Probability density functions

By Edwin Koh (P132) on Tuesday, October 31, 2000 - 10:13 am :

If x, y, z are selected independently and at random from the interval [0,1], then what is the probability that x is greater or equal to yz?

Let x and y be uniformly distributed, independent random variables on [0,1]. What is the probability that the distance between x and y is less than 1/2?

This isn't homework. It's a GRE problem but I haven't done a course on probability before. I'm just practising some GRE problems before the GRE which is this Saturday - I'll be truly grateful if you can help me.

By Dave Sheridan (Dms22) on Tuesday, October 31, 2000 - 05:56 pm :

Well, I don't see how you're going to understand much of the solution if you haven't done probability before, but I'll give it a try.

This is quite a simple problem in that we're dealing with independent uniform random variables, so the joint density function is 1. To work out the probabilities, simply work out the measure of the set concerned.

Thus, for P(X> YZ) you need to work out what the set {X> YZ} actually is, and work out its area. One way to think about this is to say that whatever Y and Z are, X must be smaller than this. So we consider {X> YZ} to be {Y,Z in [0,1] and X in [YZ,1]} so the integral is

$begin{displaymath}int\_{y=0,z=0,x=yz}^{y=1,z=1,x=1} dx dz dy end{displaymath}$

which is equal to 3/4.

Similarly, {|X-Y| < 1/2} is the set
{X < 1/2, Y < 1/2 or X> 1/2, Y> 1/2 or X < 1/2, 1/2 < Y < 1/2+X or X> 1/2, x-1/2 < Y < 1/2}
which is easier to deal with if you just draw it, although you can integrate over this set to get the same answer, which is 3/4.

I can explain any of this in more detail but I think it's beyond our remit to give you a whole course on probability theory!

-Dave

By Edwin Koh (P132) on Wednesday, November 1, 2000 - 03:58 am :

I think I get it. In both problems, the variables are independent - so we can view the variables as different axes of Rⁿ . To find the probability, we just have to find the fraction of the "volume" where the variables satisfy the given conditions over the "volume" of all possible tuples of the variables. Is this correct?

By Dave Sheridan (Dms22) on Wednesday, November 1, 2000 - 10:26 am :

Yes, that's right. It gets much more complicated when things aren't uniform or if they're not independent. Generally, there's a density function

p (x, y, z, \dots)

corresponding to the random variables concerned. This is non-negative and integrates to 1. Then for any set

A

(technically, a measurable set but basically something involving

X

Y

and/or

Z

) we have

$P (A) = \int_{A} p (x, y, z) dx dy dz$

For uniform random variables, $p = 1$ so this is much simpler.

The pdf may not be integrable in a closed algebraic form (for example, the normal distribution) and some are rather horrendous. I suggest you have a look back through some past discussions on probability to understand more.

If you have any specific questions, I can help there...

-Dave