Index: The Book of Statistical ProofsProbability Distributions ▷ Univariate continuous distributions ▷ Gamma distribution ▷ Logarithmic expectation

Theorem: Let $X$ be a random variable following a gamma distribution:

$\label{eq:gam} X \sim \mathrm{Gam}(a, b) \; .$

Then, the expectation of the natural logarithm of $X$ is

$\label{eq:gam-logmean} \mathrm{E}(\ln X) = \psi(a) - \ln(b)$

where $\psi(x)$ is the digamma function.

Proof: Let $Y = \ln(X)$, such that $\mathrm{E}(Y) = \mathrm{E}(\ln X)$ and consider the special case that $b = 1$. In this case, the probability density function of the gamma distribution is

$\label{eq:X-pdf-s1} f_X(x) = \frac{1}{\Gamma(a)} \, x^{a-1} \, \mathrm{exp} [-x] \; .$

Multiplying this function with $\mathrm{d}x$, we obtain

$\label{eq:X-pdf-s2} f_X(x) \, \mathrm{d}x = \frac{1}{\Gamma(a)} \, x^a \, \mathrm{exp} [-x] \, \frac{\mathrm{d}x}{x} \; .$

Substituting $y = \ln x$, i.e. $x = e^y$, such that $\mathrm{d}x/\mathrm{d}y = x$, i.e. $\mathrm{d}x/x = \mathrm{d}y$, we get

$\label{eq:Y-pdf-s1} \begin{split} f_Y(y) \, \mathrm{d}y &= \frac{1}{\Gamma(a)} \, \left( e^y \right)^a \, \mathrm{exp} [-e^y] \, \mathrm{d}y \\ &= \frac{1}{\Gamma(a)} \, \mathrm{exp}\left[ ay - e^y \right] \, \mathrm{d}y \; . \end{split}$

Because $f_Y(y)$ integrates to one, we have

$\label{eq:Y-pdf-s2} \begin{split} 1 &= \int_{\mathbb{R}} f_Y(y) \, \mathrm{d}y \\ 1 &= \int_{\mathbb{R}} \frac{1}{\Gamma(a)} \, \mathrm{exp}\left[ ay - e^y \right] \, \mathrm{d}y \\ \Gamma(a) &= \int_{\mathbb{R}} \mathrm{exp}\left[ ay - e^y \right] \, \mathrm{d}y \; . \end{split}$

Note that the integrand in \eqref{eq:Y-pdf-s2} is differentiable with respect to $a$:

$\label{eq:dfy-da} \begin{split} \frac{\mathrm{d}}{\mathrm{d}a} \mathrm{exp}\left[ ay - e^y \right] \, \mathrm{d}y &= y \, \mathrm{exp}\left[ ay - e^y \right] \, \mathrm{d}y \\ &\overset{\eqref{eq:Y-pdf-s1}}{=} \Gamma(a) \, y \, f_Y(y) \, \mathrm{d}y \; . \end{split}$

Now we can calculate the expected value of $Y = \ln(X)$:

$\label{eq:E-Y-s1} \begin{split} \mathrm{E}(Y) &= \int_{\mathbb{R}} y \, f_Y(y) \, \mathrm{d}y \\ &\overset{\eqref{eq:dfy-da}}{=} \frac{1}{\Gamma(a)} \int_{\mathbb{R}} \frac{\mathrm{d}}{\mathrm{d}a} \mathrm{exp}\left[ ay - e^y \right] \, \mathrm{d}y \\ &= \frac{1}{\Gamma(a)} \frac{\mathrm{d}}{\mathrm{d}a} \int_{\mathbb{R}} \mathrm{exp}\left[ ay - e^y \right] \, \mathrm{d}y \\ &\overset{\eqref{eq:Y-pdf-s2}}{=} \frac{1}{\Gamma(a)} \frac{\mathrm{d}}{\mathrm{d}a} \Gamma(a) \\ &= \frac{\Gamma'(a)}{\Gamma(a)} \; . \end{split}$

Using the derivative of a logarithmized function

$\label{eq:log-der} \frac{\mathrm{d}}{\mathrm{d}x} \ln f(x) = \frac{f'(x)}{f(x)}$

and the definition of the digamma function

$\label{eq:psi} \psi(x) = \frac{\mathrm{d}}{\mathrm{d}x} \ln \Gamma(x) \; ,$

we have

$\label{eq:E-Y-s2} \mathrm{E}(Y) = \psi(a) \; .$

Finally, noting that $1/b$ acts as a scaling parameter on a gamma-distributed random variable,

$\label{eq:gam-sgam} X \sim \mathrm{Gam}(a,1) \quad \Rightarrow \quad \frac{1}{b} X \sim \mathrm{Gam}(a,b) \; ,$

and that a scaling parameter acts additively on the logarithmic expectation of a random variable,

$\label{eq:logmean} \mathrm{E}\left[\ln(cX)\right] = \mathrm{E}\left[\ln(X) + \ln(c)\right] = \mathrm{E}\left[\ln(X)\right] + \ln(c) \; ,$

it follows that

$\label{eq:E-Y-s3} X \sim \mathrm{Gam}(a,b) \quad \Rightarrow \quad \mathrm{E}(\ln X) = \psi(a) - \ln(b) \; .$
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Metadata: ID: P110 | shortcut: gam-logmean | author: JoramSoch | date: 2020-05-25, 21:28.