I would like to attempt to solve the given problem below. I am sorry that I dont know whether such a question is solvable at all, even for just one specific example. If the solution set is empty, I am also helpless about how to show it. I asked this question two years ago at math.stackexchange with no answer or helpful comment.

Given two distinct and continuous probability density functions on real numbers, $f_0$ and $f_1$ consider the following set of density functions:

$$\mathscr{G}_0=\left\{g_0:\int_{\mathbb{R}}\log\left(\frac{g_0(y)}{f_0(y)}\right)g_0(y)\mathrm{d}y\leq \epsilon_0\right\} $$ and $$\mathscr{G}_1=\left\{g_1:\int_{\mathbb{R}}\log\left(\frac{g_1(y)}{f_1(y)}\right)g_1(y)\mathrm{d}y\leq \epsilon_1\right\} $$

for some sufficiently small $\epsilon_0$ and $\epsilon_1$ such that $\mathscr{G}_0$ and $\mathscr{G}_1$ are also distinct. In other words any density $g_0\in \mathscr{G}_0$ is not an element of $\mathscr{G}_1$ and vice verse. Note that $f_0$ and $f_1$ are known and assumed to be positive so that the terms $\log(g_i/f_i)$, $i\in\{0,1\}$, are well defined.

Question: Are there a pair of density functions $g^*_0\in \mathscr{G}_0$ and $g_1^*\in \mathscr{G}_1$ such that

$$g^*_0=\arg\sup_{g_0\in \mathscr{G}_0}\left(\inf_{u\in \mathbb{R}}\ln \int_{\mathbb{R}}\left(\frac{g_1^*(y)}{g_0^*(y)}\right)^u g_0(y)\mathrm{d}y\right)$$ and $$g^*_1=\arg\sup_{g_1\in \mathscr{G}_1}\left(\inf_{u\in \mathbb{R}}\ln \int_{\mathbb{R}}\left(\frac{g_1^*(y)}{g_0^*(y)}\right)^u g_1(y)\mathrm{d}y\right)?$$

  • $\begingroup$ (I.t.N.o.G.), In the definition of $g_0^*$, shouldn't the fraction inside the integrand be inverted, e.g. to $(\dfrac{g^*_0(y)}{g^*_1(y)})$ ? $\endgroup$ – Toughee Feb 23 '17 at 21:39
  • 1
    $\begingroup$ @Toughee this is a good point. Originally, they are not inverted. But if there exist $g^*_0$ and $g^*_1$, for the solution of the two equations above $u$ wont be the same in equation 1 and equation 2. I think it is possible to take $-u$ in the second equation and make infimum over $-u$. Then the $u$ of both equations will be the same. $\endgroup$ – Seyhmus Güngören Feb 23 '17 at 21:50

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