Assume two independent and continuous variables $A$ and $B$, with densities $f_A$ and $f_B$. Let $F_A$ and $F_B$ denote the their respective cumulative distribution functions, with $F(t) = \int_{-\infty}^t f(x) \;\; dx$. Finally, assume that $\mathbb{E}(A) \geq \mathbb{E}(B)$.
Now, consider the probability that a sample from $A$ is larger than a sample from $B$, conditional on both samples being below a value $\kappa$; i.e., $P(A > B \mid A, B < \kappa)$. This probability is given by $$R(\kappa) = \displaystyle\int_{-\infty}^\kappa \frac{f_A(x)F_B(x)}{F_A(\kappa)F_B(\kappa)} \;\; dx.$$
I am trying to show that $R(\kappa)$ is monotonically increasing function. When differentiating over $\kappa$, I end up with $$R(\kappa)^\prime = \frac{f_A(\kappa)}{F_A(\kappa)} - R(\kappa)\times\left(\frac{f_A(\kappa)}{F_A(\kappa)} + \frac{f_B(\kappa)}{F_B(\kappa)}\right).$$
So far so good, now I just need to show that $R(\kappa)^\prime$ is strictly non-negative. But this where I am stuck.
The reason why I am trying to show this is that I found this pattern with different families of distributions, like the Normal (same $\sigma^2$ for both $A$ and $B$) and the exponential. See the Figure attached below.
Any help is greatly appreciated! Thanks!
Fundamental theorem of calculus states the following,
Let $f$ be a continuous real valued function defined on $[a,b]$ and let $F$ be a function defined on $[a,b]$ by $$F(x) = \int_a^xf(t)dt.$$ Then $F$ is uniformly continuous on $[a,b]$, differentiable on $(a,b)$, and $$F'(x) = f(x), x \in(a,b)$$
Therefore I think, with your $$R(k)=\int_{-\infty}^k {f_A(x)F_B(x) \over F_A(k)F_B(k)}dx$$ $$R'(k) = {f_A(k)F_B(k)\over F_A(k)F_B(k)}={f_A(k)\over F_A(k)} > 0 ,$$ since $f_A(x)\ge0$ for all $x$.
Could you explain how you arrived at your derivative of $R(k)$?