Suppose X has a uniform distribution on the interval (0,a) where a > 0, find P(X>X^2)?

Question

I mainly get stuck with setting up the proper integrals.

Answer 1

First consider the inequality... $X > X^2 \Leftrightarrow 1 > X$.

Therefore, this inequality is only satisfied when $X < 1$.

Then, note that for X to have a uniform distribution on $\left(0, a\right)$, the probability distribution function must be $f\left(x\right) = \frac{1}{a}, 0 \le x \le a$.

Therefore, $P\left(X > X^2\right) = \int_0^1 \frac{1}{a}dx = \frac{1}{a} \int_0^1dx$.

This means that $P\left(X > X^2\right) = \frac{1}{a}\left[x\right]_0^1 = \frac{1}{a}$.

Answer 2

Matt's answer is correct if $a>1$. Otherwise the answer should be 1. So the correct answer is $\min(1/a,1)$.