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If $x, y, z$ are selected independently and at random from the interval $[0, 1]$, then what is the probability that $x \ge yz$?

This comes from a practice GRE math test.

Did
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user70443
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2 Answers2

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Since $yz$ is in $[0,1]$ almost surely and $x$ is independent of $(y,z)$, $$P[x\geqslant yz\mid y,z]=\left.P[x\geqslant t]\right|_{t=yz}=\left.1-t\right|_{t=yz}=1-yz\ \text{almost surely}$$ hence, by the law of total expectation, $$P[x\geqslant yz]=E[P[x\geqslant yz\mid y,z]]=E[1-yz]=1-E[y]\cdot E[z]=1-\tfrac12\cdot\tfrac12$$ that is,

$$P[x\geqslant yz]=\tfrac34$$

All this assumes that the random variables $x$, $y$ and $z$ are (independent, as made explicit in the post, but also) uniformly distributed on $[0,1]$ (as was not made explicit in the post).

More generally, if $x$, $y$ and $z$ are i.i.d. with density $at^{a-1}$ on $(0,1)$, for some positive $a$, then the CDF of $x$ is $P[x\leqslant t]=t^a$ hence $P[x\geqslant yz\mid y,z]=1-(yz)^a$ almost surely, and $E[y^a]=E[z^a]=\frac12$, hence one gets the same result that $$P[x\geqslant yz]=\tfrac34$$ This coincidence with the first result (when $a=1$) is due to the fact that $[x\geqslant yz]=[x^a\geqslant y^az^a]$ and that the random variables $x^a$, $y^a$ and $z^a$ are i.i.d. uniformly distributed on $[0,1]$.

Did
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  • Could you explain how it is that $E[1-yz]$ is the probability of $x\geq yz$ – Vigneshwaren Oct 15 '14 at 15:01
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    @Vigneshwaren Sure, this is explained just before that (starting with "Since" and ending with a comma). Recall that $P(A)=E(P(A\mid \xi))$, for every event $A$ and every random variable $\xi$. – Did Oct 15 '14 at 15:56
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Let's call $$R=\bigl\{\langle x,y,z\rangle\in[0,1]^3:x\ge yz\bigr\}.$$ We want to know what fraction of the volume of the cube $[0,1]^3$ is occupied by $R$, but the cube has volume $1$, so the problem amounts to determining the volume of $R.$ Given any $y,z\in[0,1]$, the only $x\in[0,1]$ such that $x\ge yz$ lie in the interval $[yz,1].$ Our problem then reduces to evaluating the triple integral $$\begin{align}\int_0^1\int_0^1\int_{yz}^11\,dx\,dy\,dz &= \int_0^1\int_0^1(1-yz)\,dy\,dz\\ &= \int_0^1\int_0^11\,dy\,dz-\int_0^1z\int_0^1y\,dy\,dz\\ &= \left(\int_0^11\,dy\right)\left(\int_0^11\,dz\right)-\left(\int_0^1y\,dy\right)\left(\int_0^1z\,dz\right)\\ &= 1\cdot 1-\frac12\cdot\frac12\\ &=\frac34.\end{align}$$ As Did pointed out in the comments below, this is the same approach as Did's answer, put into (marginally) different terms. (A tip of my proverbial hat to Did for pointing out the reason that my original approach was so tedious.)

Cameron Buie
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  • A derivation of the probability that $x\geqslant yztuv$ (all i.i.d. and uniform on $(0,1)$) along these lines could become quite lengthy although the result $31/32$ is straightforward using other approaches. – Did Apr 01 '13 at 18:20
  • @Did: True enough (and if that had been the question, I wouldn't have bothered to answer it). My experience with probability has been almost entirely related to combinatorics or measure theory, so this was the only approach I knew of. I like yours much better (and upvoted it), but I think this approach is still useful, though tedious. – Cameron Buie Apr 01 '13 at 18:29
  • The problem is not about using measure theory or not using measure theory, but that you choose to first integrate over the distribution of $z$. If one first integrates over the distribution of $x$, things are much easier (and one falls back on the approach in my post, thinly disguised). – Did Apr 01 '13 at 18:34
  • @Did: Hmmm. Changing the order of integration certainly makes it vastly simpler. And in comparing that with your answer, I understand expected value better than I did before. Thanks! – Cameron Buie Apr 01 '13 at 18:42
  • You are welcome. – Did Apr 01 '13 at 18:43
  • @CameronBuie This is a bit late, but I was wondering if it is helpful to always think about the probability of events as the ratio of volumes of certain spaces? I have not had a formal class in probability theory in a long time (the last one was an introductory undergraduate course) and really appreciate the more geometric approach you take here. What are the conditions to keep in mind and when can I use this approach? I am mostly interested just to study for the GRE. – Kevin Sheng Aug 09 '15 at 01:49
  • @Kevin: Certainly not always. Off the top of my head, I'd say it's definitely the way to go when you're dealing with 3 uniform independent random variables. Of course, we can take a similar approach with any number of uniform independent random variables. More generally, though, it might be better to think of it in terms of ratios of masses instead of volumes, with the integrand being a density function of sorts. – Cameron Buie Aug 09 '15 at 03:28