Timeline for Solve $\inf_{ X: |X| \le a \text{ a.s.}} E \left[ \frac{1}{1+(X-X^\prime)^2} \right] $
Current License: CC BY-SA 3.0
16 events
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| Sep 11, 2017 at 21:33 | comment | added | Mateusz Kwaśnicki | @Boby: Arguing in as in Christian's answer, the convolution $F(x)$ of $\exp(-|x|)$ and the minimising distribution $\mu$ must be constant on $[-a,a]$. (By the way, this is a fundamental result in potential theory). Therefore, $F(x)=c$ for $x\in[-a,a]$, $F(x)=c\exp(a+x)$ for $x<-a$ and $F(x)=c\exp(a-x)$ for $x>a$. Now you can either guess the measure $\mu$, or evaluate it as $\mu=\tfrac{1}{2}(-\Delta+1)F$ in the sense of distributions ($\tfrac{1}{2}(\Delta-1)$ is the inverse of the convolution operator with kernel $\exp(-|x|)$). The constant $c$ is now determined by $\mu([-a,a])=1$. | |
| Sep 11, 2017 at 21:20 | comment | added | Boby | @MateuszKwaśnicki Yes, I would love to get the explicit answer. Should I post a new question? | |
| Sep 11, 2017 at 20:02 | history | edited | Christian Remling | CC BY-SA 3.0 |
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| Sep 11, 2017 at 20:01 | comment | added | Christian Remling | @MateuszKwaśnicki: You're right, $(1/2)(\delta_{-a}+\delta_a)$ is optimal for small $a$. I miscalculated originally (I thought $F(0)<F(\pm a)$ always, but that's not true). The scenario you describe sounds plausible to me. | |
| Sep 11, 2017 at 19:48 | comment | added | Mateusz Kwaśnicki | @Boby: On the other hand, minimisers of $\mathbb{E}\exp(-|X-X'|)$ can be found explicitly, as this corresponds to the energy functional of the Brownian motion killed at a uniform rate. The minimising measure can be evaluated to be $(2+a)^{-1}(\delta_{-a}(dx)+\delta_a(dx)+dx)$, a mixture of uniform and two-point distributions. I can provide the details if you are interested. | |
| Sep 11, 2017 at 19:40 | comment | added | Mateusz Kwaśnicki | @Boby: For $\exp(-(X-X')^2)$ and $a$ small enough, so that $\exp(-x^2)$ is concave on $[-2a,2a]$, the minimiser is a two-point distribution: this is true for any cost function which is a concave and decreasing function of $|X-X'|$. I expect that for large $a$ the minimiser will again be in some sense close to the uniform distribution. | |
| Sep 11, 2017 at 19:37 | comment | added | Mateusz Kwaśnicki | @ChristianRemling: Out of curiosity, I did some experiments: apparently the minimiser is a two-point distribution for $a \in [0, \sqrt{2}/2]$; then a third atom at zero appears for $a$ between $\sqrt{2}/2$ and roughly $1.2$; and then further atoms emerge. I believe this behaviour can be proved rigorously, but I did not attempt to do that. | |
| Sep 11, 2017 at 19:25 | comment | added | Christian Remling | @Boby: I don't know really off the top of my head, right now my feeling is this is going to have the same general features as the question you asked. | |
| Sep 11, 2017 at 19:06 | comment | added | Boby | Do you think $E[ e^{-(X-X^\prime)^2}]$ has an easier solution? Should I ask a new question on this? | |
| Sep 11, 2017 at 17:36 | comment | added | Mateusz Kwaśnicki | Oh, that is of course correct, sorry! | |
| Sep 11, 2017 at 15:48 | comment | added | Christian Remling | @MateuszKwaśnicki: The two terms are equal to one another, they are both double integrals $\int d\sigma(x)\int d\mu(t)\, 1/(1+(x-t)^2)$ with just the variables swapped in the second term. | |
| Sep 9, 2017 at 21:09 | comment | added | Mateusz Kwaśnicki | I think there is an error in your evaluation of the first variation: if $F_\mu(x) = \int (1+x^2)^{-1} \mu(dx)$, then $\int F_{\mu+h\sigma}(x) (\mu+\sigma)(dx) \approx \int F_\mu(x) \mu(dx) + h(\int F_\mu(x)\sigma(dx)+\int F_\sigma(x)\mu(dx)) + o(h^2)$, so apparently you are missing $\int F_\sigma(x)\mu(dx)$. I am not sure, though, if it changes the result. | |
| Sep 9, 2017 at 18:50 | history | edited | Christian Remling | CC BY-SA 3.0 |
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| Sep 9, 2017 at 17:32 | history | edited | Christian Remling | CC BY-SA 3.0 |
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| Sep 9, 2017 at 17:25 | history | edited | Christian Remling | CC BY-SA 3.0 |
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| Sep 9, 2017 at 17:15 | history | answered | Christian Remling | CC BY-SA 3.0 |