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!
Best Answer
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.