Distance between two random variables

[vvaann]

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

A. 1/4
B. 1/3
C. 1/2
D. 2/3
E. 3/4

Answer key:

SPOILER: E

[matroid]

Try to "visualize" it in a geometric probability space. (I tried to draw it here in"ASCII-art" using Courier font, but it didnt work. ) You can draw a unit square, each point corresponds to a random (x, y) pair. The "good" points are between the following lines:

y=x+1/2
y=x-1/2

(Because |x-y|=1/2 is what we need.) The area of the "good" points is 3/4.

Hope it's understandable, sy might draw a real figure...

Cheers, Md.

[vvaann]

Great explanation, matroid.

[matroid]

Wheew, I'm glad that the message "got through" despite the absence of the figure and my command of English...

Md.

[magicwand]

I still dont get this one. Matroid, could you please elaborate on your explanation a little more. Thanks

[matroid]

Sorry, I'm still not ready with the figure...

So, let's see what we can do : imagine a 1*1 (unit) square, each point of which represents a random (x, y) pair - that is, instead of picking 2 random numbers independently from [0,1] u can pick a random point from this square.

What we are looking for, is the probability of abs(x-y) < 1/2, which means: we have to calculate the measure of those (x, y) points of the square which satisfy the condition abs(x-y) < 1/2. These points are between the two lines given below. If you really draw the square and the two lines you should immeadiately see that 1/4 part of the square is "outside" the lines (two right triangles in the upper left and lower right corner are "cut off" - these are the "bad points"). The remaining 3/4 part of the square are the "good points", so the prob. we are looking for is exactly 3/4.

I'm sorry, I feel that I wrote practically the same as I did before... but I can't really put it in another way.

Still in the hope it helps, cheers,

Md.

[TokaBoy]

Matroid, great, great !!!
I will try to do it using integral calculus.

Did you take the GRE math test ?

Keep in touch...

[TokaBoy]

Ok I got it using integral!

Just draw a line segment [0, 1]
Put the point x between 0 and 1/2 (due to the symetry, we will just get the value for 0 to 1/2, and we will multiply by 2 to get the total).
What is the probability that the distance between x and y is less than 1/2 :
- If y is at the right of x, we have p(x)*p(y) = dx * 1/2
- If y is at the left of x, we have p(x)*p(y) = dx * x
Nowing that :
- the probability for x to be at the position x is p(x)=dx, the differential increment
- the probability for y to be at the right of x is that y is between x and x+1/2, p(y)=1/2
- the probability for y to be at the left of x is that y is between 0 and x, p(y)=x

Thus, the sum is dx * ( x +1/2 )

So the total is I = 2 * Integral [ (x+1/2) * dx], x:0->1/2
I = 3/4

I guess, it is a little bit complex, but I'm use to that kind of integral solutions.

Keep in touch

[matroid]

If you prefer integral calculus than elementary geometry, I have no objections to your solution. I think, you practically did same calculations to mine using a more formal method.

To answer your question: yes, I have already taken the GRE math test. I've got 92%, and I'm still waiting for my first admit.

You'll surely have better luck,

Md.

[Dragonfinity]

??? you got 92% and no admits yet? did you only apply to top 5 schools?