감마 랜덤 변수의 로그 왜곡

고려 감마 확률 변수 . 평균, 분산 및 왜곡에 대한 깔끔한 수식이 있습니다. $X\sim\Gamma(\alpha, \theta)$

\begin{aligned} E [X] & = α θ \\ Var [X] & = α θ^{2} = 1 / α \cdot E [X]^{2} \\ Skewness [X] & = 2 / \sqrt{α} \end{aligned}

$\begin{align} \mathbb E[X]&=\alpha\theta\\ \operatorname{Var}[X]&=\alpha\theta^2=1/\alpha\cdot\mathbb E[X]^2\\ \operatorname{Skewness}[X]&=2/\sqrt{\alpha} \end{align}$

이제 로그 변환 된 랜덤 변수 . Wikipedia는 평균과 분산에 대한 공식을 제공합니다. $Y=\log(X)$

\begin{aligned} E [Y] & = ψ (α) + \log (θ) \\ Var [Y] & = ψ_{1} (α) \end{aligned}

$\begin{align} \mathbb E[Y]&=\psi(\alpha)+\log(\theta)\\ \operatorname{Var}[Y]&=\psi_1(\alpha)\\ \end{align}$

감마 함수의 로그의 1 차 및 2 차 미분으로 정의되는 디 감마 및 트리 감마 함수를 통해.

왜도의 공식은 무엇입니까?

테트라 감마 기능이 나타 납니까?

(이것에 대해 궁금한 점은 로그 정규 분포와 감마 분포 사이의 선택 입니다. 감마 대 로그 정규 분포를 참조하십시오 . 무엇보다도 왜곡 특성이 다릅니다. 특히, 로그 정규 로그의 왜곡은 거의 0과 같습니다. 감마 로그의 왜도는 음수이지만 얼마나 음입니까? ..)

gamma-distribution skewness logarithm

— 아메바의 말에 따르면 복원 모니카
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합니까 이 도움을? 아니면 이것 ?

— S. Kolassa-복원 모니카

로그 감마 분포가 무엇인지 잘 모르겠습니다. lognormal이 normal과 관련되어 있기 때문에 감마와 관련이 있다면 다른 것에 대해 묻고 있습니다 ( "lognormal"은 혼란 스럽기 때문에 log (normal)가 아닌 exp (normal)의 분포이기 때문에).

— amoeba는 Reinstate Monica가

@Glen_b : 솔직히 말하면, 정규의 지수를 "로그 정규"라고 부르는 것이 훨씬 더 일관성이없고 혼란 스럽다고 말하고 싶습니다. 불행히도, 더 확립되었습니다.

— S. Kolassa-복원 모니카

@Stephan은 log-logistic, log-Cauchy, log-Laplace 등을 참조하십시오. 반대보다 더 명확하게 확립 된 규칙

— Glen_b -Reinstate Monica

네; 이런 이유로이 분포와 관련하여 "log-gamma"라고 말하지 않도록주의했습니다. (나는 과거에 로그 정규에 일관된 방식으로 그것을 사용했다)

— Glen_b-복지국 Monica

답변:

순간 발생 함수 $M(t)$ 의 $Y=\ln X$ 그것이 간단한 대수 형식을 갖기 때문에,이 경우에 유용하다. mgf의 정의에 따라

\begin{aligned} 미디엄 (티) & = 이자형 [{이자형}^{티 \ln 엑스}] = 이자형 [{엑스}^{티}] \\ = \frac{1}{Γ (α) θ^{α}} \int_{0}^{\infty} {엑스}^{α + 티 - 1} {이자형}^{- 엑스 / θ} 디 엑스 \\ = \frac{θ^{티}}{Γ (α)} \int_{0}^{\infty} {와이}^{α + 티 - 1} {이자형}^{- 와이} 디 와이 \\ = \frac{θ^{티} Γ (α + 티)}{Γ (α)} . \end{aligned}

$\begin{aligned}M(t)&=\operatorname{E}[e^{t\ln X}]=\operatorname{E}[X^t]\\ &=\frac{1}{\Gamma(\alpha)\theta^\alpha}\int_0^\infty x^{\alpha+t-1}e^{-x/\theta}\,dx\\ &=\frac{\theta^{t}}{\Gamma(\alpha)}\int_0^\infty y^{\alpha+t-1}e^{-y}\,dy\\ &=\frac{\theta^t\Gamma(\alpha+t)}{\Gamma(\alpha)}.\end{aligned}$

당신이 준 기대와 분산을 확인합시다. 도함수를 취하면, 우리는 과

{미디엄}^{'} (티) = \frac{Γ^{'} (α + 티)}{Γ (α)} θ^{티} + \frac{Γ (α + 티)}{Γ (α)} θ^{티} \ln (θ)

$M'(t)=\frac{\Gamma'(\alpha+t)}{\Gamma(\alpha)}\theta^t+\frac{\Gamma(\alpha+t)}{\Gamma(\alpha)}\theta^t\ln(\theta)$

따라서,

{미디엄}^{″} (티) = \frac{Γ^{″} (α + 티)}{Γ (α)} θ^{티} + \frac{2 Γ^{'} (α + 티)}{Γ (α)} θ^{티} \ln (θ) + \frac{Γ (α + 티)}{Γ (α)} θ^{티} \ln^{2} (θ) .

