유한 분산을 테스트 하시겠습니까?

표본이 주어진 랜덤 변수의 분산의 유한성 (또는 존재)을 테스트 할 수 있습니까? 널 (null)로서 {분산이 존재하고 유한함} 또는 {분산이 존재하지 않거나 무한함}이 허용됩니다. 철학적으로 (그리고 계산적으로), 유한 분산이없는 모집단과 매우 큰 분산이있는 모집단 (> ) 사이에 차이가 없어야하기 때문에 이것은 매우 이상하게 보입니다 . 해결되었습니다.$10^{400}$

나에게 제안 된 한 가지 접근법은 중앙 한계 정리를 통해 이루어졌습니다. 샘플이 iid이고 모집단이 유한 평균이라고 가정하면 어떻게 든 표본 평균이 증가함에 따라 표본 평균에 올바른 표준 오차가 있는지 확인할 수 있습니다. 그래도이 방법이 효과가 있다고 확신하지 않습니다. (특히, 적절한 테스트로 만드는 방법을 모르겠습니다.)

크기 의 유한 표본은 $n$예를 들어 오염 된 정규 모집단과 정규 모집단을 확실하게 구별 할 수 없기 때문에 이것은 불가능합니다.N >>$1/N$ 때 Cauchy 분포의 1 / N 양. (물론 전자는 유한 분산이고 후자는 무한 분산입니다.) 따라서 완전 비모수 적 검정은 그러한 대안에 대해 임의로 낮은 검정력을 갖습니다.$N$$n$

분포를 모르면 확신 할 수 없습니다. 그러나 그러한 당신이 크기의 샘플이있는 경우 즉, "부분적인 변화"라고 할 수 있습니다 무엇을보고 같은 당신이 할 수있는 어떤 일, 거기에 , 당신은 처음부터 추정 분산 그릴 n 개의 , 용어를 N 에 2에서 실행은 N .$N$$n$$n$$N$

유한 모집단 분산을 사용하면 부분 분산이 곧 모집단 분산에 가까워 지길 바랍니다.

모집단 분산이 무한하면 부분 분산이 급증한 후 다음으로 큰 값이 표본에 나타날 때까지 천천히 감소합니다.

이것은 Normal 및 Cauchy 랜덤 변수 (및 로그 스케일)가있는 그림입니다

분포의 모양이 충분한 신뢰도로 식별하기 위해 필요한 것보다 훨씬 더 큰 표본 크기가 필요한 경우 (예 : 유한 분산 분포의 경우 매우 큰 값은 드물지만) 무한 분산 분포의 경우 매우 드 rare니다. 주어진 분포에 대해 그 특성을 밝히지 않는 것보다 더 큰 표본 크기가있을 것입니다. 반대로, 주어진 표본 크기에 대해, 그 표본의 크기에 대해 그 특성을 위장하지 않을 가능성이 더 큰 분포가 있습니다.

Here's another answer. Suppose you could parametrize the problem, something like this:

$$H_{0}:\ X \sim t(\mathtt{df}=3)\mathrm{\ versus\ } H_{1}:\ X \sim t(\mathtt{df}=1).$$

Then you could do an ordinary Neyman-Pearson likelihood ratio test of $H_{0}$ versus $H_{1}$. Note that $H_{1}$ is Cauchy (infinite variance) and $H_{0}$ is the usual Student's $t$ with 3 degrees of freedom (finite variance) which has PDF:

$$f(x|\nu) = \frac{\Gamma\left(\frac{\nu + 1}{2}\right)}{\sqrt{\nu\pi}\Gamma\left(\frac{\nu}{2}\right)}\left(1 + \frac{x^{2}}{\nu} \right)^{-\frac{\nu + 1}{2}},$$

for $-\infty < x < \infty$. Given simple random sample data $x_{1},x_{2},\ldots,x_{n}$, the likelihood ratio test rejects $H_{0}$ when

$$\Lambda(\mathbf{x}) = \frac{\prod_{i=1}^{n}f(x_{i}|\nu = 1)}{\prod_{i=1}^{n}f(x_{i}|\nu = 3)} > k,$$ where $k \geq 0$ is chosen such that $$P(\Lambda(\mathbf{X}) > k\,|\nu = 3) = \alpha.$$

It's a little bit of algebra to simplify

$$\Lambda(\mathbf{x}) = \left(\frac{\sqrt{3}}{2}\right)^{n}\prod_{i = 1}^{n}\frac{\left(1 + x_{i}^{2}/3 \right)^{2}}{1 + x_{i}^{2}}.$$

So, again, we get a simple random sample, calculate $\Lambda(\mathbf{x})$, and reject $H_{0}$ if $\Lambda(\mathbf{x})$ is too big. How big? That's the fun part! It's going to be hard (impossible?) to get a closed form for the critical value, but we could approximate it as close as we like, for sure. Here's one way to do it, with R. Suppose $\alpha = 0.05$, and for laughs, let's say $n = 13$.

