help me please

**ms&ms** · 10-24-2010 04:30 PM

[Hello,

I would like to know if someone can help for the question as below?

Suppose that P(E) is 0 or 1, show that E is independent of all events A.

Suppose that E is independent of itself, show that P(E) is 0 or 1.

**BGM** · 10-25-2010 06:00 AM

First note that $E\cap A \subseteq E ~ \forall A \in \mathcal{F}$ . Hence,
if $P(E) = 0$ , then $P(E \cap A) \leq P(E) = 0 \Rightarrow P(E \cap A) = 0 = 0 \times P(A) = P(E)P(A)$

Then using the inclusion-exclusion principle, consider

$\forall A \in \mathcal{F}, P(E \cup A) = P(E) + P(A) - P(E \cap A)$

If $P(E) = 1$ , then $1 + P(A) - P(E \cap A) = P(E \cup A) \leq 1$

$\Rightarrow P(A) \leq P(E \cap A)$

But again we know that $E\cap A \subseteq A ~ \forall A \in \mathcal{F}$ , hence

$P(E \cap A) = P(A) = 1 \times P(A) = P(E)P(A)$

$E$ is independent from itself if and only if $P(E \cap E) = P(E)P(E)$

$\Rightarrow P(E) = P(E)^2$

Result follows.

**SadieKhan** · 10-25-2010 06:14 AM

Off the topic,but can you tell me how do you use all those symbols?
I use word and I could never find sqrt sign :|
Also integral.
Sorry to hijack your thread.

**SadieKhan** · 10-25-2010 06:18 AM

Oh its an html code perhaps, I clicked and got the tags.:P

**SmoothJohn** · 10-25-2010 10:45 AM

Hi Sadie. I think the answer you are after is in the General Discussions, under Math Typesetting in LaTex.

http://www.talkstats.com/showthread.php?t=11122

**ms&ms** · 10-25-2010 11:38 AM

Thanks a lot, it's very helpful

BGM

[/CENTER]

**SadieKhan** · 10-25-2010 02:22 PM

$coooool$
$;)$

**real123** · 10-27-2010 10:57 AM

:shakehea help me please

Suppose that (Ω ,F, P) is a probability space and E belong F satisfies p(E)>0 .
Let Q: F ----> [0,1] be defined by Q(A)=P(A/E) . Show that (Ω ,F, Q) is a probability space

**BGM** · 10-27-2010 12:57 PM

To check $(\Omega, \mathcal{F}, Q)$ is a probability space,
it is sufficient to check $Q$ is a probability measure,
by checking the three axioms of probability.

http://en.wikipedia.org/wiki/Probability_axioms

Since $P$ is a valid probability measure, it satisfy
all 3 axioms of probability.

1. Note $\forall A \in \mathcal{F}, A\cap E \in \mathcal{F}$ as $E \in \mathcal{F}$

By the axiom of probability,
$\forall A \in \mathcal{F}, P(A\cap E) \geq 0 ~ \mbox{and} ~ P(E) > 0$
$\RightarrowQ(A) = P(A|E) = \frac {P(A\cap E)} {P(E)} \geq 0$

2. Note $E \subseteq \Omega \Rightarrow \Omega \cap E = E$
Hence, $Q(\Omega) = P(\Omega|E) = \frac {P(\Omega \cap E)} {P(E)} = \frac {P(E)} {P(E)} = 1$

3. Note $\forall A_1, A_2, ... \in \mathcal{F} ~ \mbox{with} ~A_i \cap A_j = \varnothing ~ \forall i \neq j,$
$A_1 \cap E, A_2 \cap E, ... \in \mathcal{F}$
$(A_i \cap E) \cap (A_j \cap E) = (A_i \cap A_j) \cap E= \varnothing \cap E = \varnothing ~ \forall i \neq j$

Hence,
$Q\left(\bigcup_{i=1}^{+\infty} A_i\right)= P\left(\left. \bigcup_{i=1}^{+\infty} A_i\right|E\right) = \frac {{\displaystyle P\left( \left (\bigcup_{i=1}^{+\infty} A_i\right) \cap E\right) }} {P(E)}$

$= \frac {{\displaystyle P\left( \bigcup_{i=1}^{+\infty} (A_i \cap E) \right) }} {P(E)}$

$= \frac {{\displaystyle \sum_{i=1}^{+\infty} P(A_i \cap E) }} {P(E)}$ (by the axiom of probability)

$= \sum_{i=1}^{+\infty} \frac {P(A_i \cap E)} {P(E)} = \sum_{i=1}^{+\infty} P(A_i|E) = \sum_{i=1}^{+\infty} Q(A_i)$

Q.E.D.

**jamesmartinn** · 10-27-2010 03:33 PM

BGM, you are a machine.

**real123** · 10-27-2010 03:54 PM

Thank you very much wish you a good life

**ms&ms** · 11-01-2010 12:25 PM

thank uuuuu

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