$M''(t)=\frac{\Gamma''(\alpha+t)}{\Gamma(\alpha)}\theta^t+\frac{2\Gamma'(\alpha+t)}{\Gamma(\alpha)}\theta^t\ln(\theta)+\frac{\Gamma(\alpha+t)}{\Gamma(\alpha)}\theta^t\ln^2(\theta).$

그런 다음

E [Y] = ψ^{(0)} (α) + \ln (θ), E [Y^{2}] = \frac{Γ^{″} (α)}{Γ (α)} + 2 ψ^{(0)} (α) \ln (θ) + \ln^{2} (θ) .

$\operatorname{E}[Y]=\psi^{(0)}(\alpha)+\ln(\theta),\qquad\operatorname{E}[Y^2]=\frac{\Gamma''(\alpha)}{\Gamma(\alpha)}+2\psi^{(0)}(\alpha)\ln(\theta)+\ln^2(\theta).$

Var (Y) = E [Y^{2}] - E [Y]^{2} = \frac{Γ^{″} (α)}{Γ (α)} - {(\frac{Γ^{'} (α)}{Γ (α)})}^{2} = ψ^{(1)} (α) .

$\operatorname{Var}(Y)=\operatorname{E}[Y^2]-\operatorname{E}[Y]^2=\frac{\Gamma''(\alpha)}{\Gamma(\alpha)}-\left(\frac{\Gamma'(\alpha)}{\Gamma(\alpha)}\right)^2=\psi^{(1)}(\alpha).$

비대칭도를 찾기 위해 (줘서 고마워요 @probabilityislogic)를 cumulant 발생 기능을 유의하는 것은 따라서 제 1 누적량은 간단히 이다. 기억해

K (t) = \ln M (t) = t \ln θ + \ln Γ (α + t) - \ln Γ (α) .

$K(t)=\ln M(t)=t\ln\theta+\ln\Gamma(\alpha+t)-\ln\Gamma(\alpha).$

K^{'} (0) = ψ^{(0)} (α) + \ln (θ)

$K'(0)=\psi^{(0)}(\alpha)+\ln(\theta)$

ψ^{(n)} (x) = d^{n + 1} \ln Γ (x) / d x^{n + 1}

$\psi^{(n)}(x)=d^{n+1}\ln\Gamma(x)/dx^{n+1}$ , so the subsequent cumulants are

K^{(n)} (0) = ψ^{(n - 1)} (α)

$K^{(n)}(0)=\psi^{(n-1)}(\alpha)$ ,

n \geq 2

$n\geq2$ . The skewness is therefore

\frac{E [(Y - E [Y])^{3}]}{Var (Y)^{3 / 2}} = \frac{ψ^{(2)} (α)}{[ψ^{(1)} (α)]^{3 / 2}} .

$\frac{\operatorname{E}[(Y-\operatorname{E}[Y])^3]}{\operatorname{Var}(Y)^{3/2}}=\frac{\psi^{(2)}(\alpha)}{[\psi^{(1)}(\alpha)]^{3/2}}.$

As a side note, this particular distribution appeared to have been thoroughly studied by A. C. Olshen in his Transformations of the Pearson Type III Distribution, Johnson et al.'s Continuous Univariate Distributions also has a small piece about it. Check those out.

— Francis
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You should differentiate

K (t) = \log [M (t)] = t \log [θ] + \log [Γ (α + t)] - \log [Γ (α)]

$K (t)=\log [M (t)]=t\log [\theta]+\log [\Gamma (\alpha+t)]-\log [\Gamma (\alpha)]$ instead of

M (t)

$M (t)$ as this is the cumulant generating function - more directly related to central moments -

s k e w = K^{(3)} (0) = ψ^{(2)} (α)

$skew=K^{(3)}(0)=\psi^{(2)}(\alpha)$ where

ψ^{(n)} (z)

