Friday, 1 April 2016

probability - Absolute error loss for a gamma random variable




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



(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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