गैर-सामान्य नमूने के नमूना विचरण का असममित वितरण

यह इस प्रश्न द्वारा उत्पन्न समस्या का अधिक सामान्य उपचार है । नमूना विचरण के स्पर्शोन्मुख वितरण को प्राप्त करने के बाद, हम मानक विचलन के लिए इसी वितरण पर आने के लिए डेल्टा विधि को लागू कर सकते हैं।

आइडल गैर-सामान्य यादृच्छिक चर के आकार $n$ का एक नमूना दें $\{X_i\},\;\; i=1,...,n$ , इनकी औसत $\mu$ और विचरण $\sigma^2$ । सेट नमूना माध्य और के रूप में नमूना प्रसरण

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

$\bar x = \frac 1n \sum_{i=1}^nX_i,\;\;\; s^2 = \frac 1{n-1} \sum_{i=1}^n(X_i-\bar x)^2$

हम जानते हैं कि

E (s^{2}) = σ^{2}, Var (s^{2}) = \frac{1}{n} (μ_{4} - \frac{n - 3}{n - 1} σ^{4})

$E(s^2) = \sigma^2, \;\;\; \operatorname {Var}(s^2) = \frac{1}{n} \left(\mu_4 - \frac{n-3}{n-1}\sigma^4\right)$

जहां $\mu_4 = E(X_i -\mu)^4$ , और हम अपने ध्यान को उन वितरणों तक सीमित रखते हैं जिनके लिए कुछ क्षणों का अस्तित्व और परिमित होना आवश्यक है, मौजूद हैं और परिमित हैं।

क्या यह पकड़ है कि

\sqrt{n} (s^{2} - σ^{2}) \to_{d} N (0, μ_{4} - σ^{4}) ?

$\sqrt n(s^2 - \sigma^2) \rightarrow_d N\left(0,\mu_4 - \sigma^4\right)\;\; ?$

— एलेकोस पापाडोपोलोस
स्रोत

हे। मैंने सिर्फ दूसरे धागे पर पोस्ट किया है, यह महसूस नहीं किया कि आप इसे पोस्ट करेंगे। सीएलटी पर विचरण के लिए लागू की जाने वाली कई चीजें हैं (उदाहरण के लिए यहाँ p3-4 )। अच्छा जवाब btw।

— Glen_b -Reinstate Monica

धन्यवाद। हां मुझे यह मिल गया है। लेकिन वे मामला याद करते हैं @whuber ने बताया। वे भी सामान्य

साथ एक बर्नौली उदाहरण प्रदान करते हैं

p

$p$ ! (पी। 4 का आधार)। मैं कवर करने के लिए मेरा उत्तर विस्तार कर रहा हूँ

p = 1 / 2

$p=1/2$ मामले में भी।

— एलेकोस पापाडोपोलोस

हां, मैंने देखा कि उन्होंने बर्नौली पर विचार किया था लेकिन अभी तक उस विशेष मामले पर विचार नहीं किया है। मुझे लगता है कि स्केल किए गए बर्नौली (समान संभावना। द्विअर्थी मामले) के लिए अंतर का उल्लेख एक कारण है (दूसरों के एक जोड़े के बीच) क्यों यह यहां उत्तर में चर्चा करने के लिए मूल्यवान है (केवल एक टिप्पणी के बजाय) - कम से कम ऐसा नहीं है यह खोजने योग्य है।

— Glen_b -Reinstate Monica

जवाबों:

जब हम नमूने के विचरण पर विचार करते हैं, तो हम इसके लिए निर्भर करते हैं

(n - 1) s^{2} = \sum_{i = 1}^{n} ((X_{i} - μ) - (\bar{x} - μ))^{2}

$(n-1)s^2 = \sum_{i=1}^n\Big((X_i-\mu) -(\bar x-\mu)\Big)^2$

= \sum_{i = 1}^{n} (X_{i} - μ)^{2} - 2 \sum_{i = 1}^{n} ((X_{i} - μ) (\bar{x} - μ)) + \sum_{i = 1}^{n} (\bar{x} - μ)^{2}

$=\sum_{i=1}^n\Big(X_i-\mu\Big)^2-2\sum_{i=1}^n\Big((X_i-\mu)(\bar x-\mu)\Big)+\sum_{i=1}^n\Big(\bar x-\mu\Big)^2$

