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 thoughts on “Probability Theory, Homework 3, due Friday Oct 3.

  1. 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).

  2. 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.

  3. 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?

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

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