Asimmetria del logaritmo di una variabile gamma casuale


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Considera la variabile gamma random XΓ(α,θ) . Esistono formule precise per media, varianza e asimmetria:

E[X]=αθVar[X]=αθ2=1/αE[X]2Skewness[X]=2/α

Considera ora una variabile casuale trasformata in logY=log(X) . Wikipedia fornisce formule per la media e la varianza:

E[Y]=ψ(α)+log(θ)Var[Y]=ψ1(α)

tramite funzioni digamma e trigamma che sono definite come la prima e la seconda derivata del logaritmo della funzione gamma.

Qual è la formula per l'asimmetria?

Apparirà la funzione tetragamma?

(Ciò che mi ha fatto meravigliare di questo è una scelta tra le distribuzioni lognormale e gamma, vedere le distribuzioni Gamma vs. lognormale . Tra le altre cose, differiscono nelle loro proprietà di asimmetria. In particolare, l'asimmetria del log di lognormale è banalmente uguale a zero. Considerando che l'asimmetria del registro di gamma è negativa. Ma quanto è negativo? ..)



Non sono sicuro di cosa sia la distribuzione log-gamma. Se è correlato alla gamma come lognormale è correlato alla normale, allora sto chiedendo qualcos'altro (perché "lognormale", confusamente, è la distribuzione di exp (normale) non di log (normale)).
ameba dice di reintegrare Monica il

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@Glen_b: A dire il vero, direi che chiamare esponenziale del normale un "lognormale" è molto più incoerente e confuso. Sebbene, sfortunatamente, più consolidato.
S. Kolassa - Ripristina Monica il

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@Stephan vedi anche log-logistico, log-Cauchy, log-Laplace ecc. Ecc. È una convenzione più chiaramente stabilita rispetto al contrario
Glen_b -Reinstate Monica

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Sì; Per questo motivo sono stato attento a non dire "log-gamma" da nessuna parte in relazione a questa distribuzione. (L'ho usato in passato in modo coerente con il log-normale)
Glen_b -Reinstate Monica

Risposte:


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Il momento che genera la funzione di Y = ln X è utile in questo caso, poiché ha una forma algebrica semplice. Secondo la definizione di mgf, abbiamo M ( t )M(t)Y=lnX

M(t)=E[etlnX]=E[Xt]=1Γ(α)θα0xα+t1ex/θdx=θtΓ(α)0yα+t1eydy=θtΓ(α+t)Γ(α).

Verifichiamo le aspettative e la varianza che hai dato. Prendendo derivati, abbiamo eM(t)=Γ(α+t)

M(t)=Γ(α+t)Γ(α)θt+Γ(α+t)Γ(α)θtln(θ)
Quindi,E[Y]=ψ(0)(α)+ln(
M(t)=Γ(α+t)Γ(α)θt+2Γ(α+t)Γ(α)θtln(θ)+Γ(α+t)Γ(α)θtln2(θ).
Segue quindiVar(Y)=E[Y2]-E[Y]2=Γ(α)
E[Y]=ψ(0)(α)+ln(θ),E[Y2]=Γ(α)Γ(α)+2ψ(0)(α)ln(θ)+ln2(θ).
Var(Y)=E[Y2]E[Y]2=Γ(α)Γ(α)(Γ(α)Γ(α))2=ψ(1)(α).

K(t)=lnM(t)=tlnθ+lnΓ(α+t)lnΓ(α).
K(0)=ψ(0)(α)+ln(θ)ψ(n)(x)=dn+1lnΓ(x)/dxn+1K(n)(0)=ψ(n1)(α)n2
E[(YE[Y])3]Var(Y)3/2=ψ(2)(α)[ψ(1)(α)]3/2.

Come nota a margine, questa particolare distribuzione sembrava essere stata accuratamente studiata da AC Olshen nelle sue Transformations of the Pearson Type III Distribution , la Continuous Univariate Distributions di Johnson et al. Ha anche un piccolo pezzo al riguardo. Dai un'occhiata a quelli.


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K(t)=log[M(t)]=tlog[θ]+log[Γ(α+t)]log[Γ(α)] instead of M(t) as this is the cumulant generating function - more directly related to central moments - skew=K(3)(0)=ψ(2)(α) where ψ(n)(z) is the polygamma function
probabilityislogic

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@probabilityislogic: very good call, changed my answer
Francis

@probabilityislogic This is a great addition, thanks a lot. I just want to note, lest some readers be confused, that skewness is not directly given by the third cumulant: it's the third standardized moment, not the third central moment. Francis has it correct in his answer, but the last formula in your comment is not quite right.
amoeba says Reinstate Monica

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I. Direct computation

Gradshteyn & Ryzhik [1] (sect 4.358, 7th ed) list explicit closed forms for

0xν1eμx(lnx)pdx
for p=2,3,4 while the p=1 case is done in 4.352 (assuming you regard expressions in Γ,ψ and ζ functions as closed form) -- from which it is definitely doable up to kurtosis; they give the integral for all p as a derivative of a gamma function so presumably it's feasible to go higher. So skewness is certainly doable but not especially "neat".

Details of the derivation of the formulas in 4.358 are in [2]. I'll quote the formulas given there since they're slightly more succinctly stated and put 4.352.1 in the same form.

