Problem:
I am struggling for some time to show that a particular inequality holds or to find a counterexample to disprove it:
Suppose you have two continuous random variables $X_1, X_2$ with densities $f_1$ and $f_2$, respectively, which are both distributed on the same compact interval $[\underline{x},\overline{x}]\subset \mathbb{R}_+$ and have the following properties:
1) Both densities are log-concave.
2) $f_i(\underline{x})\cdot\underline{x} \le 1$ $\hspace{2ex}$ $i \in \{1,2\}$
3) $X_1$ stochastically dominates $X_2$ in the Likelihood Ratio Order, i.e. $\forall x <x'$, $x,x' \in [\underline{x},\overline{x}]$ it holds that: \begin{equation} \frac{f_1(x)}{f_2(x)} \le \frac{f_1(x')}{f_2(x')} \end{equation} Note that the third property implies that $X_1$ stochastically dominates $X_2$ also in the Hazard Rate-, Reverse Hazard Rate- and First Order.
Next, consider two particular realizations $\hat{x}_1$ and $\hat{x}_2$ which are implicitly defined in the following way:
\begin{align*} \hat{x}_1 = \frac{1-2\cdot F_1(\hat{x}_1)}{f(\hat{x}_1)} \hspace{4ex} \hat{x}_2 = \frac{1-2\cdot F_2(\hat{x}_2)}{f(\hat{x}_2)} \end{align*} Properties 1) and 2) are sufficient for existence of such values in $[\underline{x},\overline{x}]$ and property 3) implies that $\hat{x}_1 > \hat{x}_2$.
With all this information, I wanted to show that $F_1(\hat{x}_1) \le F_2(\hat{x}_2)$. Since I miserably failed in doing it so far, I also tried to find counterexamples and haven't found any.
Approach:
Start with the implicit equations defining $\hat{x}_1$ and $\hat{x}_2$. Rearranging them yields: \begin{align} &F_1(\hat{x}_1) = \frac{1-\hat{x}_1 \cdot f_1(\hat{x}_1)}{2}\\ &F_2(\hat{x}_2) = \frac{1-\hat{x}_2 \cdot f_2(\hat{x}_2)}{2} \end{align} Now, in order for the desired inequality to hold, namely $F_1(\hat{x}_1) \le F_2(\hat{x}_2)$, it suffices to show that: \begin{align} \frac{1-\hat{x}_1 \cdot f_1(\hat{x}_1)}{2} \le \frac{1-\hat{x}_2 \cdot f_2(\hat{x}_2)}{2} \Leftrightarrow \hat{x}_1 \cdot f_1(\hat{x}_1) \ge \hat{x}_2 \cdot f_2(\hat{x}_2) \end{align} Now, since $\hat{x}_1 > \hat{x}_2$, by property 3) we know that: \begin{align} \frac{f_1(\hat{x}_1)}{f_2(\hat{x}_1)} \ge \frac{f_1(\hat{x}_2)}{f_2(\hat{x}_2)} \Leftrightarrow f_1(\hat{x}_1) \ge \frac{f_2(\hat{x}_1)\cdot f_1(\hat{x}_2)}{f_2(\hat{x}_2)} \end{align} Using this for the original inequality, it now sufficies to show that: \begin{align} \hat{x}_1 \cdot \frac{f_2(\hat{x}_1)\cdot f_1(\hat{x}_2)}{f_2(\hat{x}_2)} \ge \hat{x}_2 \cdot f_2(\hat{x}_2) \Leftrightarrow \hat{x}_1 \cdot f_2(\hat{x}_1)\cdot f_1(\hat{x}_2) \ge \hat{x}_2 \cdot f_2(\hat{x}_2)^2 \end{align} And this is the step in this approach where I am stuck...
I would really appreciate some help. Thanks!
