So let's say I have two independent random variables $X \sim U(0,1)$ and $Y \sim U(0,1)$. I want to find the probability $P(Y>2X)$
For a uniformly distributed random variable $P(X<a) = F(x) = \frac{x-a}{b-a}$
However, I'm not sure how to calculate the probability when $a$ is a random variable.
Best Answer
Consider the RV $Z=Y-2X$ (which takes values from $[-2,1]$). The probability that $Z<0$ is the probability that $2X> 1$ (probability $1/2$) plus the probability that $2X\le 1$ (i.e., $X\le 1/2$ - with probability $1/2$) and $Y\le 2X$, which is the probability that one random uniform $[0,1]$ distributed RV is smaller or equal than another. The latter event has probability $1/2$ since $P(X>Y)+P(X<Y)=1$ and then use symmetry.
So, $P(Z<0)=\frac{1}{2}+\frac{1}{2}\frac{1}{2}=\frac{3}{4}$, so
$$P(Y-2X\ge 0)=P(Z\ge 0)=1-\frac{3}{4}=\frac{1}{4}.$$