और थोड़ा हेरफेर करने के बाद,

= \sum_{i = 1}^{n} (X_{i} - μ)^{2} - n (\bar{x} - μ)^{2}

$=\sum_{i=1}^n\Big(X_i-\mu\Big)^2 - n\Big(\bar x-\mu\Big)^2$

इसलिये

\sqrt{n} (s^{2} - σ^{2}) = \frac{\sqrt{n}}{n - 1} \sum_{i = 1}^{n} (X_{i} - μ)^{2} - \sqrt{n} σ^{2} - \frac{\sqrt{n}}{n - 1} n (\bar{x} - μ)^{2}

$\sqrt n(s^2 - \sigma^2) = \frac {\sqrt n}{n-1}\sum_{i=1}^n\Big(X_i-\mu\Big)^2 -\sqrt n \sigma^2- \frac {\sqrt n}{n-1}n\Big(\bar x-\mu\Big)^2$

जोड़ तोड़,

\sqrt{n} (s^{2} - σ^{2}) = \frac{\sqrt{n}}{n - 1} \sum_{i = 1}^{n} (X_{i} - μ)^{2} - \sqrt{n} \frac{n - 1}{n - 1} σ^{2} - \frac{n}{n - 1} \sqrt{n} (\bar{x} - μ)^{2}

$\sqrt n(s^2 - \sigma^2) = \frac {\sqrt n}{n-1}\sum_{i=1}^n\Big(X_i-\mu\Big)^2 -\sqrt n \frac {n-1}{n-1}\sigma^2- \frac {n}{n-1}\sqrt n\Big(\bar x-\mu\Big)^2$

= \frac{n \sqrt{n}}{n - 1} \frac{1}{n} \sum_{i = 1}^{n} (X_{i} - μ)^{2} - \sqrt{n} \frac{n - 1}{n - 1} σ^{2} - \frac{n}{n - 1} \sqrt{n} (\bar{x} - μ)^{2}

$=\frac {n\sqrt n}{n-1}\frac 1n\sum_{i=1}^n\Big(X_i-\mu\Big)^2 -\sqrt n \frac {n-1}{n-1}\sigma^2- \frac {n}{n-1}\sqrt n\Big(\bar x-\mu\Big)^2$

= \frac{n}{n - 1} [\sqrt{n} (\frac{1}{n} \sum_{i = 1}^{n} (X_{i} - μ)^{2} - σ^{2})] + \frac{\sqrt{n}}{n - 1} σ^{2} - \frac{n}{n - 1} \sqrt{n} (\bar{x} - μ)^{2}

$=\frac {n}{n-1}\left[\sqrt n\left(\frac 1n\sum_{i=1}^n\Big(X_i-\mu\Big)^2 -\sigma^2\right)\right] + \frac {\sqrt n}{n-1}\sigma^2 -\frac {n}{n-1}\sqrt n\Big(\bar x-\mu\Big)^2$

The term $n/(n-1)$ becomes unity asymptotically. The term $\frac {\sqrt n}{n-1}\sigma^2$ is determinsitic and goes to zero as $n \rightarrow \infty$ .

We also have $\sqrt n\Big(\bar x-\mu\Big)^2 = \left[\sqrt n\Big(\bar x-\mu\Big)\right]\cdot \Big(\bar x-\mu\Big)$

\sqrt{n} (\bar{x} - μ)^{2} \overset{p}{\to} 0

$\sqrt n\Big(\bar x-\mu\Big)^2\xrightarrow{p} 0$

हम शब्द के साथ रह गए हैं

[\sqrt{n} (\frac{1}{n} \sum_{i = 1}^{n} (X_{i} - μ)^{2} - σ^{2})]

$\left[\sqrt n\left(\frac 1n\sum_{i=1}^n\Big(X_i-\mu\Big)^2 -\sigma^2\right)\right]$

Alerted by a lethal example offered by @whuber in a comment to this answer, we want to make certain that $(X_i-\mu)^2$ is not constant. Whuber pointed out that if $X_i$ is a Bernoulli $(1/2)$ then this quantity is a constant. So excluding variables for which this happens (perhaps other dichotomous, not just $0/1$ binary?), for the rest we have

