Asymptotyczny rozkład wariancji próbki w próbce nienormalnej

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Jest to bardziej ogólne podejście do problemu postawionego przez to pytanie . Po uzyskaniu asymptotycznego rozkładu wariancji próbki, możemy zastosować metodę Delta, aby uzyskać odpowiedni rozkład dla odchylenia standardowego.

Niech próbka wielkości n iid nietypowych zmiennych losowych {Xi},i=1,...,n , ze średniąμ i wariancjąσ2 . Ustaw średnią próbki i wariancję próbki na

x¯=1ni=1nXi,s2=1n1i=1n(Xix¯)2

Wiemy, że

E(s2)=σ2,Var(s2)=1n(μ4n3n1σ4)

gdzie μ4=E(Xiμ)4 , a my ograniczamy naszą uwagę do rozkładów, dla których które momenty muszą istnieć i być skończone, istnieją i są skończone.

Czy to tak trzyma?

n(s2σ2)dN(0,μ4σ4)?
Alecos Papadopoulos
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Heh Właśnie napisałem w innym wątku, nie zdając sobie sprawy, że to opublikowałeś. Istnieje wiele rzeczy, które można znaleźć na CLT zastosowanym do wariancji (na przykład p3-4 tutaj ). Dobra odpowiedź btw.
Glen_b
Dzięki. Tak, znalazłem to. Ale tęsknią za sprawą wskazaną przez @whuber. Podają nawet przykład Bernoulliego z ogólnym ! (podstawa s. 4). Mam rozszerzenie moją odpowiedź na pokrycie p = 1 / 2 przypadku również. pp=1/2
Alecos Papadopoulos
Tak, widziałem, że rozważali Bernoulli, ale nie wzięli pod uwagę tego szczególnego przypadku. Myślę, że wzmianka o rozróżnieniu dla skalowanego Bernoulliego (przypadek dychotomiczny równego prawdopodobieństwa) jest jednym z powodów (wśród kilku innych), dlaczego warto omówić to tutaj w odpowiedzi (a nie tylko w komentarzu) - nie tylko to można wyszukiwać.
Glen_b

Odpowiedzi:

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Do bocznych zależności pojawiających się, gdy weźmiemy pod uwagę wariancję próbki, piszemy

(n1)s2=i=1n((Xiμ)(x¯μ))2

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

and after a little manipualtion,

=i=1n(Xiμ)2n(x¯μ)2

Therefore

n(s2σ2)=nn1i=1n(Xiμ)2nσ2nn1n(x¯μ)2

Manipulating,

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

The term n/(n1) becomes unity asymptotically. The term nn1σ2 is determinsitic and goes to zero as n.

We also have 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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+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
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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
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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