What is the probability $P(Y \leq X^2)$ given that $X$ and $Y$ are i.i.d. and uniformly distributed?

Question

I stumbled upon this exercise while studying for an upcoming exam:

Suppose that $X$ and $Y$ are i.i.d. and uniformly distributed on $[0,1]$. Calculate the probability $P(Y \leq X^2)$.

First, I thought that it had something to do with the convolution of $X^2$ and $Y$. Then, however, I ran into the problem that $P(Y - X^2 \leq 0)$ doesn't seem to fit into the formula for convolution.

Therefore, I sadly have no idea where to even start on this one.

Any help is appreciated, thank you in advance!

Answer 1

In my opinion, the best approach to such problems about uniform distribution is to use geometric meaning. Random vector $(X, Y)^T$ is uniformly distributed in the square $S=[0,1] \times [0,1]$, so $P(Y \leq X^2)$ equals to $$ \frac{\mathrm{Area}( \{(x, y) \in S : y \leq x^2 \})}{\mathrm{Area}(S)} $$ As $\mathrm{Area}([0,1]\times [0,1]) = 1$ and area in numerator equals $\int_0^1 x^2 dx = \frac{1}{3}$, you get $P(Y\leq x^2) = \frac{1}{3}$.

Answer 2

easy do a drawing of your problem

and immediately realize that the requested probability is

$$\mathbb{P}[Y\leq X^2]=\int_0^1x^2dx=\frac{1}{3}$$