E (X_{i} - μ)^{2} = σ^{2}, Var [(X_{i} - μ)^{2}] = μ_{4} - σ^{4}

$\mathrm{E}\Big(X_i-\mu\Big)^2 = \sigma^2,\;\; \operatorname {Var}\left[\Big(X_i-\mu\Big)^2\right] = \mu_4 - \sigma^4$

and so the term under investigation is a usual subject matter of the classical Central Limit Theorem, and

\sqrt{n} (s^{2} - σ^{2}) \overset{d}{\to} N (0, μ_{4} - σ^{4})

$\sqrt n(s^2 - \sigma^2) \xrightarrow{d} N\left(0,\mu_4 - \sigma^4\right)$

Note: the above result of course holds also for normally distributed samples -but in this last case we have also available a finite-sample chi-square distributional result.

— Alecos Papadopoulos
स्रोत

+1 There's no reason to check general dichotomous distributions because they are all scale and location versions of the Bernoulli: the analysis for the Bernoulli suffices. My simulations (out to sample sizes of

10^{1000}

$10^{1000}$ ) confirm the

χ_{1}^{2}

$\chi^2_1$ result.

— whuber

@whuber Thanks for checking. You' re right of course about the Benroulli being the mother of them all.

— Alecos Papadopoulos

You already have a detailed answer to your question but let me offer another one to go with it. Actually, a shorter proof is possible based on the fact that the distribution of

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

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

does not depend on $E(X) = \xi$ , say. Asymptotically, it also does not matter whether we change the factor $\frac{1}{n-1}$ to $\frac{1}{n}$ , which I will do for convenience. We then have

\sqrt{n} (S^{2} - σ^{2}) = \sqrt{n} [\frac{1}{n} \sum_{i = 1}^{n} X_{i}^{2} - {\bar{X}}^{2} - σ^{2}]

$\sqrt{n} \left(S^2 - \sigma^2 \right) = \sqrt{n} \left[ \frac{1}{n} \sum_{i=1}^n X_i^2 - \bar{X}^2 - \sigma^2 \right]$

And now we assume without loss of generality that $\xi = 0$ and we notice that

\sqrt{n} {\bar{X}}^{2} = \frac{1}{\sqrt{n}} {(\sqrt{n} \bar{X})}^{2}

$\sqrt{n} \bar{X}^2 = \frac{1}{\sqrt{n}} \left( \sqrt{n} \bar{X} \right)^2$

has probability limit zero, since the second term is bounded in probability (by the CLT and the continuous mapping theorem), i.e. it is $O_p(1)$ . The asymptotic result now follows from Slutzky's theorem and the CLT, since

\sqrt{n} [\frac{1}{n} \sum X_{i}^{2} - σ^{2}] \overset{D}{\to} N (0, τ^{2})

$\sqrt{n} \left[ \frac{1}{n} \sum X_i^2 - \sigma^2 \right] \xrightarrow{D} \mathcal{N} \left(0, \tau^2 \right)$

where $\tau^2 = Var \left\{ X^2\right\} = \mathbb{E} \left(X^4 \right) - \left( \mathbb{E} \left(X^2\right) \right)^2$ . And that will do it.

— JohnK
स्रोत

This is certainly more economical. But please reconsider how innocuous is the

E (X) = 0

$E(X) =0$ assumption. For example, it excludes the case of a Bernoulli (

p = 1 / 2

$p=1/2$ ) sample, and as I mention at the end of my answer, for such a sample, this asymptotic result does not hold.

— Alecos Papadopoulos

@AlecosPapadopoulos Indeed but the data can always be centered, right? I mean

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

$\sum_{i=1}^n \left(X_i - \mu - ( \bar{X}-\mu) \right)^2 = \sum_{i=1}^n \left(X_i - \bar{X} \right)^2$ and we can work with the these variables. For the Bernoulli case, is there something stopping us from doing so?

— JohnK

@AlecosPapadopoulos Oh yeah, I see the problem.

— JohnK

I have written a small piece on the matter, I think it is time to upload it in my blog. I will notify you in case you are interested to read it. The asymptotic distribution of the sample variance in this case is interesting, and even more the asymptotic distribution of the sample standard deviation. These results hold for any

p = 1 / 2

$p=1/2$ dichotomous random variable.

