# Find probability from uniform distribution

Let $X$, $Y$ be two independent random variables from $U(0,1)$. Then find $P[Y>(X-1/2)^2]$.

I initially tried drawing the figure but that seemed complicated. I then took expectation on both sides and got $P[E(Y)>V(X)]$. Am I right?

• What did you attempt to draw? The drawing is very simple so the area of integration should be quite simple to figure out once you sketch the x-y plane. – StatsStudent Apr 22 '17 at 5:50
• Also, why are you doing anything with expectation? You are simply asked to find the probability, not the expectation. – StatsStudent Apr 22 '17 at 5:59
• @Analyst1 Probabilities are expectations (of indicator variables). Often they can readily be found using techniques to find expectations. The mistake made here is to suppose that $E[Y\gt (X-\mu_X)^2] = E[Y \gt E[(X-\mu_X)^2]]$: this is rarely true and there are no generally applicable rules of expectation, probability, or integration that would justify such an equation. – whuber Apr 22 '17 at 12:02
• Good point @whuber, but I don't see where the OP was headed with regard to his/her approach by finding an expectation. The most straightforward approach to solving this problem seemed to be simply integrating the area bounded below by 0 and 1 and above the parabola $Y = (X-1/2)^2$. – StatsStudent Apr 22 '17 at 15:48

This ends up being the area above the curve $$Y=(X−\frac{1}{2})^2$$
This can be found by integration $$P[Y>(X−\frac{1}{2})^2] = \int_{0}^{1}\int_{(X−\frac{1}{2})^2}^{1} 1\times1 \,dydx$$ $$= \int_{0}^{1}{1-(X−\frac{1}{2})^2} \,dx$$ $$= \Big[X-\frac{1}{3}\times(X−\frac{1}{2})^3\Big]_0^1$$ $$= \Big(1-\frac{1}{3}\times(1−\frac{1}{2})^3\Big) - \Big(0-\frac{1}{3}\times(0−\frac{1}{2})^3\Big)$$ $$= \frac{23}{24}-\frac{1}{24}$$ $$= \frac{22}{24}$$