Absolute error loss for a gamma random variable

Question

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.

Answer 1

(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)

Answer 2

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 $a=0.374^-$.