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Fix constant reals $A>1$ and $D>0$. Let $f:\mathbb{R}\to[0,\infty)$ be a probability density function on $\mathbb{R}$, i.e. $\int_{-\infty}^\infty f(x)\, dx=1$, that is continuous almost everywhere in $\mathbb{R}$ and satisfying the condition:

$\forall x,x^\prime$ where $|x−x′|\leq D$, it holds $f(x)\leq A \,f(x^\prime)$

Let $S$ be the set of these functions. Is there an optimal one $\hat{f}$ which gives the minimum expected error:

$E_f[|x|]=\int_{-\infty}^\infty |x|\,f(x)dx$

P.S. If it helps, we can assume that $f$ is symmetric i.e. $f(x)=f(−x)$. Note that the expected value is lower bounded by $0$, i.e. there is an infimum. I just need to show that there is a minimum (not necessary to find it). Thanks for any advice !

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    $\begingroup$ Do you want to tell us give us some idea of why you want to know this? $\endgroup$ Aug 7, 2013 at 4:26
  • $\begingroup$ Yes, simply I want to know if there is a minimal function, such that I start looking for it. $\endgroup$
    – ems
    Aug 7, 2013 at 9:48
  • $\begingroup$ The set of these functions can be shown to be convex. Does it help ? $\endgroup$
    – ems
    Aug 7, 2013 at 11:48
  • $\begingroup$ math.stackexchange.com/questions/461410 $\endgroup$
    – Did
    Aug 9, 2013 at 0:58
  • $\begingroup$ Thanks @MartinHairer ! How do you guarantee that moving masses from further points to close points results in a piecewise constant function with some a ? why not for example add more mass to the close point to further improve the expectation ? Also, it seems that you have an implicit assumption that there is an optimal f which I want to guarantee in the first place. $\endgroup$
    – ems
    Sep 18, 2013 at 15:58

1 Answer 1

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OK, first instead of using $A$, I will use $\kappa = D/(A-1)$ which is a more natural parameter for the problem. (If $D$ is small, $\kappa$ is essentially a bound on the logarithmic derivative.) Now the natural guess is that the optimiser will saturate the constraint at every point. Indeed, if it fails in the neighbourhoods of two points, I can transfer a bit of mass from the one further from the origin to the one closer to it and I improve the expectation of $|x|$. This means that I can restrict myself to piecewise constant functions, which I can assume to be symmetric.

Take now $f$ such that $f(x) = b$ for $|x| < a$. Then, in order to saturate the constraint everywhere, I have no more choice: I have to set $f(x) = b/A$ for $a \le |x| < a+D$, $f(x) = b/A^2$ for $a+D \le |x| < a+2D$, etc. Now if I want to get a probability measure, it turns out that I have to choose $$ b = {1\over 2(a+\kappa)}\;. $$ It remains to optimise over $a$ to minimise the expectation. The answer to this exercise is $$ a = \sqrt{\kappa(D+\kappa)}-\kappa\;. $$ With this choice, one then obtains $$ \int |x| f(x)\,dx = \sqrt{\kappa(D+\kappa)}\;. $$ I haven't double-checked the calculations, but this should be about right...

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  • $\begingroup$ PS: In the limit $D \to 0$ with $\kappa$ fixed, you get the exponential distribution, which isn't surprising... $\endgroup$ Aug 13, 2013 at 19:29

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