Re:The Man Who Invented Modern Probability
by tzs on Thursday August 22, 2013 @06:06PM
(#6260415)
Nope. To see where the subtlety lies in this problem, simplify it. Change it to a discrete problem: the side can be 1, 3, or 5, each with probability 1/3. The expected value of the side length is still 3.
1/3 of the cubes have volume 1, 1/3 have volume 27, and 1/3 have volume 125. The expected value of the volume is 51. Think about why it is 51, not 27, and you'll see why it is not 27 for the continuous case.
In general, if you have a random variable X, and some function of that random variable, f(X), then it is NOT true that expected value of f(X) = f(expected value of X).
1/3 of the cubes have volume 1, 1/3 have volume 27, and 1/3 have volume 125. The expected value of the volume is 51. Think about why it is 51, not 27, and you'll see why it is not 27 for the continuous case.
In general, if you have a random variable X, and some function of that random variable, f(X), then it is NOT true that expected value of f(X) = f(expected value of X).