Let $\mathscr P \_0$ and $\mathscr P \_1$ be two non-overlapping sets of probability distributions defined on $(\Omega,\mathcal{A})$. Consider the distance defined as $$D\_u(P\_0,P\_1)=\int\_\Omega \left(\frac{p\_1}{p\_0}\right)^u p\_0 \mathrm{d}\mu<\infty.$$ Two distributions are chosen from each set $Q\_0\in\mathscr P \_0$ and $Q\_1\in\mathscr P \_1$ such that $$D\_u(Q\_0,Q\_1)\geq D\_u(P\_0,P\_1)\quad \forall (P\_0,P\_1)\in \mathscr P \_0\times \mathscr P \_1,\forall u\in[0,1]$$



> 
> Is it true that (**A:**)$$\int \min(q\_0,t q\_1)\mathrm{d}\mu\geq \int \min(p\_0,t p\_1)\mathrm{d}\mu\quad \forall (P\_0,P\_1)\in \mathscr P \_0\times \mathscr P \_1,\forall t$$or equaivalently (**B:**) $$Q\_0\left[\frac{q\_1}{q\_0}>t\right]\geq P\_0\left[\frac{q\_1}{q\_0}>t\right]\quad\ \forall P\_0\in \mathscr P \_0,\forall t$$
>  $$Q\_1\left[\frac{q\_1}{q\_0}\leq t\right]\geq P\_1\left[\frac{q\_1}{q\_0}\leq t\right]\quad\ \forall P\_1\in \mathscr P \_1,\forall t$$ 
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**Notes:**


$\bullet$ One can consider **A** or **B** since both conditions are equivalent.


$\bullet$ $p\_0$ and $p\_1$ are densities of $P\_0$ and $P\_1$ and the same goes to $q\_0$ and $q\_1$ with $Q\_0$ and $Q\_1$.


**What I know:** 


From [Huber's paper (pages 260-261) Theorem 6.1](https://projecteuclid.org/download/pdf_1/euclid.aos/1176342363) I know that if the distance is the $f$-divergence, i.e. $D\_f$, then **A** and **B** are correct. Additionally, if **A** and **B** are correct, then $Q\_0$ and $Q\_1$ minimize $D\_f$ (iff condition).


Huber considers $$Q\_{jt}=(1-t)Q\_{0t}+t Q\_{1t}\\q\_{jt}=(1-t)q\_{0t}+t q\_{1t}$$
and finds the first and second derivatives of $D\_f(Q\_{0t},Q\_{1t})$. He then shows that the second derivative is $\geq 0$ (convex) and hence $(Q\_{00},Q\_{10})$ minimizes $D\_f$ **if and only if** the first derivative evaluated at $t=0$ is $\geq 0$ for all $(Q\_{01},Q\_{11})\in(\mathscr P \_0\times\mathscr P \_1)$. He shows that this is really the case, hence the claim is true.


I think that this result can be strenghtened, i.e. if $(Q\_0,Q\_1)$ maximizes $D\_u$ for all $u\in[0,1]$, then it should satisfy **A** or equivalently **B**. I dont know how to proceed.


**Addendum:** It seems that the question eventually boils down to finding $(Q\_0,Q\_1)$ which maximizes $D\_u$ for all $u\in[0,1]$ and fails to minimize $D\_f$ for at least one $f$. This will be a counterexample to the claim (of course if there exists such a pair).



Does maximizing $D_u$ imply stochastic ordering?
Answer: this does NOT extends to other cases than f-divergence.


For the case $u=0$, there are many counter examples since $D\_u$ is not separating any densities. So I think it suffices to consider $u\in (0,1]$.


Let $U^n=Unif(0,\frac{1}{n})$ the uniform measure on unit interval. And $\mathscr{P}\_0$ corresponds to those with odd $n$, $\mathscr{P}\_1$ contains exactly those with even $n$.
Let $Q\_0=U^n,Q\_1=U^m$,
 thus $D\_u(Q\_0,Q\_1)< \infty$ if $n<m$; and $=\infty$ otherwise due to the fact that $$\frac{q\_{1}}{q\_{0}}=\frac{m\boldsymbol{1}\_{[0,\frac{1}{m}]}(\omega)}{n\boldsymbol{1}\_{[0,\frac{1}{n}]}(\omega)}=\begin{cases}
\frac{m}{n}>1 & \omega\in[0,\frac{1}{m}]\\
0 & \omega\in[\frac{1}{m},\frac{1}{n}]\\
\infty & \omega\in[\frac{1}{n},1]
\end{cases},n<m$$.(It suffices to consider the integration over the last segment w.r.t. $Q\_0$'s density $q\_0$.) 


Therefore $D\_u(Q\_0,Q\_1)\geq D\_u(P\_0,P\_1)\quad \forall (P\_0,P\_1)\in \mathscr P \_0\times \mathscr P \_1,\forall u\in[0,1]$ as long as $Q\_0=U^n,Q\_1=U^m$ satisfying $n\geq m$. Now for any pair $Q\_0=U^n,Q\_1=U^m\text{ such that },n\geq m$, $$\frac{q\_{1}}{q\_{0}}=\frac{m\boldsymbol{1}\_{[0,\frac{1}{m}]}(\omega)}{n\boldsymbol{1}\_{[0,\frac{1}{n}]}(\omega)}=\begin{cases}
\frac{m}{n}\leq1 & \omega\in[0,\frac{1}{n}]\\
\infty & \omega\in[\frac{1}{n},1]
\end{cases},n\geq m$$ 
Let $t=1$ now and$ \left\{ \omega\in[0,1]\mid\frac{q\_{1}}{q\_{0}}>t\right\} =[\frac{1}{n},1]$
$$Q\_{1}\left[\frac{q\_{1}}{q\_{0}}>t\right]=m\cdot\mu\left[[\frac{1}{n},1]\cap[0,\frac{1}{m}]\right]=m\cdot\mu\left[\frac{1}{n},\frac{1}{m}\right]=m[{\frac{1}{m}-\frac{1}{n}}]>0$$


Choose $P\_{1}=U^{k}$ for any $k\geq n$ $$P\_{1}\left[\frac{q\_{1}}{q\_{0}}>t\right]=k\cdot\mu\left[[\frac{1}{n},1]\cap[0,\frac{1}{k}]\right]=0$$


