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Let $X \sim \operatorname{Gamma}(2,1)$, I would like to minimize with respect to $a$ $E|aX-1|=\int_0^{1/a}(1-ax)xe^{-x}dx+\int_{1/a}^\infty (ax-1)xe^{-x}dx$

Is there some neat way to do this? The only way I know is to use calculus on the RHS to find the minimum with respect to $a$. By neat, I mean a way that use facts from probability or gamma function? Thanks.

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    @Nicolas: Actually the rate parameter is just the inverse of the scale parameter. So they encode the same quantity. But, in this particular case they are both equal to 1, which is what makes the notation ambiguous *even* after you've given the density. :)2011-10-01

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(This is a not answer -see Didier's- rather a comment). For $a>0$, $E( | a X - 1 | ) = a E( | X - a^{-1}| )$ so the problem is equivalent to find $b>0 $ ($b= 1/a$) that minimizes $ g(b) = \frac{E(|X-b|)}{b} = \frac{h(b)}{b} $

All we know from "facts from probability" is that the median of $X$ minimizes $h(b)$, but this does not lead to a solution of $g(b)$ (all that we can expect is that the minimum happens for some $b_0 > med(X) \approx 1.67$ ) but this is not very useful (the median of a gamma variable has not a closed form and the bound is not tight)

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    I think both o$f$ your post excellent. I mark the post o$f$ leonbloy as answer, because he elaborates more on the "facts from probability". But, Didier post is very good too. Thanks.2011-09-30
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Differentiate $\mathrm E(\,\mid aX-1\mid\, )$ with respect to $a$. The result is $ E(X\,;\,aX>1)-E(X\,;\,aX<1)=E(X)-2E(X\,;\,aX<1). $ If $X$ is Gamma$(2,1)$, this is zero when $t=1/a$ solves $ t^2+2t+2=\mathrm e^t, $ that is, for $a=0.374^-$.

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    @NicolasEssis-Breton: Yes, that equation can only be solved numerically (same as the median) http://www.wolframalpha.com/input/?i=solve+t%5E2%2B2+t+%2B2+%3D+exp%28t%292011-09-30