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).
If Farmer George starts painting in the morning, and only has 3/4 of a chicken coop left to paint: 3-3/4= 12/4-3/4=9/4 painting in the morning To put it in a mixed number, 2 1/4 (aka 2.25)
