Two random variables $X$ and $Y$ are i.i.d. with probability density function $f$ (unknown). Two variables are nonnegative. What is the probability $P(X>2Y)$ ?
My question is if there is a deterministic answer.
Here is my attempt:
$$Pr(X>2Y) = Pr(Y<X/2) = \int \int f_{X,Y}(u,v)dvdu=\int_{0}^{\infty} \int_{0}^{u/2} f(u)f(v) dvdu = $$ $$\int_{0}^{\infty} f(u) \big[\int_{0}^{u/2}f(v)dv \big] du =\int_{0}^{\infty}f(u)F(u/2)du$$ I attempted to use integration by parts but was not able to work out a deterministic answer.
Note that $P(X>Y) = P(Y>X) = 1/2$ by symmetry. I do not know if there is a connection between these two problems.
There is no deterministic answer without more information about the distribution
You could say $P(X>2Y) = P(Y>2X)$ since they are i.i.d.; these are mutually exclusive events and $P(\frac12 Y \le X \le 2Y)$ will be positive, so you can bound with $$0 \le P(X>2Y) \lt \tfrac12$$
You can achieve the lower bound with for example a uniform distribution on the interval $[2,3]$
You can approach the upper bound with for example a Pareto-type density $f_{\alpha}(x)= \frac{\alpha}{x^{1+\alpha}}$ and cumulative probability $F_{\alpha}(x) = 1-\frac1{x^{\alpha}}$ for $x \gt 1$: your expression would give $\displaystyle P(X>2Y)= \int_2^\infty\left( \tfrac{\alpha}{x^{1+\alpha}}-\tfrac{\alpha2^{\alpha}}{x^{1+2\alpha}}\right)\, dx = \tfrac1{2^{1+\alpha}} \to \tfrac12$ from below as $\alpha \to 0$ from above