Distribusi asimptotik dari varians sampel dari sampel non-normal

Ini adalah perlakuan yang lebih umum dari masalah yang ditimbulkan oleh pertanyaan ini . Setelah menurunkan distribusi asimptotik dari varians sampel, kita dapat menerapkan metode Delta untuk sampai pada distribusi yang sesuai untuk deviasi standar.

Biarkan sampel berukuran dari variabel acak tidak normal iid , dengan mean dan varians . Tetapkan mean sampel dan varians sampel sebagai { X i } $n$$\{X_i\},\;\; i=1,...,n$$\mu$$\sigma^2$

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

Kita tahu bahwa

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

di mana , dan kami membatasi perhatian kami pada distribusi yang momen-momen apa yang perlu ada dan terbatas, ada, dan terbatas.$\mu_4 = E(X_i -\mu)^4$

Apakah itu berlaku?

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

Untuk dependensi sisi-langkah yang timbul ketika kami mempertimbangkan varians sampel, kami menulis

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

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

dan setelah sedikit manipulasi,

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

Karena itu

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

Memanipulasi,

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

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

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

Istilah menjadi kesatuan tanpa gejala. Istilah $n/(n-1)$bersifat determinitik dan menjadi nol sebagain→∞.$\frac {\sqrt n}{n-1}\sigma^2$$n \rightarrow \infty$

Kami juga punya . Komponen pertama menyatu dalam distribusi ke Normal, konvergen kedua dalam probabilitas ke nol. Kemudian oleh teorema Slutsky produk konvergen dalam probabilitas ke nol,$\sqrt n\Big(\bar x-\mu\Big)^2 = \left[\sqrt n\Big(\bar x-\mu\Big)\right]\cdot \Big(\bar x-\mu\Big)$

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

Kita dibiarkan dengan istilah itu

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

Diperingatkan oleh contoh mematikan yang ditawarkan oleh @whuber dalam komentar atas jawaban ini , kami ingin memastikan bahwa tidak konstan. Whuber menunjukkan bahwa jika X i adalah Bernoulli ( 1 / 2 ) maka kuantitas ini adalah konstan. Jadi tidak termasuk variabel yang ini terjadi (mungkin lain dikotomis, bukan hanya 0 / 1 biner?), Untuk sisa kita memiliki$(X_i-\mu)^2$$X_i$$(1/2)$$0/1$

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

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

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

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.

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

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

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

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

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

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

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

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

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.

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:

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

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.

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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:

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

Re-arranging this result we obtain the approximation:

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

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

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

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.