# Expected value of $\ln X$ if $X$ is $\Gamma(a,b)$ distributed.

I'm new here and hope you can help.

It's really late here in South Africa, maybe my mind just doesn't want to function now! But I need to figure out how to get a closed form expression hopefully for $E(\ln X)$ and even $E(\ln (X^2))$ if $X$ is $\Gamma(a,b)$ distributed. Any help would be greatly appreciated!!

Scrofungulus

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Is it $E(\ln (X^2))$ or $E(\ln^2 (X))$ ? –  Pedro Tamaroff Apr 28 '12 at 23:01
$\ln (X^2)$ is the same as $2\ln(X)$, so its expected value is just $2$ times that of $\ln X$. If you meant $(\ln X)^2$, that should be written differently. –  Michael Hardy Apr 29 '12 at 0:00
Wikipedia says, and Maple confirms, that $E(\ln X) = -\ln b + \psi(a)$, where $\psi(t) = \frac{d}{dt} \ln \Gamma(t)$ is the digamma function. –  Nate Eldredge Apr 29 '12 at 0:10

The probability density function for the gamma distribution is $$f(x;a,b) = \frac{1}{b^a\Gamma(a)} x^{a-1}e^{-x/b},$$ so the integral we must consider is $$\mathbb{E}(\ln (X^n)) = \frac{1}{b^a \Gamma(a)} \int_0^\infty dx\, x^{a-1} e^{-x/b} \ln x^n.$$ Let $z=x/b$. We find $$\begin{eqnarray*} \mathbb{E}(\ln (X^n)) &=& \frac{n}{\Gamma(a)} \int_0^\infty dz\, z^{a-1} e^{-z}\ln (b z) \\ &=& n\ln b + \frac{n}{\Gamma(a)} \int_0^\infty dz\, z^{a-1} e^{-z}\ln z \\ &=& n\ln b + \frac{n}{\Gamma(a)} \frac{d}{d a} \int_0^\infty dz\, z^{a-1} e^{-z} \\ &=& n\left(\ln b + \frac{\Gamma'(a)}{\Gamma(a)}\right) \\ &=& n\left(\ln b + \psi(a)\right), \end{eqnarray*}$$ where we have used the definition of the gamma function, $\Gamma(a) = \int_0^\infty d x \, x^{a-1} e^{-x}$, and the digamma function, $\psi(z) = \Gamma'(z)/\Gamma(z)$.

Addendum 1: It is possible to show (using the fact that $\ln x\leq z^s/s$ for all $s>0$) that $\int_0^\infty dz\, z^{a-1} e^{-z}\ln z$ converges uniformly for $1< a < \infty$. This justifies the trick of differentiating $\Gamma(a)$.

Addendum 2: There is a possibility that the OP is interested in $\mathbb{E}((\ln X)^n)$. @Sasha dealt with this below. It can also be handled with the derivative trick, $$\begin{eqnarray*} \mathbb{E}((\ln X)^n) &=& \frac{1}{\Gamma(a)} \int_0^\infty dz\, z^{a-1} e^{-z}(\ln (b z))^n \\ &=& \frac{1}{\Gamma(a)} \int_0^\infty dz\, z^{a-1} e^{-z}(\ln b + \ln z)^n \\ &=& \frac{1}{\Gamma(a)} \int_0^\infty dz\, z^{a-1} e^{-z} \sum_{k=0}^n {n \choose k} (\ln b)^{n-k} (\ln z)^k \\ &=& \frac{1}{\Gamma(a)} \sum_{k=0}^n {n \choose k} (\ln b)^{n-k} \int_0^\infty dz\, z^{a-1} e^{-z} (\ln z)^k \\ &=& \frac{1}{\Gamma(a)} \sum_{k=0}^n {n \choose k} (\ln b)^{n-k} \frac{d^k}{d a^k} \int_0^\infty dz\, z^{a-1} e^{-z} \\ &=& \frac{1}{\Gamma(a)} \sum_{k=0}^n {n \choose k} (\ln b)^{n-k} \Gamma^{(k)}(a). \end{eqnarray*}$$ This is equivalent to the solution given by @Sasha. For $n=2$ we arrive at $$\mathbb{E}((\ln X)^2) = (\ln b + \psi(x))^2 + \psi^{(1)}(a),$$ where $\psi^{(m)}(z) = d^{m+1}(\ln \Gamma(z))/d z^{m+1}$ is the polygamma function.

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The exponential generating function for expectations of the powers of $\log(X)$ is $$\begin{eqnarray} \sum_{r=0}^\infty \frac{t^r}{r!} \mathbb{E}(\log^r(X)) &=& \mathbb{E}(\sum_{r=0}^\infty \frac{t^r}{r!} \log^r(X)) = \mathbb{E}\left(X^t\right) = \int_0^\infty x^t \frac{b^{-a}}{\Gamma(a)} x^{a-1} \mathrm{e}^{-\frac{x}{b}} \mathrm{d} x = b^{t} \frac{\Gamma(t+a)}{\Gamma(a)} \end{eqnarray}$$ Hence: $$\mathbb{E}\left(\log(X)\right) = \left. \frac{\mathrm{d}}{\mathrm{d} t} b^{t} \frac{\Gamma(t+a)}{\Gamma(a)} \right|_{t=0} = \log(b) + \psi(a)$$ $$\mathbb{E}\left(\log^2(X)\right) = \left. \frac{\mathrm{d}^2}{\mathrm{d} t^2} b^{t} \frac{\Gamma(t+a)}{\Gamma(a)} \right|_{t=0} = \log^2(b) + 2 \log(b) \psi(a) + \psi(a)^2 + \psi^{(1)}(a)$$ where $\psi(a)$ denotes the digamma function, and $\psi^{(1)}(a)$ denotes the trigamma function.
Thanks for the edit. Looking at the generating function for $\mathbb{E}(\log^r X)$ is very clever. (+1) –  user26872 Apr 29 '12 at 6:33