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Thread: Probability measure on Borel-sigma

  1. #1
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    Probability measure on Borel-sigma

    Let $\displaystyle (\Omega , \mathcal{F} , \mathbb{P} ) $ be a probabilty space and $\displaystyle X: \ \Omega \rightarrow \ \bar{\mathbb{R}} $ be a random variable with distribution $\displaystyle \mu $. Show that $\displaystyle \mu$ is a probabilty measure on $\displaystyle \bar{\mathcal{B}} $. That is, verify the three axioms of probabilty for $\displaystyle \mu $, using the collection of "events" on $\displaystyle \bar{\mathcal{B}} $.
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  2. #2
    Moo
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    Hello,

    But how is $\displaystyle \mathcal{\overline B}$ defined ?
    Is it the Borel algebra over $\displaystyle \mathbb{\overline R}$ ?
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  3. #3
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    Hi Moo,
    Yes, $\displaystyle \mathcal{\overline B} $ is the borel algebra over $\displaystyle \mathbb{\overline R} $
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    I've become stuck with the question. My teacher said that I need to use the fact that $\displaystyle \mathbb{P} $ is a probability measure and that $\displaystyle X $ is a random variable. I've racked my brain but can't work it out Whats more annoying is that the answer is staring me right in the face.
    I know that the 3 axioms of probability are

    (1) $\displaystyle 0 \leq \mathbb{P} \leq 1 $ for some event A

    (2) $\displaystyle \mathbb{P} (\Sigma) = 1 $

    (3) For any sequence $\displaystyle A_1, A_2, ...$ which are disjoint
    $\displaystyle \mathbb{P} ( \bigcup_i A_i ) = \sum_i \mathbb{P} (A_i). $

    I just can't get my head around this and any help would be greatly appreciated.
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  5. #5
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    I'm just learning about probability and measure theory myself, and I am not a mathematician. So here's my best shot at it:

    We have that $\displaystyle X: (\Omega,\mathcal{F}) \rightarrow (\mathbb{\bar{R}},\mathcal{\bar{B}})$, and I suppose that makes $\displaystyle X$ an extended random variable. Now, $\displaystyle X$ induces a probability measure on $\displaystyle \bar{\mathcal{B}}$, namely$\displaystyle \mathbb{P}_X(B) = \mathbb{P}\{\omega: X(\omega) \in B\},~~B \in \bar{\mathcal{B}}$. The distribution function of$\displaystyle X$ is then $\displaystyle \mu(x) = \mathbb{P}\{\omega: X(\omega) \leq x\}$. Now, because for $\displaystyle a < b$, $\displaystyle \mu(b) - \mu(a) = \mathbb{P}\{\omega: a < X(\omega) \leq b\} = \mathbb{P}_X(a,b]$, the distribution $\displaystyle \mu$ corresponds to the measure $\displaystyle \mathbb{P}_X$. We should have $\displaystyle \mu(x) = 1$ as $\displaystyle x \rightarrow \infty$ and $\displaystyle \mu = 0$ as $\displaystyle x \rightarrow -\infty$. Is that of any help?

    By the way, which book do you use to study measure theory?
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  6. #6
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    If X is a random variable then you can think of $\displaystyle X^{-1}:\mathcal{\overline{B}}\rightarrow \mathcal{F}$. Now $\displaystyle \mu=\mathbb{P}\circ X^{-1}$.

    Quote Originally Posted by funnyinga View Post
    (1) $\displaystyle 0 \leq \mathbb{P} \leq 1 $ for some event A
    This is immediately obvious.
    (2) $\displaystyle \mathbb{P} (\Sigma) = 1 $
    We have that $\displaystyle X^{-1}(\overline{\mathbb{R}})=\Omega$, and so this is obvious as well.
    (3) For any sequence $\displaystyle A_1, A_2, ...$ which are disjoint
    $\displaystyle \mathbb{P} ( \bigcup_i A_i ) = \sum_i \mathbb{P} (A_i). $
    Take $\displaystyle A_1,A_2,... \in \overline{\mathcal{B}}$ disjoint, i.e. $\displaystyle \bigcap_{i \in \mathbb{N}}A_i=\emptyset$. This implies that $\displaystyle X^{-1}\left(\bigcap_{i \in \mathbb{N}}A_i\right)=\bigcap_{i \in \mathbb{N}}X^{-1}(A_i)=\emptyset$. As for each i, we have $\displaystyle X^{-1}(A_i) \in \mathcal{F}$ the result follows.
    Last edited by Focus; Aug 12th 2009 at 09:53 AM. Reason: Nonsensical assumption removed
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