Ask Question
11 May, 22:56

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).

+5
Answers (1)
  1. 12 May, 01:35
    0
    Step-by-step explanation:

    idek
Know the Answer?
Not Sure About the Answer?
Find an answer to your question ✅ “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 ...” in 📘 Mathematics if you're in doubt about the correctness of the answers or there's no answer, then try to use the smart search and find answers to the similar questions.
Search for Other Answers