Let $X, Y, Z$ be independant standard uniform random variables, find $P(X/Y < Z)$

Question

My understanding of uniform random variables is that they are all P(n) = 1 but I am unclear on when I build the triple integral for $P(X/Y < Z)$ on how I should create my boundary conditions?

https://math.stackexchange.com/q/185501/321264, https://math.stackexchange.com/q/348225/321264 — StubbornAtom, Oct 02 '21 at 06:25
No, Mark's answer below was a lot cleaner and helped me understand how to answer some other questions in my pset — goldenlinx, Oct 03 '21 at 21:23

score 1 · Accepted Answer · answered Oct 01 '21 at 23:29

1

You want to find the volume of the region described by the following inequalities:

$$0 \leq x, y, z \leq 1$$ $$x/y < z$$

One way to solve this is to note that $x / y < z$ is equivalent to $x < yz$. Note that since $y, z \leq 1$, we have $yz \leq 1$.

$$0 \leq x < yz$$ $$0 \leq y, z \leq 1$$

So the integral in question will be $\int_0^1 \int_0^1 \int_0^{yz} dx dy dz = \frac{1}{4}$.

answered Oct 01 '21 at 23:29

Mark Saving

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You should consider answering the duplicate question. – Peter O. Oct 18 '21 at 16:30

Let $X, Y, Z$ be independant standard uniform random variables, find $P(X/Y < Z)$

1 Answers1