Was ist der erwartete Wert des Logarithmus der Gammaverteilung?

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Wenn der erwartete Wert von G a m m a ( α , β )Gamma(α,β) ist αβαβ , was ist der erwartete Wert vonlog(Gamma(α,β))log(Gamma(α,β))? Kann es analytisch berechnet werden?

Die Parametrisierung, die ich verwende, ist die Formrate.

Stefano Vespucci
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4
Wenn X Gamma ( a , b ) istXGamma(a,b) , dann ist nach mathStatica / Mathematica E [ log ( X ) ] = log ( b )E[log(X)]=log(b) + PolyGamma [a], wobei PolyGamma die Digammafunktion bezeichnet
Wölfe
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Ich sollte hinzufügen, dass Sie die PDF-Form Ihrer Gamma-Variablen nicht angeben, und da Sie angeben, dass der Mittelwert α / β istα/β (während es für mich a b wäreab , scheint es, dass Sie eine andere Notation verwenden als ich, wo Ihre β = 1 / bβ=1/b
Wolfies
Tut mir leid. Die Parametrisierung, die ich verwende, ist die Formrate. β αΓ ( α ) xα-1e-βxβαΓ(α)xα1eβx Ich werde versuchen, es für diese Parametrisierung zu finden. Könnten Sie bitte die Abfrage für Mathematica / WolframAlpha vorschlagen?
Stefano Vespucci
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Siehe auch Johnson, Lotz und Balakrishna (1994), kontinuierliche univariate Verteilungen, Band 1, 2. Aufl. S. 337-349.
Björn
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Siehe auch Wikipedia: Gamma Distribution # Logarithmische Erwartung und Varianz
Glen_b -Reinstate Monica

Antworten:

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Dies kann (vielleicht überraschend) mit einfachen Elementaroperationen durchgeführt werden (unter Verwendung von Richard Feynmans Lieblingstrick der Differenzierung unter dem Integralzeichen in Bezug auf einen Parameter).


We are supposing XX has a Γ(α,β)Γ(α,β) distribution and we wish to find the expectation of Y=log(X).Y=log(X). First, because ββ is a scale parameter, its effect will be to shift the logarithm by logβ.logβ. (If you use ββ as a rate parameter, as in the question, it will shift the logarithm by logβ.logβ.) This permits us to work with the case β=1.β=1.

After this simplification, the probability element of XX is

fX(x)=1Γ(α)xαexdxx

fX(x)=1Γ(α)xαexdxx

where Γ(α)Γ(α) is the normalizing constant

Γ(α)=0xαexdxx.

Γ(α)=0xαexdxx.

Substituting x=ey,x=ey, which entails dx/x=dy,dx/x=dy, gives the probability element of YY,

fY(y)=1Γ(α)eαyeydy.

fY(y)=1Γ(α)eαyeydy.

The possible values of YY now range over all the real numbers R.R.

Because fYfY must integrate to unity, we obtain (trivially)

Γ(α)=Reαyeydy.

Γ(α)=Reαyeydy.(1)

Notice fY(y)fY(y) is a differentiable function of α.α. An easy calculation gives

ddαeαyeydy=yeαyeydy=Γ(α)yfY(y).

ddαeαyeydy=yeαyeydy=Γ(α)yfY(y).

The next step exploits the relation obtained by dividing both sides of this identity by Γ(α),Γ(α), thereby exposing the very object we need to integrate to find the expectation; namely, yfY(y):yfY(y):

E(Y)=RyfY(y)=1Γ(α)Rddαeαyeydy=1Γ(α)ddαReαyeydy=1Γ(α)ddαΓ(α)=ddαlogΓ(α)=ψ(α),

E(Y)=RyfY(y)=1Γ(α)Rddαeαyeydy=1Γ(α)ddαReαyeydy=1Γ(α)ddαΓ(α)=ddαlogΓ(α)=ψ(α),

the logarithmic derivative of the gamma function (aka "polygamma"). The integral was computed using identity (1).(1).

Re-introducing the factor ββ shows the general result is

E(log(X))=logβ+ψ(α)

E(log(X))=logβ+ψ(α)

for a scale parameterization (where the density function depends on x/βx/β) or

E(log(X))=logβ+ψ(α)

E(log(X))=logβ+ψ(α)

for a rate parameterization (where the density function depends on xβxβ).

whuber
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With polygamma function do you mean of which order (e.g., 0,1) being a digamma (As @wolfies pointed out), trigamma?
Stefano Vespucci
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@Stefano I mean the logarithmic derivative of gamma, as stated. That means ψ(z)=Γ(z)/Γ(z).ψ(z)=Γ(z)/Γ(z).
whuber
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The answer by @whuber is quite nice; I will essentially restate his answer in a more general form which connects (in my opinion) better with statistical theory, and which makes clear the power of the overall technique.

Consider a family of distributions {Fθ:θΘ}{Fθ:θΘ} which consitute an exponential family, meaning they admit a density fθ(x)=exp{s(x)θA(θ)+h(x)}

fθ(x)=exp{s(x)θA(θ)+h(x)}
with respect to some common dominating measure (usually, Lebesgue or counting measure). Differentiating both sides of
fθ(x) dx=1
fθ(x) dx=1
with respect to θθ we arrive at the score equation fθ(x)=fθ(x)fθ(x)fθ(x)=uθ(x)fθ(x) dx=0
fθ(x)=fθ(x)fθ(x)fθ(x)=uθ(x)fθ(x) dx=0()
where uθ(x)=ddθlogfθ(x)uθ(x)=ddθlogfθ(x) is the score function and we have defined fθ(x)=ddθfθ(x)fθ(x)=ddθfθ(x). In the case of an exponential family, we have uθ(x)=s(x)A(θ)
uθ(x)=s(x)A(θ)
where A(θ)=ddθA(θ)A(θ)=ddθA(θ); this is sometimes called the cumulant function, as it is evidently very closely related to the cumulant-generating function. It follows now from ()() that Eθ[s(X)]=A(θ)Eθ[s(X)]=A(θ).

We now show this helps us compute the require expectation. We can write the gamma density with fixed ββ as an exponential family fθ(x)=βαΓ(α)xα1eβx=exp{log(x)α+αlogβlogΓ(α)βx}.

fθ(x)=βαΓ(α)xα1eβx=exp{log(x)α+αlogβlogΓ(α)βx}.
This is an exponential family in α alone with s(x)=logx and A(α)=logΓ(α)αlogβ. It now follows immediately by computing ddαA(α) that E[logX]=ψ(α)logβ.

guy
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+1 Thank you for pointing out this nice generalization.
whuber