Let δ=ψ(a)lnμ. Then:

0xa1eμxlnxdx=Γ(a)μa{δ}0xa1eμxln2xdx=Γ(a)μa{δ2+ζ(2,a)}0xa1eμxln3xdx=Γ(a)μa{δ3+3ζ(2,a)δ2ζ(3,a)}0xa1eμxln4xdx=Γ(a)μa{δ4+6ζ(2,a)δ28ζ(3,a)δ+3ζ2(2,a)+6ζ(4,a))}

where ζ(z,q)=n=01(n+q)z is the Hurwitz zeta function (the Riemann zeta function is the special case q=1).

Now on to the moments of the log of a gamma random variable.

Noting firstly that on the log scale the scale or rate parameter of the gamma density is merely a shift-parameter, so it has no impact on the central moments; we may take whichever one we're using to be 1.

If XGamma(α,1) then

E(logpX)=1Γ(α)0logpxxα1exdx.

We can set μ=1 in the above integral formulas, which gives us raw moments; we have E(Y), E(Y2), E(Y3), E(Y4).

Since we have eliminated μ from the above, without fear of confusion we're now free to re-use μk to represent the k-th central moment in the usual fashion. We may then obtain the central moments from the raw moments via the usual formulas.

Then we can obtain the skewness and kurtosis as μ3μ23/2 and μ4μ22.


A note on terminology

It looks like Wolfram's reference pages write the moments of this distribution (they call it ExpGamma distribution) in terms of the polygamma function.

By contrast, Chan (see below) calls this the log-gamma distribution.


II. Chan's formulas via MGF

Chan (1993) [3] gives the mgf as the very neat Γ(α+t)/Γ(α).

(A very nice derivation for this is given in Francis' answer, using the simple fact that the mgf of log(X) is just E(Xt).)

Consequently the moments have fairly simple forms. Chan gives:

E(Y)=ψ(α)

and the central moments as

E(YμY)2=ψ(α)E(YμY)3=ψ(α)E(YμY)4=ψ(α)

and so the skewness is ψ(α)/(ψ(α)3/2) and kurtosis is ψ(α)/(ψ(α)2). Presumably the earlier formulas I have above should simplify to these.

Conveniently, R offers digamma (ψ) and trigamma (ψ) functions as well as the more general polygamma function where you select the order of the derivative. (A number of other programs offer similarly convenient functions.)

Consequently we can compute the skewness and kurtosis quite directly in R:

skew.eg <- function(a) psigamma(a,2)/psigamma(a,1)^(3/2)
kurt.eg <- function(a) psigamma(a,3)/psigamma(a,1)^2

Trying a few values of a (α in the above), we reproduce the first few rows of the table at the end of Sec 2.2 in Chan [3], except that the kurtosis values in that table are supposed to be excess kurtosis, but I just calculated kurtosis by the formulas given above by Chan; these should differ by 3.

(E.g. for the log of an exponential, the table says the excess kurtosis is 2.4, but the formula for β2 is ψ(1)/ψ(1)2 ... and that is 2.4.)

Simulation confirms that as we increase sample size, the kurtosis of a log of an exponential is converging to around 5.4 not 2.4. It appears that the thesis possibly has an error.

Consequently, Chan's formulas for central moments appear to actually be the formulas for the cumulants (see the derivation in Francis' answer). This would then mean that the skewness formula was correct as is; because the second and third cumulants are equal to the second and third central moments.

Nevertheless these are particularly convenient formulas as long as we keep in mind that kurt.eg is giving excess kurtosis.

References

[1] Gradshteyn, I.S. & Ryzhik I.M. (2007), Table of Integrals, Series, and Products, 7th ed.
Academic Press, Inc.

[2] Victor H. Moll (2007)
The integrals in Gradshteyn and Ryzhik, Part 4: The gamma function
SCIENTIA Series A: Mathematical Sciences, Vol. 15, 37–46
Universidad Técnica Federico Santa María, Valparaíso, Chile
http://129.81.170.14/~vhm/FORM-PROOFS_html/final4.pdf

[3] Chan, P.S. (1993),
A statistical study of log-gamma distribution,
McMaster University (Ph.D. thesis)
https://macsphere.mcmaster.ca/bitstream/11375/6816/1/fulltext.pdf


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Cool. Thanks a lot! According to the encyclopedia entry that Stephan linked to above, the final answer for skewness is ψ(α)/ψ(α)3/2 (which almost qualifies as "neat"!). So it seems that all the scary zetas will have to cancel out.
amoeba says Reinstate Monica

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Sorry only just now saw your comment (I've been editing for about an hour or so); that's correct, though if the Encyclopedia gives kurtosis the way Chan gives it in his thesis, it seems that it's wrong (as given above), but readily corrected. The neat formulas appear to be for cumulants rather than standardized central moments.
Glen_b -Reinstate Monica

Yes, the Encyclopedia does give the same formula for kurtosis.
amoeba says Reinstate Monica

Hmm, I mean to refer to the things normally denoted γ1 and γ2. I will fix.
Glen_b -Reinstate Monica

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I should probably add the note that the Hurwitz zeta function can be expressed in terms of the polygamma function, and vice versa:
ψ(n)(z)=(1)n+1Γ(n+1)ζ(n+1,z)
So, the answer to the @amoeba's question of "will the tetragamma function appear?" is YES.
J. M. is not a statistician
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