# $x,y,z$ independent and uniform random on $[0,1]$, $P(x \geq yz)$?

$x,y,z$ are independent and uniform random on $[0,1]$.

I know that $P(x \geq yz) = 3/4$ by triple integrating the triple density function.

However, is there a easier way to get this solution? Without triple integration.

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I don't know, performing $\int_0^1\int_0^1 (1 - yz) \,dy\,dz$ is pretty easy... what sort of method would you consider to be easier? –  Zarrax Aug 22 '12 at 16:50
I was trying to avoid triple integration. Can I reason this way: P(x >= yz) = 1 - yz. So we're just double integrating the function 1-yz over the unit square region? Totally not mentioning triple integrals at all... –  Legendre Aug 22 '12 at 16:57
Given that $y = y_0$ and $z = z_0$, the chance that $x > yz$ is $1 - y_0z_0$. So since that $y$ and $z$ are uniformly distributed in $[0,1]$ you end out with the double integral I wrote above... so you don't have to do any triple integrals (but it's somewhat implicit in what I'm saying here). –  Zarrax Aug 22 '12 at 17:01
Thank you so much! –  Legendre Aug 22 '12 at 17:07

The conditional probability given $z \in [0,1]$, $P(x \ge yz | z)$, is the area of the region of the square $[0,1]^2]$ below the line $y = x/z$. Since the region above the line is a right triangle with sides $1$ and $z$, we have $P(x \ge y z | z) = 1 - z/2$.
Since $E[z] = 1/2$, the unconditional probability is $E[P(x \ge y z | z)] = E[1 - z/2] = 1 - E[z]/2 = 3/4$.

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Nice answer! Thanks! –  Legendre Aug 22 '12 at 17:19

$$\mathrm P(X\leqslant YZ\mid Y,Z)=YZ\qquad\quad \mathrm E(YZ)=\mathrm E(Y)\cdot\mathrm E(Z)=1/4$$

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As explained in @Zarrax's comment. –  Did Aug 22 '12 at 19:29
Are you saying that $E(YZ) = P(X \leq YZ)$? If so, why? –  Legendre Aug 23 '12 at 15:48
Yes I am. Because (1.) $P(X\leqslant x)=x$ for every $x$in $(0,1)$, (2.) $YZ\in(0,1)$ almost surely, (3.) $YZ$ is independent of $X$. –  Did Aug 23 '12 at 16:10

Here is an alternative way to see it.

If $U$ is uniformly distributed on $(0,1)$ random variable, $X= -\log(U)$ follows the exponential distribution with unit rate. Thus: $$\mathbb{P}\left( U_1 \geqslant U_2 U_3 \right) = \mathbb{P}\left(-\log(U_1) \leqslant -\log(U_2) -\log(U_3) \right) = \mathbb{P}\left(X_1 \leqslant X_2 + X_3 \right) = \mathbb{P}\left(\frac{X_1}{X_1+X_2+X_3} \leqslant \frac{1}{2} \right)$$ The ratio $\frac{X_1}{X_1+X_2+X_3}$ follows Beta distribution with parameters $(1,2)$ (e.g., see here, or in "Question about order statistic") meaning that $$\mathbb{P}\left( U_1 \geqslant U_2 U_3 \right) = \int_0^{\frac{1}{2}} \frac{1-x}{\operatorname{B}(1,2)} \mathrm{d} x \stackrel{\operatorname{B}(1,2)=\frac{1}{2}}{=} \int_\frac{1}{2}^1 2 x \mathrm{d} x = 1^2 - \left(\frac{1}{2}\right)^2 = \frac{3}{4}$$

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The beta(1,2) distribution is also the distribution of the minimum of two independent uniform[0,1] random variables $X$ and $Y$. (This is a standard result in the theory of order statistics.) In order to have $min(X,Y) \le 1/2$ we must have $X \le 1/2$ or $Y \le 1/2$, which obviously has probability 3/4. –  Michael Lugo Aug 22 '12 at 18:37
Whoever cast the down vote, would you please explain what you would like improved about the post? –  Sasha Aug 22 '12 at 19:28
Cute, but maybe more complicated than necessary. –  Michael Hardy Aug 22 '12 at 21:54