cortejosni

## Open question

2022-08-16

Probability of geometric mean being higher than 1/6
If we choose 2 numbers from [0,2], independent from each other,what is the probability of their geometric mean being higher than 1/6?

Kyle George

Step 1
Note that $\sqrt{XY}>1/6$ iff $Y>1/\left(36X\right)$ where X, Y are i.i.d uniform [0,2] random variables. Since X,Y are i.i.d uniform [0,2] it follows that $P\left(\left(X,Y\right)\in A\right)=\frac{m\left(A\cap \left[0,2{\right]}^{2}\right)}{4}$ i.e. (X,Y) is uniformly distributed on $\left[0,2{\right]}^{2}$ (here m means area) and $A\subset {\mathbb{R}}^{2}$). Letting $A=\left\{\left(x,y\right)\in {\mathbb{R}}^{\mathbb{2}}\mid 36xy>1\right\}$ we have that $P\left(\sqrt{XY}>1/6\right)=P\left(Y>1/\left(36X\right)\right)=\frac{1}{4}{\int }_{0}^{2}{\int }_{1/36x}^{2}\phantom{\rule{thinmathspace}{0ex}}dy\phantom{\rule{thinmathspace}{0ex}}dx$ which you can compute.

janine83fz

Explanation:
The values of x and y, you are interested in, lie in a square $\left[0;2\right]×\left[0;2\right]$. So you have a chart of $y=\frac{1}{36x}$, and an answer to your problem equals to an area of intersection of an epigraph and a given square divided by an area of a square. Which equals to $\frac{4-{\int }_{1/72}^{2}\frac{1}{36x}-\frac{2}{72}}{4}$

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