# Homework Help: Problem with density function

1. May 1, 2013

### trenekas

Hello. I have problem with one task of probability theory. Hope that someone will be able to help me!

The rod is dividing in two parts accidentally. (Division is evenly distributed, dont know if its good saying in english). Find the density function of random variable X which is equal to longer part of rod divided by smaller part of rod.

So X can get infinity values i think. The smaller is 1, when rod is divided 0.5/0.5=1. And coming up to infinity when smaller part of rod coming up to 0. for example 0.99/0.01=99... and so on.

I dont know but i think that density function of this X does not exist. Or im wrong?

2. May 1, 2013

### HallsofIvy

If the shorter length is y, the longer length is 1- y so the function is X= y/(1- y). Further, you seem to be saying that "y" is equally likely to be any number between 0 and 1/2.

3. May 1, 2013

### haruspex

It's often clearer conceptually to deal with the CDF and deduce the pdf later. HoI has given you the formula for X, so you want F(x) = P[X < x] = P[Y/(1-Y) < x]. Can you proceed from there?

4. May 1, 2013

### trenekas

I think the task ask to do not the same as you say. I think we need 1-y/y... where 1-y is equal to longer part and y shorter...If i need dividing shorter from longer it much easier. or i dont understand you?
Maybe my opinion that this function does not exist is true?

5. May 1, 2013

### Ray Vickson

The parts (u,1-u) have zero probability for u = 0 or u = 1 because the endpoints of the interval [0,1] have zero probability for a continuous distribution. Therefore, we are (with probability 1) never dividing by zero, so a value X = ∞ has probability 0. We can even change to the open interval u ε (0,1), and so division by zero never happens---not even with probability zero. So, X does have a well-defined density function. Having X unbounded above is not a problem---lots of random variables are unbounded above, below, or both, and we deal with them with no problem. Of course, there is the issue of whether or not X has finite mean and/or variance, etc, but that is a separate question from the existence of a density for X.

Last edited: May 1, 2013
6. May 1, 2013

### trenekas

ok understand.

7. May 1, 2013

### trenekas

But cant invent what this function shuold be. Maybe some hints? I thought more than an hour and have no idea.

8. May 1, 2013

### haruspex

Did you understand what I suggested? Let Y be the r.v. representing the distance from the nearer end. Y is uniformly distributed on [0, 0.5]. The CDF of X is F(x) = P[X < x] = P[(1-Y)/Y < x]. Can you evaluate that?

9. May 2, 2013