# uniform random multivariate on interval (0,1)

• Feb 5th 2011, 11:27 AM
nikie1o2
uniform random multivariate on interval (0,1)
Hi guys,

Im given X, Y, Z are independent uniform random variables on the interval (0,1).
The question is find P(x<y<z).
Im thinking its a triple integration of f(x,y,z) dxdydz ? not too sure of the bounds though..

thanks for your time
• Feb 5th 2011, 01:05 PM
theodds
Quote:

Originally Posted by nikie1o2
Hi guys,

Im given X, Y, Z are independent uniform random variables on the interval (0,1).
The question is find P(x<y<z).
Im thinking its a triple integration of f(x,y,z) dxdydz ? not too sure of the bounds though..

thanks for your time

No need to integrate. You can solve this by counting, since
$\displaystyle (X < Y < Z) \cup (X < Z < Y) \cup \cdots \cup (Z < Y < X)$
is the whole space (except for a set of measure 0) and each of those events is disjoint and equally likely. Hence, the probability is 1/6.

If you did want to integrate, which I wouldn't recommend, the integral would be of this form:

$\displaystyle \displaystyle \int_0 ^ 1 \int_0 ^ z \int_0 ^ y \ dx \ dy \ dz$
• Feb 6th 2011, 12:03 PM
nikie1o2
nice. so your saying those 6 possibilities x<y<z,x<z<y,z<x<y,z<y<x,y<x<z,y<z<x equals our sample space and there's a 1/6 probability that x<y<z. I get it.!

The next question i have to answer is whats the p(xy<z) can i solve that in a similar way ?
• Feb 6th 2011, 02:53 PM
theodds
Quote:

Originally Posted by nikie1o2
nice. so your saying those 6 possibilities x<y<z,x<z<y,z<x<y,z<y<x,y<x<z,y<z<x equals our sample space and there's a 1/6 probability that x<y<z. I get it.!

The next question i have to answer is whats the p(xy<z) can i solve that in a similar way ?

That one I integrate, although if you apply the same idea I initially used it makes things a bit easier. First, I would calculate 1 - P(XY > Z) and note that
$\displaystyle (0 < Z < XY < X < Y) \cup (0 < Z < XY < Y < X)$
is the equivalent to (XY > Z) (except on a set of probability 0); moreover we have a disjoint union of equally likely sets, so it suffices to calculate
$\displaystyle 1 - 2 \cdot \int_0 ^ 1 \int_0 ^ x \int_0 ^ {xy} \ dz \ dy \ dx.$

Another approach would be to take the -log of both sides. You end up needing to find P(Gamma(2, 1) > Exponential(1)) with this method, which is easy enough. There may be an easier way that I'm not seeing that lets you get this immediately.
• Feb 7th 2011, 05:36 PM
nikie1o2
ok, im just confused when you take the complement of p(XY <Z) why it = 1-P(Z<XY) is that a typo? or im just not seeing why that's correct
• Feb 7th 2011, 07:04 PM
theodds
Quote:

Originally Posted by nikie1o2
ok, im just confused when you take the complement of p(XY <Z) why it = 1-P(Z<XY) is that a typo? or im just not seeing why that's correct

No typo, that is what it is. Clearly $\displaystyle XY < Z$ or $\displaystyle XY \ge Z$ correct? I can make the inequality strict because equality happens on a set of probability 0; it doesn't matter at all, but I made the inequalities strict so that I could get a disjoint union.

Both the methods I listed give the same answer, which leads me to believe I'm not making a stupid mistake.