$\psi^{(n)}(z)$ is the polygamma function

— probabilityislogic

@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

I. Direct computation

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

\int_{0}^{\infty} x^{ν - 1} e^{- μ x} (\ln x)^{p} d x

$\int_0^\infty x^{\nu-1}e^{-\mu x}(\ln x)^p dx$ for

p = 2, 3, 4

$p=2,3,4$ while the

p = 1

$p=1$ case is done in 4.352 (assuming you regard expressions in

Γ, ψ

$\Gamma, \psi$ and

ζ

$\zeta$ functions as closed form) -- from which it is definitely doable up to kurtosis; they give the integral for all

p

$p$ 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 $\delta= \psi(a)-\ln \mu$ . Then:

\begin{aligned} \int_{0}^{\infty} x^{a - 1} e^{- μ x} \ln x d x & = \frac{Γ (a)}{μ^{a}} {δ} \\ \int_{0}^{\infty} x^{a - 1} e^{- μ x} \ln^{2} x d x & = \frac{Γ (a)}{μ^{a}} {δ^{2} + ζ (2, a)} \\ \int_{0}^{\infty} x^{a - 1} e^{- μ x} \ln^{3} x d x & = \frac{Γ (a)}{μ^{a}} {δ^{3} + 3 ζ (2, a) δ - 2 ζ (3, a)} \\ \int_{0}^{\infty} x^{a - 1} e^{- μ x} \ln^{4} x d x & = \frac{Γ (a)}{μ^{a}} {δ^{4} + 6 ζ (2, a) δ^{2} - 8 ζ (3, a) δ + 3 ζ^{2} (2, a) + 6 ζ (4, a))} \end{aligned}

$\begin{align} \int_0^\infty x^{a-1} e^{-\mu x} \ln x \,dx &=\frac{\Gamma(a)}{\mu^a}\left\{ \delta \right\} \\ \int_0^\infty x^{a-1} e^{-\mu x} \ln^2\!x \,dx &=\frac{\Gamma(a)}{\mu^a}\left\{ \delta^2+\zeta(2,a) \right\} \\ \int_0^\infty x^{a-1} e^{-\mu x} \ln^3\!x \,dx &=\frac{\Gamma(a)}{\mu^a}\left\{ \delta^3+3\zeta(2,a)\delta-2\zeta(3,a) \right\} \\ \int_0^\infty x^{a-1} e^{-\mu x} \ln^4\!x \,dx &=\frac{\Gamma(a)}{\mu^a}\left\{ \delta^4+6\zeta(2,a)\delta^2-8\zeta(3,a)\delta + 3\zeta^2(2,a)+6\zeta(4,a)) \right\} \end{align}$

where $\zeta(z,q)=\sum_{n=0}^\infty \frac{1}{(n+q)^z}$ 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 $X\sim \text{Gamma}(\alpha,1)$ then

E (\log^{p} X) = \frac{1}{Γ (α)} \int_{0}^{\infty} \log^{p} x x^{α - 1} e^{- x} d x .

$E(\log^{p}\!X) = \frac{1}{\Gamma(\alpha)}\int_0^\infty \log^{p}\!x\, x^{\alpha-1} e^{-x} \,dx.$

We can set $\mu=1$ in the above integral formulas, which gives us raw moments; we have $E(Y)$ , $E(Y^2)$ , $E(Y^3)$ , $E(Y^4)$ .

Since we have eliminated $\mu$ from the above, without fear of confusion we're now free to re-use $\mu_k$ 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 $\frac{\mu_3}{\mu_2^{3/2}}$ and $\frac{\mu_4}{\mu_2^{2}}$ .

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 $\Gamma(\alpha+t)/\Gamma(\alpha)$ .

(A very nice derivation for this is given in Francis' answer, using the simple fact that the mgf of $\log(X)$ is just $E(X^t)$ .)

Consequently the moments have fairly simple forms. Chan gives:

E (Y) = ψ (α)

$E(Y)=\psi(\alpha)$

and the central moments as

\begin{aligned} E (Y - μ_{Y})^{2} & = ψ^{'} (α) \\ E (Y - μ_{Y})^{3} & = ψ^{″} (α) \\ E (Y - μ_{Y})^{4} & = ψ^{‴} (α) \end{aligned}

$\begin{align} E(Y-\mu_Y)^2 &= \psi'(\alpha) \\ E(Y-\mu_Y)^3 &= \psi''(\alpha) \\ E(Y-\mu_Y)^4 &= \psi'''(\alpha) \end{align}$

and so the skewness is $\psi''(\alpha)/(\psi'(\alpha)^{3/2})$ and kurtosis is $\psi'''(\alpha)/(\psi'(\alpha)^{2})$ . Presumably the earlier formulas I have above should simplify to these.

Conveniently, R offers digamma ( $\psi$ ) and trigamma ( $\psi'$ ) 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 ( $\alpha$ 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 $\beta_2$ is $\psi'''(1)/\psi'(1)^2$ ... 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

— Glen_b -Reinstate Monica
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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}

$\psi''(\alpha)/\psi'(\alpha)^{3/2}$ (which almost qualifies as "neat"!). So it seems that all the scary zetas will have to cancel out.

— amoeba says Reinstate Monica

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

γ_{1}

$\gamma_1$ and

γ_{2}

$\gamma_2$ . I will fix.

— Glen_b -Reinstate Monica

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 + 1} Γ (n + 1) ζ (n + 1, z)

$\psi^{(n)}(z)=(-1)^{n+1}\,\Gamma(n+1)\,\zeta(n+1,z)$ So, the answer to the @amoeba's question of "will the tetragamma function appear?" is YES.

— J. M. is not a statistician