On 6/3/10, Dan Asimov <dasimov@earthlink.net> wrote:
Suppose two points are chosen independently at random from [0,1], cutting it into three pieces.
What's the best guess G, in closed form, for the size of the left-hand piece, if best is defined as minimizing the average distance between G and the actual size of the left-hand piece?
On Thu, Jun 3, 2010 at 11:41 AM, Fred lunnon <fred.lunnon@gmail.com> wrote:
What's the difference between that and the mean left-hand piece size? WFL
Average error is minimized at the median value, not at the mean. Suppose the value of X is 0 with probability 2/3, and 1 with probability 1/3. If you guess 1/3, the average error is 4/9. If you guess 0, the average error is 1/3. More generally, if the minimum average error is given by the guess M, then increasing the guess by epsilon increases the error by epsilon if the actual value is less than M, and decreases the value by epsilon if the actual value is greater than M. Since the derivative must be 0 at the minimum, For a continuous distribution the chance that the value is less than M must be 1/2, so M is the median. (for a discrete distribution with an even number of equally probably points, any value between the two middle possible values does an equally good job of minimizing mean error; the choice that the median makes of the average of the two middle values is in some sense arbitrary). So returning to the original question, the chance that the length of the left-hand piece is <= x is the chance that both points are >= x, which is (1-x)^2. So the answer to Dan's question is the value where (1-x)^2 = 1/2. So the best guess is 1 - (sqrt(2) / 2). Andy -- Andy.Latto@pobox.com