Let $X$ , $Y$, and $Z$ be random variables distributed uniform(0,1). What is $\textrm{P}(YZ < X)$?

Question

Let $X$, $Y$, and $Z \sim \textrm{Uniform}(0,1)$ and be i.i.d.

How do I set up the integration for the calculation of $\textrm{P}(YZ < X)$?

Answer 1

What's wrong with simply $$P=\int_0^1 \int_0^1 \int_{zy}^1 1 \, dx\, dy \, dz =\int_0^1 \int_0^1 (1-zy) dy \, dz = \int_0^1 (1 - \frac{1}{2}z)\, dz = \frac{3}{4}$$?

Edited: As pointed out in comments (Did , achille hui ...) the integral can be expressed in such simple form because, "fortunately", $y, z \in (0,1) \implies yz \in (0,1)$. If we had instead $\textrm{P}(a YZ < X)$ for some $a>1$ the integral would be more complicated.

Answer 2

Here is a simulation that might help you set up integrals for an analytic solution. It seems $P(W < X) = 3/4.$

The histogram below is for a million simulated realizations of $W = YZ$. [Addendum: Per @Did's Comment, I have overlaid the PDF of $W$ on the histogram.]

For clarity, the bivariate plot is for 40,000 realizations of $W$ and $X$, with the appropriate 3/4 of the points plotted in green.

m = 10^6;  x = runif(m);  y = runif(m);  z=runif(m)
w = y*z;  mean(w < x)
## 0.750319    # aprx P(YZ < X)