We generate a bunch of samples under $H_{0}$, calculate $\Lambda$ for each sample, and then find the 95th quantile.

set.seed(1)
x <- matrix(rt(1000000*13, df = 3), ncol = 13)
y <- apply(x, 1, function(z) prod((1 + z^2/3)^2)/prod(1 + z^2))
quantile(y, probs = 0.95)

This turns out to be (after some seconds) on my machine to be $\approx 12.8842$, which after multiplied by $(\sqrt{3}/2)^{13}$ is $k \approx 1.9859$. Surely there are other, better, ways to approximate this, but we're just playing around.

In summary, when the problem is parametrizable you can set up a hypothesis test just like you would in other problems, and it's pretty straightforward, except in this case for some tap dancing near the end. Note that we know from our theory the test above is a most powerful test of $H_{0}$ versus $H_{1}$ (at level $\alpha$), so it doesn't get any better than this (as measured by power).

Disclaimers: this is a toy example. I do not have any real-world situation in which I was curious to know whether my data came from Cauchy as opposed to Student's t with 3 df. And the original question didn't say anything about parametrized problems, it seemed to be looking for more of a nonparametric approach, which I think was addressed well by the others. The purpose of this answer is for future readers who stumble across the title of the question and are looking for the classical dusty textbook approach.

P.S. it might be fun to play a little more with the test for testing $H_{1}:\nu \leq 1$, or something else, but I haven't done that. My guess is that it'd get pretty ugly pretty fast. I also thought about testing different types of stable distributions, but again, it was just a thought.

In order to test such a vague hypothesis, you need to average out over all densities with finite variance, and all densities with infinite variance. This is likely to be impossible, you basically need to be more specific. One more specific version of this and have two hypothesis for a sample $D\equiv Y_{1},Y_{2},\dots,Y_{N}$:

1. $H_{0}:Y_{i}\sim Normal(\mu,\sigma)$
2. $H_{A}:Y_{i}\sim Cauchy(\nu,\tau)$

One hypothesis has finite variance, one has infinite variance. Just calculate the odds:

$$\frac{P(H_{0}|D,I)}{P(H_{A}|D,I)}=\frac{P(H_{0}|I)}{P(H_{A}|I)}\frac{\int P(D,\mu,\sigma|H_{0},I)d\mu d\sigma}{\int P(D,\nu,\tau|H_{A},I)d\nu d\tau}$$

Where $\frac{P(H_{0}|I)}{P(H_{A}|I)}$ is the prior odds (usually 1)

$$P(D,\mu,\sigma|H_{0},I)=P(\mu,\sigma|H_{0},I)P(D|\mu,\sigma,H_{0},I)$$ And $$P(D,\nu,\tau|H_{A},I)=P(\nu,\tau|H_{A},I)P(D|\nu,\tau,H_{A},I)$$

Now you normally wouldn't be able to use improper priors here, but because both densities are of the "location-scale" type, if you specify the standard non-informative prior with the same range $L_{1}<\mu,\tau<U_{1}$ and $L_{2}<\sigma,\tau<U_{2}$, then we get for the numerator integral:

$$\frac{\left(2\pi\right)^{-\frac{N}{2}}}{(U_1-L_1)log\left(\frac{U_2}{L_2}\right)}\int_{L_2}^{U_2}\sigma^{-(N+1)}\int_{L_1}^{U_1} exp\left(-\frac{N\left[s^{2}-(\overline{Y}-\mu)^2\right]}{2\sigma^{2}}\right)d\mu d\sigma$$

Where $s^2=N^{-1}\sum_{i=1}^{N}(Y_i-\overline{Y})^2$ and $\overline{Y}=N^{-1}\sum_{i=1}^{N}Y_i$. And for the denominator integral:

$$\frac{\pi^{-N}}{(U_1-L_1)log\left(\frac{U_2}{L_2}\right)}\int_{L_2}^{U_2}\tau^{-(N+1)}\int_{L_1}^{U_1} \prod_{i=1}^{N}\left(1+\left[\frac{Y_{i}-\nu}{\tau}\right]^{2}\right)^{-1}d\nu d\tau$$

