Tricky joint-density function problem

[shwin]

Homework Statement

Two points are selected randomly on a line of length L so as to be on opposite sides of the midpoint of the line. This means that their joint density function is a constant over the region A = (0, L/2) x (L/2, L); normalization to 1 defines the constant.

a. Find the probability that the distance between the two points is greater than L/3.

b. Find the probability that the three line segments of (0, L) formed by the two points can form a triangle (so as to satisfy the triangle inequality)



The attempt at a solution

I drew the region on an xy plane, and integrated the constant C twice with respect to each set of limits (L/2 to L, and 0 to L). However, I end up with C = 2/L and don't know what do with this...which means I can't even do the second part of the problem.

 

[Avodyne]

The joint density function is p(x,y) = C if 0<x<L/2 and L/2<y<L, and zero otherwise. The value of C is fixed by the condition
$$\int_0^{L/2}dx\int_{L/2}^L dy\,p(x,y) = 1$$
This does not yield your value of C.

For part a, you must integrate p(x,y) over the range of x and y that corresponds to y-x > L/3.

 

[shwin]

So then how would I find those limits? I mean, the two points are randomly selected, so what limits of integration would I establish to ensure that the difference between y and x is greater than L/3?

 

[shwin]

does anyone know?

 

[Avodyne]

Suppose I give you the value of x. For what values of y is y-x > L/3 ? (The is answer is completely trivial and obvious; don't overthink it!)

So, y must be greater than THE LARGER OF whatever you got above, and L/2. This gives you the lower limit (as a function of x) on the integral over y. What is the upper limit? What are the limits on x?

 

[shwin]

So the limits of integration for dx is (0, L/2) and the limits of integration for dy is (L/3 + x, L)?

 

[Avodyne]

You're almost there. But the lower limit on y is a little more subtle. To see why, suppose x is near zero. Then your lower limit on y is near L/3. But y can't be less than L/2. So the actual lower limit on y is the larger of ...

 

[shwin]

5L/6? Sorry I am really lost on this problem, and part b is no easier...how would I satisfy the triangle inequality for this? I know they involve the three lengths being x, y - x, and L - y, but I don't know how to set up a double integral to satisfy the inequality. I think that with this and the rest of the problems I have trouble with, I am very close to actually understanding the concepts...I need a push in the right direction.

 

[Avodyne]

Well, I haven't thought about part (b). But for part (a), the lower limit on y is L/3+x, PROVIDED this is not less than L/2, because y is not allowed to be less than L/2. So the lower limit is the larger of L/2 and L/3+x. OK, now under what circumstances is L/2 larger than L/3+x? Answer: x<L/6. So the lower limit on the y integral is L/2 for 0<x<L/6, and L/3+x for L/6<x<L/2. So the integral is
$$C\int_0^{L/6}dx\int_{L/2}^L dy+C\int_{L/6}^{L/2}dx\int_{L/3+x}^L dy$$
where C is fixed by the condition in post #2.

 

[Avodyne]

OK, I know how think about part (b). The line is divided into three segments, and we want to know if the sum of the two shorter ones is longer than the longest, which would mean that the triangle inequality is satisfied.

Suppose the middle segment is one of the two shortest. Then the triangle inequality is always satisfied, because the middle plus either end is always bigger than L/2, which is longer than the other end. So the only way the triangle inequality could NOT be satisfied is if the middle segment is the longest, and is longer than the sum of the two ends. This translates into a simple condition on x and y. So it's easiest to compute the probability that the triangle inequality is NOT satisfied, and then subtract from one.

 

[shwin]

So now we are looking at P(y - x > L/2) then?

 

[Avodyne]

Yep!