(Un)Expected Geometric Mean

Average all the possibilities using limits to infinity: $\displaystyle\lim_{N\to\infty}\frac{2}{N^2} \sum_{a=0}^N \sum_{b=0}^{N-a} \sqrt{\frac{a}{N}\frac{b}{N}}$

(We need to average all possible values for $\frac{a}{N}$ and $\frac{b}{N}$ , where $0≤a≤N$ and $0≤b≤N-a$ as $N$ approaches $\infty$ . There are $\frac{N^2+N}{2}$ terms summed, which for large values of $N$ is approximately equal to $\frac{N^2}{2}$ .)

Substitute $x=\frac{a}{N}$ , $dx=\frac{1}{N}$ , $y=\frac{b}{N}$ , and $dy=\frac{1}{N}$ and use integral notation: $I=2\int_0^1\displaystyle\int_0^{1-x}\sqrt{xy}\ dy\ dx$ $=\frac{4}{3}\displaystyle\int_0^1\left[\sqrt{xy^3}\right]_{y=0}^{y=1-x}\ dx$ $=\frac{4}{3}\displaystyle\int_0^1\sqrt{x{(1-x)}^3}\ dx$ $=\frac{4}{3}\displaystyle\int_0^1(1-x)\sqrt{x(1-x)}\ dx$

Since $\displaystyle\int_a^bf(x)\ dx=\displaystyle\int_a^bf(a+b-x)\ dx$ , it can be seen that $I$ also equals $=\frac{4}{3}\displaystyle\int_0^1x\sqrt{x(1-x)}\ dx$

Add the two together and divide by two, obtaining $I=\frac{2}{3}\displaystyle\int_0^1(1-x)\sqrt{x(1-x)}\ dx+\frac{2}{3}\displaystyle\int_0^1x\sqrt{x(1-x)}\ dx$ $=\frac{2}{3}\displaystyle\int_0^1\left((1-x)\sqrt{x(1-x)}+x\sqrt{x(1-x)}\right)\ dx$ $=\frac{2}{3}\displaystyle\int_0^1\sqrt{x(1-x)}\ dx$ $=\frac{2}{3}\displaystyle\int_0^1\sqrt{x-x^2}\ dx$ $=\frac{2}{3}\displaystyle\int_0^1\sqrt{\frac{1}{4}-{\left(\frac{1}{2}-x\right)}^2}\ dx$

Substitute $\frac{1}{2}\sin{u}=\frac{1}{2}-x$ and $\frac{1}{2}\cos{u}\ du=dx$ and simplify using trigonometric identities: $=\frac{2}{3}\displaystyle\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\sqrt{\frac{1}{4}-\frac{1}{4}\sin^2{u}}\frac{1}{2}\cos{u}\ du$ $=\frac{1}{6}\displaystyle\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\cos^2{u}\ du$ $=\frac{1}{12}\displaystyle\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}(1+\cos{2u})\ du$ $=\frac{1}{12}[u+\frac{\sin{2u}}{2}]_{-\frac{\pi}{2}}^{\frac{\pi}{2}}$ $=\frac{1}{12}\pi$

Thus, the answer is $\boxed{\frac{\pi}{12}}$ .

Let the random variables $X$ and $Y$ represent the first and the second number chosen, respectively. From the question, we have that the p.d.f. of the joint distribution $(X,Y)$ is $f_{X,Y}(x,y) = \frac{1}{1-x},\quad 0<x<1,\,0<y<1-x.$ From the definition of expected value and geometric mean, we have $\begin{aligned} \mathbb{E}[\sqrt{XY}] &= \int_{0}^1\int_0^{1-x} \sqrt{xy}\,f_{X,Y}(x,y) \,\mathrm{d}y\,\mathrm{d}x \\ &= \int_{0}^1\int_0^{1-x} \frac{\sqrt{xy}}{1-x} \,\mathrm{d}y\,\mathrm{d}x \\ &= \frac{2}{3} \int_{0}^1 \sqrt{x(1-x)} \mathrm{d}x. \end{aligned}$ Using the following change of variables $u = 2x-1$ , one can show that $\mathbb{E}[\sqrt{XY}] = \frac{1}{6} \int_{-1}^1 \sqrt{1-u^2}\,\mathrm{d}u.$ Notice that the integral corresponds to the area of a semi-circle of radius 1, which is $\pi/2$ . Therefore $\boxed{\mathbb{E}[\sqrt{XY}] = \frac{\pi}{12}}$ .