And now taking the ratio we find that the important parts of the normalising constants cancel and we get:

$$\frac{P(D|H_{0},I)}{P(D|H_{A},I)}=\left(\frac{\pi}{2}\right)^{\frac{N}{2}}\frac{\int_{L_2}^{U_2}\sigma^{-(N+1)}\int_{L_1}^{U_1} exp\left(-\frac{N\left[s^{2}-(\overline{Y}-\mu)^2\right]}{2\sigma^{2}}\right)d\mu d\sigma}{\int_{L_2}^{U_2}\tau^{-(N+1)}\int_{L_1}^{U_1} \prod_{i=1}^{N}\left(1+\left[\frac{Y_{i}-\nu}{\tau}\right]^{2}\right)^{-1}d\nu d\tau}$$

And all integrals are still proper in the limit so we can get:

$$\frac{P(D|H_{0},I)}{P(D|H_{A},I)}=\left(\frac{2}{\pi}\right)^{-\frac{N}{2}}\frac{\int_{0}^{\infty}\sigma^{-(N+1)}\int_{-\infty}^{\infty} exp\left(-\frac{N\left[s^{2}-(\overline{Y}-\mu)^2\right]}{2\sigma^{2}}\right)d\mu d\sigma}{\int_{0}^{\infty}\tau^{-(N+1)}\int_{-\infty}^{\infty} \prod_{i=1}^{N}\left(1+\left[\frac{Y_{i}-\nu}{\tau}\right]^{2}\right)^{-1}d\nu d\tau}$$

The denominator integral cannot be analytically computed, but the numerator can, and we get for the numerator:

$$\int_{0}^{\infty}\sigma^{-(N+1)}\int_{-\infty}^{\infty} exp\left(-\frac{N\left[s^{2}-(\overline{Y}-\mu)^2\right]}{2\sigma^{2}}\right)d\mu d\sigma=\sqrt{2N\pi}\int_{0}^{\infty}\sigma^{-N} exp\left(-\frac{Ns^{2}}{2\sigma^{2}}\right)d\sigma$$

Now make change of variables $\lambda=\sigma^{-2}\implies d\sigma = -\frac{1}{2}\lambda^{-\frac{3}{2}}d\lambda$ and you get a gamma integral:

$$-\sqrt{2N\pi}\int_{\infty}^{0}\lambda^{\frac{N-1}{2}-1} exp\left(-\lambda\frac{Ns^{2}}{2}\right)d\lambda=\sqrt{2N\pi}\left(\frac{2}{Ns^{2}}\right)^{\frac{N-1}{2}}\Gamma\left(\frac{N-1}{2}\right)$$

And we get as a final analytic form for the odds for numerical work:

$$\frac{P(H_{0}|D,I)}{P(H_{A}|D,I)}=\frac{P(H_{0}|I)}{P(H_{A}|I)}\times\frac{\pi^{\frac{N+1}{2}}N^{-\frac{N}{2}}s^{-(N-1)}\Gamma\left(\frac{N-1}{2}\right)}{\int_{0}^{\infty}\tau^{-(N+1)}\int_{-\infty}^{\infty} \prod_{i=1}^{N}\left(1+\left[\frac{Y_{i}-\nu}{\tau}\right]^{2}\right)^{-1}d\nu d\tau}$$

So this can be thought of as a specific test of finite versus infinite variance. We could also do a T distribution into this framework to get another test (test the hypothesis that the degrees of freedom is greater than 2).

The counterexample is not relevant to the question asked. You want to test the null hypothesis that a sample of i.i.d. random variables is drawn from a distribution having finite variance, at a given significance level. I recommend a good reference text like "Statistical Inference" by Casella to understand the use and the limit of hypothesis testing. Regarding h.t. on finite variance, I don't have a reference handy, but the following paper addresses a similar, but stronger, version of the problem, i.e., if the distribution tails follow a power law.

POWER-LAW DISTRIBUTIONS IN EMPIRICAL DATA SIAM Review 51 (2009): 661--703.

One approach that had been suggested to me was via the Central Limit Theorem.

This is a old question, but I want to propose a way to use the CLT to test for large tails.

Let $X = \{X_1,\ldots,X_n\}$ be our sample. If the sample is a i.i.d. realization from a light tail distribution, then the CLT theorem holds. It follows that if $Y = \{Y_1,\ldots,Y_n\}$ is a bootstrap resample from $X$ then the distribution of:

$$Z = \sqrt{n}\times\frac{mean(Y) - mean(X)}{sd(Y)},$$

is also close to the N(0,1) distribution function.

Now all we have to do is perform a large number of bootstraps and compare the empirical distribution function of the observed Z's with the e.d.f. of a N(0,1). A natural way to make this comparison is the Kolmogorov–Smirnov test.

The following pictures illustrate the main idea. In both pictures each colored line is constructed from a i.i.d. realization of 1000 observations from the particular distribution, followed by a 200 bootstrap resamples of size 500 for the approximation of the Z ecdf. The black continuous line is the N(0,1) cdf.