expectation of log(1+x) if x is a gamma random variable - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-25T07:44:02Z http://mathoverflow.net/feeds/question/85418 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/85418/expectation-of-log1x-if-x-is-a-gamma-random-variable expectation of log(1+x) if x is a gamma random variable Jose M Del Castillo 2012-01-11T14:42:00Z 2012-01-24T22:13:06Z <p>I would like to know if there is a closed form expression for the expectation of log(1+x) when x is a gamma random variable. </p> <p>Thank you.</p> http://mathoverflow.net/questions/85418/expectation-of-log1x-if-x-is-a-gamma-random-variable/85446#85446 Answer by Robert Israel for expectation of log(1+x) if x is a gamma random variable Robert Israel 2012-01-11T19:55:53Z 2012-01-11T19:55:53Z <p>If $X$ has the gamma distribution with rate $\lambda$ and shape parameter $n$, you're asking for $$J(\lambda, n) = \frac{\lambda^n}{\Gamma(n)} \int_0^\infty t^{n-1} e^{-\lambda t} \log(1+t)\ dt = \frac{1}{\Gamma(n)} \int_0^\infty s^{n-1} e^{-s} \log(1+s/\lambda) \ ds$$</p> <p>Using Maple, I get</p> <p>$$\Psi \left( n \right) -\ln \left( \lambda \right) +{\frac { {\mbox{_2F_2}(1,1;\,2,2-n;\lambda)}\lambda}{n-1}}+{\frac { \left( -1 \right) ^{-n}\pi }{\sin \left( \pi n \right) }}-{\frac { \left( -1 \right) ^{-n}\pi \Gamma \left( n,-\lambda \right) }{\sin \left( \pi n \right) \Gamma \left( n \right) }}$$</p> <p>which seems to be correct when $n$ is a non-integer. For integer values of $n$, the result seems to be $\frac{\Gamma(n,-\lambda)}{\Gamma(n)} Ei(1,\lambda)$ plus a polynomial in $\lambda$ of degree $n-2$.</p> http://mathoverflow.net/questions/85418/expectation-of-log1x-if-x-is-a-gamma-random-variable/86579#86579 Answer by Avi for expectation of log(1+x) if x is a gamma random variable Avi 2012-01-24T22:13:06Z 2012-01-24T22:13:06Z <p>I may be mistaken, but if you are making the change of variable $s = \lambda t$, shouldn't there be an extra factor of $\lambda$ outside the integral? </p>