Let $f$ and $F$ denote, respectively, the pdf and cdf of a probability distribution on $\mathbb R$. Take any natural $n\ge3$ and any real $a$ and $c$ such that $a\le c$. 


Does it always follow that  

$$
\int\_{a}^{c}e^{\rho(b-a)}F(b)^{n-3}\left[ (n-1)F(b)-(n-2)F(c)\right] f(b)db
$$
is nondecreasing in $\rho\ge0$? 
(The displayed integral is related to order statistics.) 



Proving that an integral related to order statistics is increasing in a certain parameter
The answer is yes. The derivative in $\rho$ of the integral in question can be written as 
\begin{equation}
 J:=\int\_{a}^{c}G(b)h(b)\,db, 
\end{equation}
where 
\begin{equation}
 G(b):=(b-a)e^{\rho(b-a)},
\end{equation}
\begin{equation} 
 h(b):=F(b)^{n-3}\left[ (n-1)F(b)-(n-2)F(c)\right] f(b). 
\end{equation}
It is enough to show that $J\ge0$. 


Clearly, there is some $b\_\*\in[a,c]$ such that 
$h(b)\le0$ for $b\in[a,b\_\*)$ and $h(b)\ge0$ for $b\in[b\_\*,c]$. 
Moreover, the function $G$ is increasing, for each $\rho\ge0$. 
Therefore, 
\begin{equation}
 J=\int\_{[a,b\_\*)}G(b)h(b)\,db+\int\_{[b\_\*,c]}G(b)h(b)\,db
\end{equation}
\begin{equation} 
 \ge \int\_{[a,b\_\*)}G(b\_\*)h(b)\,db+\int\_{[b\_\*,c]}G(b\_\*)h(b)\,db
\end{equation}
\begin{equation} 
 =G(b\_\*)\int\_{[a,c]}h(b)\,db
 =G(b\_\*)F(a)^{n-2}[F(c)-F(a)]\ge0, 
\end{equation}
as desired.


