Was ist der erwartete Wert des Logarithmus der Gammaverteilung?

Wenn der erwartete Wert von $\mathsf{Gamma}(\alpha, \beta)$ ist $\frac{\alpha}{\beta}$ , was ist der erwartete Wert von$\log(\mathsf{Gamma}(\alpha, \beta))$? Kann es analytisch berechnet werden?

Die Parametrisierung, die ich verwende, ist die Formrate.

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 $X$ has a $\Gamma(\alpha,\beta)$ distribution and we wish to find the expectation of $Y=\log(X).$ First, because $\beta$ is a scale parameter, its effect will be to shift the logarithm by $\log\beta.$ (If you use $\beta$ as a rate parameter, as in the question, it will shift the logarithm by $-\log\beta.$) This permits us to work with the case $\beta=1.$

After this simplification, the probability element of $X$ is

$$f_X(x) = \frac{1}{\Gamma(\alpha)} x^\alpha e^{-x} \frac{\mathrm{d}x}{x}$$

where $\Gamma(\alpha)$ is the normalizing constant

$$\Gamma(\alpha) = \int_0^\infty x^\alpha e^{-x} \frac{\mathrm{d}x}{x}.$$

Substituting $x=e^y,$ which entails $\mathrm{d}x/x = \mathrm{d}y,$ gives the probability element of $Y$,

$$f_Y(y) = \frac{1}{\Gamma(\alpha)} e^{\alpha y - e^y} \mathrm{d}y.$$

The possible values of $Y$ now range over all the real numbers $\mathbb{R}.$

Because $f_Y$ must integrate to unity, we obtain (trivially)

$$\Gamma(\alpha) = \int_\mathbb{R} e^{\alpha y - e^y} \mathrm{d}y.\tag{1}$$

Notice $f_Y(y)$ is a differentiable function of $\alpha.$ An easy calculation gives

$$\frac{\mathrm{d}}{\mathrm{d}\alpha}e^{\alpha y - e^y} \mathrm{d}y = y\, e^{\alpha y - e^y} \mathrm{d}y = \Gamma(\alpha) y\,f_Y(y).$$

The next step exploits the relation obtained by dividing both sides of this identity by $\Gamma(\alpha),$ thereby exposing the very object we need to integrate to find the expectation; namely, $y f_Y(y):$

$$\eqalign{ \mathbb{E}(Y) &= \int_\mathbb{R} y\, f_Y(y) = \frac{1}{\Gamma(\alpha)} \int_\mathbb{R} \frac{\mathrm{d}}{\mathrm{d}\alpha}e^{\alpha y - e^y} \mathrm{d}y \\ &= \frac{1}{\Gamma(\alpha)} \frac{\mathrm{d}}{\mathrm{d}\alpha}\int_\mathbb{R} e^{\alpha y - e^y} \mathrm{d}y\\ &= \frac{1}{\Gamma(\alpha)} \frac{\mathrm{d}}{\mathrm{d}\alpha}\Gamma(\alpha)\\ &= \frac{\mathrm{d}}{\mathrm{d}\alpha}\log\Gamma(\alpha)\\ &=\psi(\alpha), }$$

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

Re-introducing the factor $\beta$ shows the general result is

$$\mathbb{E}(\log(X)) = \log\beta + \psi(\alpha)$$

for a scale parameterization (where the density function depends on $x/\beta$) or

$$\mathbb{E}(\log(X)) = -\log\beta + \psi(\alpha)$$

for a rate parameterization (where the density function depends on $x\beta$).

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 $\{F_\theta : \theta \in \Theta\}$ which consitute an exponential family, meaning they admit a density

$$f_\theta(x) = \exp\left\{s(x)\theta - A(\theta) + h(x)\right\}$$ with respect to some common dominating measure (usually, Lebesgue or counting measure). Differentiating both sides of
$$\int f_\theta(x) \ dx = 1$$ with respect to $\theta$ we arrive at the score equation $$\int f'_\theta(x) = \int \frac{f'_\theta(x)}{f_\theta(x)} f_\theta(x) = \int u_\theta(x) \, f_\theta(x) \ dx = 0 \tag{$\dagger$}$$ where $u_\theta(x) = \frac d {d\theta} \log f_\theta(x)$ is the score function and we have defined $f'_\theta(x) = \frac{d}{d\theta} f_\theta(x)$. In the case of an exponential family, we have $$u_\theta(x) = s(x) - A'(\theta)$$ where $A'(\theta) = \frac d {d\theta} A(\theta)$; this is sometimes called the cumulant function, as it is evidently very closely related to the cumulant-generating function. It follows now from $(\dagger)$ that $E_\theta[s(X)] = A'(\theta)$.

We now show this helps us compute the require expectation. We can write the gamma density with fixed $\beta$ as an exponential family

$$f_\theta(x) = \frac{\beta^\alpha}{\Gamma(\alpha)} x^{\alpha-1} e^{-\beta x} = \exp\left\{\log(x) \alpha + \alpha \log \beta - \log \Gamma(\alpha) - \beta x \right\}.$$ This is an exponential family in $\alpha$ alone with $s(x) = \log x$ and $A(\alpha) = \log \Gamma(\alpha) - \alpha \log \beta$. It now follows immediately by computing $\frac d {d\alpha} A(\alpha)$ that $$E[\log X] = \psi(\alpha) - \log \beta.$$