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Let $\mathcal P$ be the set of probability densities on $[0,1]$ with mean $1/2$, i.e. $p\in \mathcal P$ iff

$$\int_0^1 p(x)dx=1,\quad \int_0^1 xp(x)dx=\frac{1}{2}\quad \mbox{and}\quad p(x)\ge 0, ~~\forall x\in [0,1].$$

How to solve the minimization problem below ?

$$\min_{p\in\mathcal P}~ \left\{V(p) ~:=~ \int_0^1 \log\big(p(x)\big)p(x)dx + \int_0^1 \big(x\log(x)+(1-x)\log(1-x)\big)p(x)dx\right\}.$$

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  • $\begingroup$ Isn't this simply the Gibbs measure $p=\frac{1}{Z}e^{-U(x)}dx$, where $U(x)=x\log x +(1-x)\log(1-x)$ and $Z\int U(x)dx$ the corresponding normalizing constant? This would be the usual minimizer in the absence of your mean constraint, and since your "potential" $U(x)$ is symmetric w.r.t. $x=1/2$ this should do. $\endgroup$
    – leo monsaingeon
    Commented Nov 15, 2021 at 5:12
  • $\begingroup$ @leomonsaingeon Thanks for the reply. I do not know about Gibbs' measure. Do you mind to specify a bit more? I do appreciate if you are able to provide more details $\endgroup$
    – user128095
    Commented Nov 15, 2021 at 5:52

1 Answer 1

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As in the comment by leo monsaingeon, let $$p_*(x):=e^{h(x)}/c,$$ where $h(x):=-x\ln x-(1-x)\ln(1-x)$ and $c:=\int_0^1 e^{h(x)}\,dx$, so that $p$ is a pdf on $(0,1)$ with mean $1/2$, and $$V(p)=\int_0^1(p(x)\ln p(x)-h(x)p(x))\,dx.$$

For any pdf $q$ on $(0,1)$ with $V(q)<\infty$, the directional derivative of $V$ at $p_*$ in the direction of $q-p_*$ is $$\begin{aligned} &\frac d{dt}\,V(p_*+t(q-p_*))\Big|_{t=0} \\ &=\int_0^1 (1+\ln p_*(x)-h(x))(q(x)-p_*(x))\,dx \\ &=\int_0^1 (1+h(x)-\ln c-h(x))(q(x)-p_*(x))\,dx=0, \end{aligned}$$ since $q$ and $p_*$ are pdf's on $(0,1)$.

The crucial point is that the function $V$ is convex (since $u\ln u$ is convex in $u\ge0$, with $0\ln0:=0$). So, $p_*$ is indeed a minimizer of $V$.

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  • $\begingroup$ Thank you very much Iosif for the solution. Indeed, I don't know this methodology and I wish to know where comes from this idea to choose a candidate of the form of $p^*$. I also need to consider a similar optimization problem by replacing $h$ by more general function $H$ and the constraint $\int_0^1p(x)dx=1/2$ by $\int_0^1p(x)dx=m\in [0,1]$. Could you please provide some related references? Many thanks! $\endgroup$
    – user128095
    Commented Nov 15, 2021 at 14:45
  • $\begingroup$ @Neymar : If you take the directional derivatives of $V$ as in this answer, you immediately see that a necessary condition for $p$ to be a minimizer is that $p=p_*$ almost everywhere. Since $V$ is convex, this condition is also sufficient. $\endgroup$
    – Iosif Pinelis
    Commented Nov 15, 2021 at 14:49
  • $\begingroup$ Thanks a lot for the details. If, for example, the constraint is given by $\int_0^1 xp(x)dx=1/4$, is the calculus of variation still adapted? $\endgroup$
    – user128095
    Commented Nov 15, 2021 at 19:24
  • $\begingroup$ @Neymar : Yes, it should work, but the things become more complicated with this additional active restriction. With the mean $1/2$, that restriction was satisfied automatically, but here that is not the case. You may want to post a further question about this separately. $\endgroup$
    – Iosif Pinelis
    Commented Nov 15, 2021 at 19:35
  • $\begingroup$ Great. I creat a post for the further question at mathoverflow.net/questions/408598/… Could you please take a look? $\endgroup$
    – user128095
    Commented Nov 15, 2021 at 19:51

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