P.D.F. of independent/dependent Uniform R.V.'s

I am trying to solve this:

Consider a stick of length 1. You break the stick in two random places, X and Y.
a. Define the individual probability distribution functions of the breaking points X and Y.
b. Write the joint PDF of the breaking points, if X and Y are independent. Sketch its support (domain) and indicate the density values of this domain.
c. Assume that Y is such that $Y > X$. What is the joint PDF of $(X,Y)$ if it needs to be uniform on this domain. Again, sketch its support and indicate the density value.
d. The two breaking points divide the stick in three segments. What is the probability of the left-most segment is the shortest segment, when $Y > X$? Sketch the area in the $X-Y$ plane that corresponds to this event. (Hint: to be the shortest segment the leftmost segment needs to satisfy two constraints simultaneously.)

Solutions:

a/b)

$$PDF(X,Y) = 1 | (x,y) \in [0,1]\times [0,1]$$$$0|otherwise$$

c) we know that $X$~U$(0,1)$. The clause that $Y\gt X$ means that $Y$~U$(X, 1)$.

Now, Y has become dependent on X, correct? Therefore, the joint PDF isn't as simple as multiplying the PDF's together. How do I get the joint PDF in this case?

The answer I was given is that $$PDF(X,Y) = 2|(X,Y)\in [0,1]\times [0,1] \cap (Y > X)$$ but I'm having trouble understanding why this is...

I think I understand the concept that, because $Y>X$, the area in which Y can live is halved, because, when X is chosen (since it is chosen uniformly), Y's value depends on the value of X, and the uniform distribution means that the area Y can occupy is halved. but, why does the P.D.F. = 2 in this case? Wouldn't it still be equal to 1, as in: $$PDF(X,Y) = 1|(X,Y)\in [0,1]\times [0,1] \cap (Y > X)$$ I thought the only change would be in the are for which the PDF was valid (hence the distribution needing to intersect the area where $Y>X$

d) For this one, I came up with 3 constraints: $\frac{1}{2} <Y<\frac{3}{4}$ and $X<\frac{1}{4}$

thus, $X$~U$(0,\frac{1}{4} )$ and $Y$~U$(\frac{1}{2} , \frac{3}{4})$

making $$PDF(X) = \frac{1}{\frac{1}{4}} = 4$$and $$PDF(Y)= \frac{1}{\frac{3}{4} -\frac{1}{2}} = 4$$

so joint $PDF(X,Y) = 16|\frac{1}{2} <Y<\frac{3}{4}$ and $X<\frac{1}{4}$

However, this is incorrect, and I can't rightly figure out why.

Any advice as to where I'm going wrong? Thank you very much!

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The basic rule is that the integral of a (joint) PDF over the region where it is nonzero must be $1$. Since the area of the region in question (c) is $1/2$, the density (given that it is uniform) must be $2$ there.

There's another way to look at it, without being told that the density should be uniform in the region. What you have here in (c) is actually the conditional joint PDF given that $Y > X$. If you integrate this over some region $A$ of the $xy$ plane you should get the conditional probability $P((X,Y) \in A | Y > X)$ that $(X,Y)$ is in $A$, given that $Y > X$. This conditional probability is $P((X,Y) \in A \ \text{and}\ Y > X)/P(Y > X)$. In this case you know $P(Y > X) = 1/2$, so if $A$ is contained in the region where $Y > X$ that means $P((X,Y) \in A | Y > X) = 2 P((X,Y) \in A)$. Now $P((X,Y) \in A | Y > X)$ is obtained by integrating the conditional joint PDF over $A$ while $P((X,Y) \in A)$ is obtained by integrating the (unconditional) joint PDF over $A$, this can be arranged by making the conditional joint PDF $2$ times the unconditional joint PDF.

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In question (b), the area of the region in question is $1$. –  Did May 3 '12 at 5:50

Robert's answer points out where you went wrong in c).

In d), it seems you misunderstood the question. The question asks for the probability of a certain event given the probability distribution defined in c), not for a new probability distribution. Also, your constraints are wrong; unfortunately I can't say where you went wrong in obtaining them because you didn't tell us how you obtained them.

If $Y\gt X$, the lengths of the pieces are $X$, $Y-X$ and $1-Y$. The required constraints for the first to be the smallest are the two inequalities stating that the first is smaller than either of the other two. If you sketch the area as suggested in the exercise, you'll find that it's a triangle with base the left-hand edge of the unit square, whose area is simply half of the $x$ coordinate of the apex, which you can get by solving the linear system of equations you get from turning the inequalities into equalities. Then the probability you want is that area times the density $2$.

Here's a more elegant way to show that when you randomly uniformly choose $n-1$ breaking points, all the resulting $n$ pieces have the same length distribution and thus the same probability $1/n$ of being the smallest: On a circle, randomly uniformly distribute $n-1$ points, plus an $n$-th point at which to break the circle into a linear interval. By symmetry, all $n$ resulting intervals have the same length distribution, and the distribution is the same as if you first choose where to break the circle and then distribute $n-1$ points in the resulting interval.

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