Chucky

Mathematics

1 October, 22:08

You have a ruler of length 1 and you choose a place to break it using a uniform probability distribution. Let random variable X represent the length of the left piece of the ruler. X is distributed uniformly in [0, 1]. You take the left piece of the ruler and once again choose a place to break it using a uniform probability distribution. Let random variable Y be the length of the left piece from the second break. (a) Find the conditional expectation of Y given X, E (Y/X). (b) Find the unconditional expectation of Y. One way to do this is to apply the law of iterated expectation which states that E (Y) = E (E (Y/X)). The inner expectation is the conditional expectation computed above, which is a function of X. The outer expectation finds the expected value of this function. (c) Compute E (XY). This means that E (XY/X) = XE (Y/X) (d) Using the previous results, compute cov (X, Y).