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Probability Theory, Homework 3, due Friday Oct 3.

From Jacod and Protter

9.5  Let (\Omega, \mathcal{A}, \mathbb{P}) be a probability space. Suppose that X is a random variable with X \geq 0 almost surely and \mathbb{E}(X) = 1. Define Q : \mathcal{A} \to \mathbb{R} by Q(A) = \mathbb{E}(X 1_A). Show that Q defines a probability measure on (\Omega, \mathcal{A}).

9.7 Suppose that \mathbb{P}(X > 0) = 1 and let Q be defined as above. Let \mathbb{E}^Q denote expectation with respect to Q.  Show that \mathbb{E}^Q(Y) = \mathbb{E}^P(Y X).

Further exercises

Exercise 3. Suppose that X \geq 0 and Y \geq 0 are random variables and that p \geq 0.

  1. Prove \displaystyle \mathbb{E}[(X + Y)^p] \leq 2^p \big(\mathbb{E}(X^p) + \mathbb{E}(Y^p)\big).
  2. If p > 1, the factor 2^p may be replaced with 2^{p-1}.
  3. If 0 \leq p \leq 1, the factor 2^p can be replaced with 1.

Exercise 4. Suppose that \mathbb{E}(X^2) = 1 and \mathbb{E}(|X|) \geq a > 0. Prove for 0 \leq \lambda \leq 1 that

CORRECTION: \displaystyle \mathbb{P}\big(|X| \geq \lambda a\big) \geq (1 - \lambda)^2 a^2.

Hint:  Think Cauchy-Schwarz.

8 Comments Post a comment
  1. Hung #

    Prob 4. Let a=1/sqrt(1.81). Put X =0.1a wp 1/2 and 1.9a wp 1/2. Then E(X^2)=1 and E(X) = a.
    Let lamda = 0.11, then
    P(X>= 0.11a) = 1/2 and
    (1-0.11^2)/1.81=0.54 >P(X>= 0.11a).

    September 26, 2014
  2. Jason #

    For the last problem, exercise 2, should the conclusion read P( |X| < \lambda *a) instead of P( |X| \geq \lambda * a)?

    September 27, 2014
  3. Should the last problem conclusion read “P ( |X| < \lambda *a)" rather than "P ( |X| \geq \lambda * a)?

    I think I've shown the former, the later seems much harder, although I haven't given up trying, just thought I'd double check.

    September 27, 2014
  4. Appears by Hung’s Counterexample (If I’m reading it right) that we did want the inequality inside the 4th exercise to be < instead of \geq… unless someone has anything else to add?

    September 27, 2014
  5. Scott Alister McKinley #

    Hi guys, Looks like you’re right. I’ve made the adjustment to the statement of the problem. Sorry for the typo.

    September 29, 2014
  6. Scott Alister McKinley #

    Nope. Everyone was wrong. I think I’ve got a proof of the corrected statement above for problem 4.

    September 29, 2014
  7. Ray Ray #

    Should 3(c) say the factor 2^p can be replaced with 1?

    September 30, 2014
    • Scott Alister McKinley #

      Yes. That’s right. Sorry for the typo.

      October 1, 2014

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