Results 1 to 4 of 4

Thread: [solved] p(a)=1

  1. #1
    Moo
    Moo is offline
    A Cute Angle Moo's Avatar
    Joined
    Mar 2008
    From
    P(I'm here)=1/3, P(I'm there)=t+1/3
    Posts
    5,618
    Thanks
    6

    [solved] p(a)=1

    Hi

    A little question...

    Let $\displaystyle \Omega$ be the probability space.
    We know that if $\displaystyle P(A)=1$, then $\displaystyle P(\Omega \backslash A)=0$ (we don't necessarily have $\displaystyle A=\Omega$)

    Is there a proof (with formulae, not diagrams) for that :
    For any $\displaystyle X \subset \Omega$, if P(A)=1, then $\displaystyle P(A \cap X)=P(X)$
    ?

    Thanks
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Newbie
    Joined
    Jan 2009
    Posts
    11
    Since

    $\displaystyle P(A \cap X)=P(X \mid A) P(A)$
    if P(A)=1 its pretty clear right?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor

    Joined
    Aug 2008
    From
    Paris, France
    Posts
    1,174
    Quote Originally Posted by Moo View Post
    Hi

    A little question...

    Let $\displaystyle \Omega$ be the probability space.
    We know that if $\displaystyle P(A)=1$, then $\displaystyle P(\Omega \backslash A)=0$ (we don't necessarily have $\displaystyle A=\Omega$)

    Is there a proof (with formulae, not diagrams) for that :
    For any $\displaystyle X \subset \Omega$, if P(A)=1, then $\displaystyle P(A \cap X)=P(X)$
    ?

    Thanks
    Here's my proof:
    $\displaystyle P(X)=P(A\cap X)+P(A^c\cap X)$ where $\displaystyle A^c=\Omega\setminus A$,
    and $\displaystyle 0\leq P(A^c\cap X)\leq P(A^c)=1-P(A)=0$,
    so that $\displaystyle P(X)=P(A\cap X)$.

    It has the significant advantage (compared to the previous one) of avoiding to divide by 0.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Moo
    Moo is offline
    A Cute Angle Moo's Avatar
    Joined
    Mar 2008
    From
    P(I'm here)=1/3, P(I'm there)=t+1/3
    Posts
    5,618
    Thanks
    6
    Quote Originally Posted by Robert Hall View Post
    Since

    $\displaystyle P(A \cap X)=P(X \mid A) P(A)$
    if P(A)=1 its pretty clear right?
    Hmm So how would you prove that $\displaystyle P(X \mid A)=P(X)$ ? That's the problem...

    Quote Originally Posted by Laurent View Post
    Here's my proof:
    $\displaystyle P(X)=P(A\cap X)+P(A^c\cap X)$ where $\displaystyle A^c=\Omega\setminus A$,
    and $\displaystyle 0\leq P(A^c\cap X)\leq P(A^c)=1-P(A)=0$,
    so that $\displaystyle P(X)=P(A\cap X)$.

    It has the significant advantage (compared to the previous one) of avoiding to divide by 0.
    That's... erm perfect... :P

    Too bad Plato deleted his post, I can't remember it completely ><


    Thank you to all three for replying
    And special thanks to Laurent, because he can always provide the answer I'm looking for
    Follow Math Help Forum on Facebook and Google+

Search Tags


/mathhelpforum @mathhelpforum