Say $Y$ is a random variable with Laplace distribution with zero mean and variance parameter $b$. I am trying to compute the expectation of $\ln(Y+\alpha)$ ($\alpha>0$), that is: $$\int^{\infty}_{0}\frac{1}{2b}exp(-\frac{y}{b})\ln(y+\alpha)dy+\int^{0}_{-\alpha}\frac{1}{2b}exp(\frac{y}{b})\ln(y+\alpha)dy.$$ It seems difficult to obtain an exact form. I am at least hoping for a lower bound. Note by Jensen's inequality the upper bound is $\ln(\alpha)$.
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$\begingroup$ Please edit typos in your question. $\endgroup$– user64494Jan 4, 2015 at 19:10
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$\begingroup$ You are not computing the expectation of $\ln(Y+\alpha)$ but of $\ln(Y+\alpha)\mathbb{I}(Y>-\alpha)$. $\endgroup$– Xi'an ні війніOct 7, 2015 at 8:28
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1 Answer
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The closed form can be obtained with Maple 18.02 by
with(Statistics): Y := RandomVariable(Laplace(0, b)): Mean(ln(Y+alpha));
$$ 1/2\, \left( {\it Ei} \left( 1,{\frac {\alpha}{b}} \right) {{\rm e}^{2 \,{\frac {\alpha}{b}}}}+2\,\ln \left( \alpha \right) {{\rm e}^{{ \frac {\alpha}{b}}}}+{\it Ei} \left( 1,-{\frac {\alpha}{b}} \right) \right) {{\rm e}^{-{\frac {\alpha}{b}}}}, $$ where the exponential integral $Ei$ is described here.
