非均值和方差独立的正态分布

我想知道平均值和方差彼此独立（即，方差不是均值的函数）时，除正态以外是否还有其他分布。

distributions

— 沃尔夫冈
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我不确定我是否正确理解了这个问题。您是否要问，除正态分布外，是否存在均由均值和方差完全指定的分布？从某种意义上说，方差是均值的函数，因为它是均值周围散布的量度，但是我想这并不是您要考虑的。

您的意思是样本均值和样本方差是独立的。好问题！也许投影一个高斯随机变量将保持独立性？

\bar{X} = \frac{1}{n} \sum_{i = 1}^{n} X_{i}

$\bar{X}=\frac{1}{n}\sum_{i=1}^nX_i$

\frac{1}{n} \sum_{i = 1}^{n} (X_{i} - \bar{X})^{2}

$\frac{1}{n}\sum_{i=1}^n(X_i-\bar{X})^2$

— 罗宾吉拉德

Srikant是正确的。如果问题是关于“样本均值和方差”的，则答案为“否”。如果问题是关于总体均值和方差，那么答案是肯定的；否则，答案是肯定的。大卫在下面给出了很好的例子。

为了澄清，我的意思是这个。对于正态分布，均值和方差完全表征了分布，并且不是的函数。对于许多其他发行版，情况并非如此。例如，对于二项式分布，我们具有均值和方差，因此方差是均值的函数。其他示例是具有参数（比例）和（形状）的伽玛分布，其中均值为而方差为，因此方差实际上是

μ

$\mu$

σ^{2}

$\sigma^2$

σ^{2}

$\sigma^2$

μ

$\mu$

π

$\pi$

n π (1 - π)

$n\pi(1-\pi)$

θ

$\theta$

κ

$\kappa$

μ = κ θ

$\mu = \kappa \theta$

κ t h e t a^{2}

$\kappa theta^2$

μ θ

$\mu \theta$ 。

— 沃尔夫冈

请考虑修改你的问题，那么，因为你检查作为首选答案的响应并没有回答这个问题，因为它代表（和其他人做）。当前，您以特殊的方式使用“独立”一词。您的Gamma示例显示了这一点：因为我们可以恢复theta = sigma / mu和kappa = mu ^ 2 / sigma，所以可以简单地用均值（mu）和方差（sigma）重新配置Gamma。换句话说，参数的功能 “独立性”通常是没有意义的（单参数族除外）。

— ub

Answers:

注意：请阅读@G的答案。Jay Kerns，请参见Carlin和Lewis 1996或您最喜欢的概率参考，作为背景的均值和方差计算的期望值和随机变量的第二矩。

对Carlin和Lewis（1996）中附录A的快速浏览提供了以下分布，在这方面与正态相似，因为在均值和方差的计算中未使用相同的分布参数。正如@robin所指出的那样，当从样本中计算参数估计时，需要样本均值来计算sigma。

多元正态

E (X) = μ

$E(X) = \mu$

V a r (X) = Σ

$Var(X) = \Sigma$

t and multivariate t:

E (X) = μ

$E(X) = \mu$

V a r (X) = ν σ^{2} / (ν - 2)

$Var(X) = \nu\sigma^2/(\nu - 2)$

Double exponential:

E (X) = μ

$E(X) = \mu$

V a r (X) = 2 σ^{2}

$Var(X) = 2\sigma^2$

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

Carlin, Bradley P., and Thomas A. Louis. 1996. Bayes and Empirical bayes Methods for Data Analysis, 2nd ed. Chapman and Hall/CRC, New York

— David LeBauer
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In any location-scale family the mean and variance will be functionally independent in this fashion!

— whuber

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?

— Wolfgang

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.

— David LeBauer

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.

@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.

I have aske the question about other funy characterisation ... stats.stackexchange.com/questions/4364/…

— robin girard

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).

— Wolfgang

Geary, R. C. (1936), “The Distribution of ‘Student’s’ Ratio for Non-Normal Samples,” Journal of the Royal Statistical Society, Suppl. 3, 178–184.

— vqv