Distribuição assintótica da variação da amostra não normal

19

Este é um tratamento mais geral da questão colocada por esta questão . Depois de derivar a distribuição assintótica da variação da amostra, podemos aplicar o método Delta para chegar à distribuição correspondente para o desvio padrão.

Deixe uma amostra do tamanho das variáveis ​​aleatórias não normais do iid , com média e variância . Defina a média da amostra e a variação da amostra como n{Xi},i=1,...,nμσ2

x¯=1ni=1nXi,s2=1n1i=1n(Xix¯)2

Sabemos que

E(s2)=σ2,Var(s2)=1n(μ4n3n1σ4)

onde , e restringimos nossa atenção a distribuições para as quais quais momentos precisam existir e ser finitos, existem e são finitos.μ4=E(Xiμ)4

Segura isso

n(s2σ2)dN(0,μ4σ4)?
Alecos Papadopoulos
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Heh. Acabei de postar no outro segmento, sem perceber que você postou isso. Há várias coisas a serem encontradas no CLT aplicadas à variação (como p3-4 aqui, por exemplo). Boa resposta btw.
Glen_b -Reinstar Monica
Obrigado. Sim, eu encontrei isso. Mas eles sentem falta do caso @whuber apontou. Eles até fornecem um exemplo de Bernoulli com geral ! (base da p. 4). Estou estendendo minha resposta para cobrir também o caso . pp=1/2
Alecos Papadopoulos
Sim, vi que eles consideravam os Bernoulli e não consideravam esse caso especial. Penso que a menção da distinção para Bernoulli em escala (caso probic dicotômico igual) é uma das razões (entre algumas outras) por que é valioso tê-lo discutido em resposta aqui (e não apenas em um comentário) - não menos importante: é pesquisável para.
Glen_b -Reinstala Monica

Respostas:

20

Para dependências secundárias que surgem quando consideramos a variação da amostra, escrevemos

(n1)s2=i=1n((Xiμ)(x¯μ))2

=i=1n(Xiμ)22i=1n((Xiμ)(x¯μ))+i=1n(x¯μ)2

e depois de um pouco de manipulação,

=i=1n(Xiμ)2n(x¯μ)2

Portanto

n(s2σ2)=nn1i=1n(Xiμ)2nσ2nn1n(x¯μ)2

Manipular,

n(s2σ2)=nn1i=1n(Xiμ)2nn1n1σ2nn1n(x¯μ)2

=nnn11ni=1n(Xiμ)2nn1n1σ2nn1n(x¯μ)2

=nn1[n(1ni=1n(Xiμ)2σ2)]+nn1σ2nn1n(x¯μ)2

O termo torna-se unidade assintoticamente. O termo n/(n1)é determinístico e passa a zero comon.nn1σ2n

Nós também temos n(x¯μ)2=[n(x¯μ)](x¯μ). The first component converges in distribution to a Normal, the second convergres in probability to zero. Then by Slutsky's theorem the product converges in probability to zero,

n(x¯μ)2p0

We are left with the term

[n(1ni=1n(Xiμ)2σ2)]

Alerted by a lethal example offered by @whuber in a comment to this answer, we want to make certain that (Xiμ)2 is not constant. Whuber pointed out that if Xi 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(Xiμ)2=σ2,Var[(Xiμ)2]=μ4σ4

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

n(s2σ2)dN(0,μ4σ4)

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
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3
+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 101000) confirm the χ12 result.
whuber
@whuber Thanks for checking. You' re right of course about the Benroulli being the mother of them all.
Alecos Papadopoulos
10

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

S2=1n1i=1n(XiX¯)2

does not depend on E(X)=ξ, say. Asymptotically, it also does not matter whether we change the factor 1n1 to 1n, which I will do for convenience. We then have

n(S2σ2)=n[1ni=1nXi2X¯2σ2]

And now we assume without loss of generality that ξ=0 and we notice that

nX¯2=1n(nX¯)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 Op(1). The asymptotic result now follows from Slutzky's theorem and the CLT, since

n[1nXi2σ2]DN(0,τ2)

where τ2=Var{X2}=E(X4)(E(X2))2. And that will do it.

JohnK
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This is certainly more economical. But please reconsider how innocuous is the E(X)=0 assumption. For example, it excludes the case of a Bernoulli (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
i=1n(Xiμ(X¯μ))2=i=1n(XiX¯)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 dichotomous random variable.
Alecos Papadopoulos
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Dumb question, but how can we assume that S2 is ancillary if the Xi are not normal? Or is S2 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
3

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:

Sn2σ2Chi-Sq(df=DFn)DFn,

where DFn2/V(Sn2/σ2)=2n/(κ(n3)/(n1)) and κ=μ4/σ4 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 κ=3 which gives DFn=n1, 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:

n(Sn2σ2)N(0,σ4(κ1)).

Re-arranging this result we obtain the approximation:

Sn2σ2N(1,κ1n).

Since the chi-squared distribution is asymptotically normal, as DF we have:

Chi-Sq(DF)DF1DFN(DF,2DF)=N(1,2DF).

Taking DFn2/V(Sn2/σ2) (which yields the above formula) gives DFn2n/(κ1) which ensures that the chi-squared distribution is asymptotically equivalent to the normal approximation from the limiting theorem.

Reinstate Monica
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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