cortejosni

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?

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

Beginner2022-08-17Added 22 answers

Step 1

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

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

janine83fz

Beginner2022-08-18Added 2 answers

Explanation:

The values of x and y, you are interested in, lie in a square $[0;2]\times [0;2]$. 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 $\frac{4-{\int}_{1/72}^{2}\frac{1}{36x}-\frac{2}{72}}{4}$

The values of x and y, you are interested in, lie in a square $[0;2]\times [0;2]$. 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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