Consider the integral given by
$V(\Lambda) = \int_0^1 F(x) [1 - \Lambda +\Lambda F(x)] [x f(x) - 1 + F(x)] - [1-F(x)][1 + \Lambda F(x)][F(x) + f(x) x]\; dx $,
where $F$ is the cdf of some distribution on $[0,1]$, $f$ is its pdf, and $\Lambda \in [0,1]$ is some parameter. I want to show that for any continuous cdf $F$ we have $V(\Lambda)\leq 0$. It has already been shown that $V(0)<0$.
For the uniform distribution we have $V(\Lambda) = - (1 + \Lambda)/6$. Moreover, I have run simulations in R for the beta distribution using different shape parameters which indeed always yielded negative values for $V(\Lambda)$.
So far, however, I have failed to prove $V(\Lambda)\leq0$ for arbitrary, continuous $F$. My latest approach is the following. As I've mentioned, I know $V(0) <0$. Thus, to show that $V(\Lambda)\leq$ for any $\Lambda \in [0,1]$ it is sufficient to show that $V'(\Lambda)\leq0$. We have
$V'(\Lambda) = \int_0^1 F(x) [F(x)-1][2 F(x) + 2 f(x)x - 1]\; dx$.
Thus, for sufficiently small values of $x$ the integrand is positive, while it is negative for larger values. The cdf $F(x) = x^n$ with larger $n$ is such that the integrand will be positive for many values of $x$ and might thus be a suitable candidate for a counter example. However, we find that for this specification we get
$V'(\Lambda) = - \frac{n(1 + n + 2n^2)}{1 + 6n + 11n^2 + 6 n^3}<0$,
which is monotonically decreasing in $n$ and reaches $-1/3$ in the limit.
Any advice is greatly appreciated, thanks!
My latest approach has worked out. The claim is that $V(\Lambda) <0$ for all $\Lambda \in [0,1]$. As I mentioned, it was already shown that $V(0)<0$. In what follows I will show that $V'(\Lambda)<0$ which implies the claim.
Notice that we can write
$V'(\Lambda) = \int_0^1 F(x) (1- F(x))^2 - F^2(x)(1-F(x)) - 2x f(x) F(x) (1-F(x))\; dx $.
I will now consider the order statistics with sample size 3 to proceed. Recall that the cdf of the first (largest), second, and third (smallest) order statistic are given by \begin{align*} F_{1, 3}(x) &= F^3(x),\\ F_{2, 3}(x) &= 3 (1- F(x))F^2(x) + F^3(x),\\ F_{3,3}(x) &= 3(1- F(x))^2F(x) + 3 (1- F(x))F^2(x) + F^3(x). \end{align*}
Next, notice that \begin{align*} F(x) (1- F(x))^2 &= (F_{2,3}(x) - F_{1,3}(x))/3,\\ F^2(x)(1-F(x)) &= (F_{3,3}(x) - F_{2,3}(x))/3. \end{align*} Hence, we can rewrite the derivative to \begin{align*} V'(\Lambda) &= \int_0^1 (F_{2,3}(x) - F_{1,3}(x))/3 - (F_{3,3}(x) - F_{2,3}(x))/3 \; dx - \mathbb{E}[X_{2,3}]/3, \end{align*} where $\mathbb{E}[X_{2,3}]$ is the expectation of the second order statistic, because $ 2f(x) F(x) (1-F(x)) = f_{2,3}(x)/3$. Making use of the identity \begin{align*} \mathbb{E}[X] = \int_0^1 (1- F(x)) \; dx \end{align*} we can further rewrite the expression to \begin{align*} V'(\Lambda) &= \frac{1}{3}\int_0^1 \left((1- F_{1,3}(x)) - (1-F_{2,3}(x))\right) - ((1-F_{2,3}(x))-(1-F_{3,3}(x)) ) \; dx - \mathbb{E}[X_{2,3}]\\ &= \frac{1}{3} \left(\mathbb{E}[X_{2,3}]-\mathbb{E}[X_{1,3}]-\mathbb{E}[X_{3,3}]\right)\\ &<0 \end{align*} where the inequality is strict because we are not considering continuous distributions only.
I would be happy if someone could verify this.