a stick of length 1 is split at a point U that is uniformly distributed over (0,1). find the expected length of the piece that contains the point p, 0<p<1.
is p(x) = 1/ (1-0) = 1?
how do i go about solving this?
thanks!
The length of the piece that contains p is:
l = u if p<u (the piece (0,u) is chosen)
l = 1-u if p>u (the piece (u,1) is chosen)
So the length of the chosen stick as a function of u is:
l(u) = u if u>p
l(u) = 1-u if u<p
and the expected value of l(u) is:
$\displaystyle \int_{0}^{1}l(u)\ du$
To evaluate the integral, you need to split it up:
$\displaystyle \int_{0}^{1}l(u)\ du = \int_{0}^{p} l(u)\ du + \int_{p}^{1} l(u)\ du = \int_{0}^{p} u\ du + \int_{p}^{1} 1-u\ du$
Post again if you're still having trouble.
Oops! I set up the integrals wrong. In the integral from 0 to p, u is less than p, so l(u) is 1-u, and in the integral from p to 1, u is greater than p, so l(u) is u.
Maybe if I give you the right expression, it will be easier to interpret:
$\displaystyle \int_{0}^{1}l(u)\ du = \int_{0}^{p} l(u)\ du + \int_{p}^{1} l(u)\ du = \int_{0}^{p} 1-u\ du + \int_{p}^{1} u\ du$
The first integral is for when the cut is made to the left of p, so the right part of the stick is chosen, and it has length 1-u. The second integral is for when the cut is made to the right of p, so the left part of the stick is chosen, and it has length u.
So in the first integral we add up all the cases starting from when the cut is made at the left end (and we get the whole stick), up to where the cut is made just to the left of p, and we get 1-p. Along the way, we get 1 minus wherever the cut is made.
In the second integral we add up all the cases starting from where the cut is made just to the right of p, and we now get the other part: p instead of 1-p, continuing up to where the cut is made at the right end and we get the whole stick. Along the way, we get a portion equal to wherever the cut is made
If you're still having trouble, feel free to post again. Sorry about my mistake on the first post.