# 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. Hung says:

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. Jason says:

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

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.

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?

5. Scott Alister McKinley says:

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

6. Scott Alister McKinley says:

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

7. Ray Ray says:

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

1. Scott Alister McKinley says:

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