and the variance $\sigma^2$ fully characterizes the distribution and $\sigma^2$ is not a function of $\mu$. For many other distributions, this is not so. For example, for the binomial distribution, we have the mean $\pi$ and the variance $n\pi(1-\pi)$, so the variance is a function of the mean. Other examples are the gamma distribution with parameters $\theta$ (scale) and $\kappa$ (shape), where the mean is $\mu = \kappa \theta$ and the variance is $\kappa theta^2$, so the variance is actually $\mu \theta$. $\endgroup$– WolfgangCommented Nov 9, 2010 at 21:19
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8$\begingroup$ Please consider modifying your question, then, because the response you checked as your preferred answer does not answer the question as it stands (and the other one does). Currently you are using the word "independent" in an idiosyncratic way. Your example with Gamma shows this: one could simply reparameterize Gamma in terms of the mean (mu) and variance (sigma), because we can recover theta = sigma/mu and kappa = mu^2/sigma. In other words, functional "independence" of the parameters is usually meaningless (except for single-parameter families). $\endgroup$– whuber ♦Commented Nov 12, 2010 at 18:15
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2 Answers
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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.
Multivariate Normal
$$E(X) = \mu$$ $$Var(X) = \Sigma$$
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
Reference
answered Nov 9, 2010 at 18:52
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9$\begingroup$ In any location-scale family the mean and variance will be functionally independent in this fashion! $\endgroup$– whuber ♦Commented Nov 9, 2010 at 20:29
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1$\begingroup$ David, the double exponential is an excellent example. Thanks! I did not think of that one. The t-distribution is also a good example, but isn't E(X) = 0 and Var(X) = v/(v-2)? Or does Carlin et al. (1996) define a generalized version of the t-distribution that is shifted in its mean and scaled by sigma^2? $\endgroup$– WolfgangCommented Nov 9, 2010 at 21:30
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$\begingroup$ You are correct, the t-distribution appears to be frequently characterized with a mean = 0 and variance = 1, but the general pdf for t provided by Carlin and Louis explicitly includes both sigma and mu; the nu parameter accounts for the difference between the normal and the t. $\endgroup$ Commented Nov 9, 2010 at 21:48
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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.
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4$\begingroup$ @onestop I guess it is an unfortunate artifact of my age. It is not an understatement to say that Feller's books revolutionized how probability was done - worldwide. A large part of our modern notation is due to him. For decades, his books were the probability books to study. Maybe they still should be. BTW: I've added the title for those that haven't heard of his books. $\endgroup$– user1108Commented Nov 9, 2010 at 19:39
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1$\begingroup$ I have aske the question about other funy characterisation ... stats.stackexchange.com/questions/4364/… $\endgroup$ Commented Nov 9, 2010 at 20:20
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2$\begingroup$ Jay, thanks for the reference to the paper by Lukacs, who nicely shows that the sampling distributions of the sample mean and variance are only independent for the normal distribution. As for second central moment, there are some distributions where it is not a function of the first moment (David gave some nice examples). $\endgroup$– WolfgangCommented Nov 9, 2010 at 21:42
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1$\begingroup$ Geary, R. C. (1936), “The Distribution of ‘Student’s’ Ratio for Non-Normal Samples,” Journal of the Royal Statistical Society, Suppl. 3, 178–184. $\endgroup$– vqvCommented Dec 20, 2010 at 9:11