Analyzing an Integral I have an integral of the form 
$$\int_{0}^{1} \int_{0}^{y} F\left(\frac{x+y}{2}\right) f(x)f(y)\;dx\; dy $$
where $f$ is a pdf and $F$ is the corresponding cdf. Form of $f$ is not known and could be any continuous distribution. Is there any way I can simplify this expression or analyze it? I have sort of obtained its bounds, but it would be great if there is some other way to analyze it. I have already tried integration by parts.  
 A: This integral equals to 
$$
\mathbb P(2Z \leq X+Y, 0<X<Y<1)=\int_{0}^{1} \int_{0}^{y} F\left(\frac{x+y}{2}\right) f(x)f(y)\;dx\; dy:= I$$
where $X,Y,Z$ are independent random values with the same pdf $f(x)$ and cdf $F(x)$. This is the same probability as 
$$\mathbb P(2Z \leq X+Y, 0<Y<X<1)=\mathbb P(2Z \leq X+Y, 0<Y\leq X<1),$$ therefore we can add this two equal probabilities and obtain 
$$2\mathbb P(2Z \leq X+Y, 0<X<Y<1)= 2\mathbb P(2Z \leq X+Y, 0<X<1, 0<Y<1) = $$ 
$$=\int_{0}^{1} \int_{0}^{1} F\left(\frac{x+y}{2}\right) f(x)f(y)\;dx\; dy.$$
So, the initial integral $I$ is one half of the integral over unit square:
$$I=\frac12\mathbb P(2Z \leq X+Y, 0<X<1, 0<Y<1) = \frac12 \int_{0}^{1} \int_{0}^{1} F\left(\frac{x+y}{2}\right) f(x)f(y)\;dx\; dy.$$
Then $0\leq I\leq \frac12$
No more precise conclusions about the integral can be made in general case. Draw the unit cube with the X, Y, Z axes and the area $Z<\tfrac{X+Y}{2}$. One need to integrate joint density of $X,Y,Z$ through this area. If pdf in a neighborhood of unity is large, then the integral can be made arbitrarily close to 1. Conversely, if pdf is large near zero, then the integral can be arbitrarily small.
