# A question on uniformly distributed random variables over the interval (0, 1).

How to find the approach to solve the following problem:

Let $(X, Y, Z)$ be independent uniformly distributed random variables over the interval (0, 1). Find $P(Z≥XY^2)$ . No idea how to start this question. I have knowledge on uniform distribution, but this seems new to me. Need little bit of hint/help. Thanks for the help.

• Well, to get a probability for a continuous random variable, you need to integrate the density function over an appropriate region, in this case $\{ (x,y,z) \in (0,1) \times (0,1) \times (0,1) : z \geq xy^2 \}$. What is the density in this case? After that, how can you parametrize the region? (A hint for that: you'll want $dz$ to be the innermost integration.) – Ian Mar 17 '17 at 11:55
• To condition $x,y,z$ are independent uniformly distributed random variables on the interval $(0,1)$ is equivalent to saying that $(x,y,z)$ is a random point in the standard unit cube, $0 \le x,y,z \le 1$. So the probability of a specified event is just the volume of the region in the cube corresponding to that event, divided by the volume of the cube (which is clearly just $1$). – quasi Mar 17 '17 at 11:59
• And in this case, the region you want is the subset $S$ of the standard unit cube such that $z \ge xy^2$. How would you find the volume of $S$? – quasi Mar 17 '17 at 12:03
• quite confusing. Do we need to find marginal distributions for each X, Y, and Z? – monalisa Mar 17 '17 at 12:04
• yeah I got it now. thanks a lot. – monalisa Mar 17 '17 at 12:30

1) if V has values in [0,1], then $P(U<V)=E(V)$
2) $E(UV)= \frac{E(V)}{2}$
3) $E(U^2)=\frac{1}{3}$
Then you have $P(Z>XY^2)=1-E(XY^2)=1-\frac{E(Y^2)}{2}=\frac{5}{6}$