— Alecos Papadopoulos

Dumb question, but how can we assume that

S^{2}

$S^2$ is ancillary if the

X_{i}

$X_i$ are not normal? Or is

S^{2}

$S^2$ always ancillary (w.r.t. mean parametrization I guess) but only independent of the sample mean when the sample mean is a complete sufficient statistic (i.e. normally distributed) by Basu's theorem?

— Chill2Macht

The excellent answers by Alecos and JohnK already derive the result you are after, but I would like to note something else about the asymptotic distribution of the sample variance.

It is common to see asymptotic results presented using the normal distribution, and this is useful for stating the theorems. However, practically speaking, the purpose of an asymptotic distribution for a sample statistic is that it allows you to obtain an approximate distribution when $n$ is large. There are lots of choices you could make for your large-sample approximation, since many distributions have the same asymptotic form. In the case of the sample variance, it is my view that an excellent approximating distribution for large $n$ is given by:

\frac{S_{n}^{2}}{σ^{2}} \sim \frac{Chi-Sq (df = D F_{n})}{D F_{n}},

$\frac{S_n^2}{\sigma^2} \sim \frac{\text{Chi-Sq}(\text{df} = DF_n)}{DF_n},$

where $DF_n \equiv 2 / \mathbb{V}(S_n^2 / \sigma^2) = 2n / ( \kappa - (n-3)/(n-1))$ and $\kappa = \mu_4 / \sigma^4$ is the kurtosis parameter. This distribution is asymptotically equivalent to the normal approximation derived from the theorem (the chi-squared distribution converges to normal as the degrees-of-freedom tends to infinity). Despite this equivalence, this approximation has various other properties you would like your approximating distribution to have:

Unlike the normal approximation derived directly from the theorem, this distribution has the correct support for the statistic of interest. The sample variance is non-negative, and this distribution is has non-negative support.
In the case where the underlying values are normally distributed, this approximation is actually the exact sampling distribution. (In this case we have $\kappa = 3$ which gives $DF_n = n-1$ , which is the standard form used in most texts.) It therefore constitutes a result that is exact in an important special case, while still being a reasonable approximation in more general cases.

Derivation of the above result: Approximate distributional results for the sample mean and variance are discussed at length in O'Neill (2014), and this paper provides derivations of many results, including the present approximating distribution.

This derivation starts from the limiting result in the question:

\sqrt{n} (S_{n}^{2} - σ^{2}) \sim N (0, σ^{4} (κ - 1)) .

$\sqrt{n} (S_n^2 - \sigma^2) \sim \text{N}(0, \sigma^4 (\kappa - 1)).$

Re-arranging this result we obtain the approximation:

\frac{S_{n}^{2}}{σ^{2}} \sim N (1, \frac{κ - 1}{n}) .

$\frac{S_n^2}{\sigma^2} \sim \text{N} \Big( 1, \frac{\kappa - 1}{n} \Big).$

Since the chi-squared distribution is asymptotically normal, as $DF \rightarrow \infty$ we have:

\frac{Chi-Sq (D F)}{D F} \to \frac{1}{D F} N (D F, 2 D F) = N (1, \frac{2}{D F}) .

$\frac{\text{Chi-Sq}(DF)}{DF} \rightarrow \frac{1}{DF} \text{N} ( DF, 2DF ) = \text{N} \Big( 1, \frac{2}{DF} \Big).$

Taking $DF_n \equiv 2 / \mathbb{V}(S_n^2 / \sigma^2)$ (which yields the above formula) gives $DF_n \rightarrow 2n / (\kappa - 1)$ which ensures that the chi-squared distribution is asymptotically equivalent to the normal approximation from the limiting theorem.

— Reinstate Monica
स्रोत

One empirically interesting question is that which of these two asymptotic results works better in finite sample cases under various underlying data distributions.

— lzstat

Yes, I think that would be a very interesting (and publishable) simulation study. Since the present formula is based on kurtosis-correction of the variance of the sample variance, I would expect that the present result would work best when you have an underlying distribution with a kurtosis parameter that is far from mesokurtic (i.e., when the kurtosis-correction matters most). Since the kurtosis would need to be estimated from the sample, it is an open question as to when there would be a substantial improvement in overall performance.

— Reinstate Monica