Problem: In scenario 1 you choose $x$ and $y$ at random from $[0,b]$. In scenario 2 you choose $x$ at random from $[0,b]$ and then $y$ at random from $[0,2x]$. In both cases you find the average value of $f(x) + f(y)$ where $f$ is positive. Can you find any simple general conditions on the function $f$ which will allow you to say that one scenario has a higher average value than the other?
Example: Let $f(x) = x^2$ so that $f(x)+f(y) = x^2 + y^2$, and let $b=10$ so that we are on the interval $[0,10]$.
In scenario 1 we have that the average of $x^2 + y^2$ on $[0,10]\times[0,10]$ is
$$ \begin{aligned} \frac{1}{100}\int_0^{10}\int_0^{10} (x^2 + y^2)\, dx\, dy &= \frac{1}{100}\int_0^{10} \left[\frac{1}{3}x^3 + xy^2\right]_0^{10}\, dy\\ &= \frac{1}{100}\int_0^{10} \left(\frac{1000}{3} + 10y^2\right)\, dy\\ &= \frac{1}{100}\left[\frac{1000}{3}y + \frac{10}{3}y^3\right]_0^{10}\\ &= \frac{200}{3}. \end{aligned} $$
In scenario 2 we have that the average value of $x^2 + y^2$ is
$$ \begin{aligned} \frac{1}{10}\int_0^{10}\frac{1}{2x}\int_0^{2x} (x^2 + y^2)\, dy\, dx &= \frac{1}{10}\int_0^{10} \frac{1}{2x}\left[yx^2 + \frac{1}{3}y^3\right]_0^{2x}\, dx\\ &= \frac{1}{10}\int_0^{10} \left(x^2 + \frac{4}{3}x^2\right)\, dy\\ &= \frac{1}{10}\int_0^{10} \frac{7}{3}x^2\,dx\\ &= \frac{1}{10}\left[\frac{7}{9}x^3\right]_0^{10}\\ &= \frac{700}{3}. \end{aligned} $$
So in this example, scenario 2 has the higher average value.
Generalizing this example to $f(x)=x^n$ for $n\geq1$ and any $b>0$, the average value would be higher in scenario 2.
But it there a better generalization?
My work so far: My strategy was to compute the averages in terms of $f$ and $F$, where $F$ is an antiderivative of $f$, and see what conditions might make one higher. But computing the average in scenario 2 has me stumped.
In scenario 1 we have that the average value is
$$ \begin{aligned} \frac{1}{b^2}\int_0^{b}\int_0^{b} \left(f(x) + f(y)\right) dx dy &= \frac{1}{b^2}\int_0^{b}\left(F(b) - F(0) + b f(y)\right) dy\\ &= \frac{1}{b^2} \left[bF(b)-bF(0) + bF(b)-bF(0)\right] \\ &= \frac{2}{b}\left[F(b)-F(0)\right]. \end{aligned} $$
But in scenario 2 I get stuck on the second/outside integral. The first/inside integral conditional on $x$ is
$$ \begin{aligned} \frac{1}{2x}\int_0^{2x} \left(f(x)+f(y)\right) dy &= \frac{1}{2x}\left[2xf(x)+F(2x)-F(0)\right]\\ &= f(x)+\frac{F(2x)}{2x}-\frac{F(0)}{2x} \end{aligned} $$
so that the average value would be
$$ \frac{1}{b}\int_0^b \left(f(x)+\frac{F(2x)}{2x}-\frac{F(0)}{2x}\right) dx $$
Question: What conditions on $f$ would allow the second integral to exist in the first place
If I require that $F(0)$ equals $0$, then all else being equal it seems to boil down to comparing
$$ \frac{1}{b^2}\int_0^{b} F(b)\, dx $$
with
$$ \frac{1}{b}\int_0^{b} \frac{F(2x)}{2x} \,dx $$
My intuition says that scenario 2 will usually lead to a higher average value since the interval $[0,2x]$ could be larger than the interval $[0,b]$.