Probability Theory, Homework 3, due Friday Oct 3.

Scott Alister McKinleySeptember 25, 2014October 1, 2014Probability Assignments

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 $p \geq 0$ .

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

Hung says:

September 26, 2014 at 5:14 pm

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).

Reply
Jason says:

September 27, 2014 at 2:32 pm

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

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kalarack says:

September 27, 2014 at 2:35 pm

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.

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kalarack says:

September 27, 2014 at 6:04 pm

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?

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Scott Alister McKinley says:

September 29, 2014 at 11:40 am

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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Scott Alister McKinley says:

September 29, 2014 at 9:51 pm

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

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Ray Ray says:

September 30, 2014 at 9:33 pm

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

Reply
1. Scott Alister McKinley says:
  
  October 1, 2014 at 12:00 am
  
  Yes. That’s right. Sorry for the typo.
  
  Reply

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