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Math Help - probability condition F (x | y)

  1. #1
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    Question probability condition F (x | X<K)

    I know that maybe this issue can be pre-college, but hey, an apology in advance.
    This is the problem:

    "The random variable X is defined on the interval [0,2]. It is known that:
    P (X = 1) = \frac{1}{4}
    F (x | X <1) = x^2
    F (x | X> 1) = x-1
    E (X) = 1

    calculate: F (1)."



    then, i did this:

    (i) F(x|X<1)=\sum\limits_{0}^{2} p(x|X<1) .... I apply the Baye’s theorem:  F(x|X<1)=\sum\limits_{0}^{2}\frac{p(X<1|x) *p(x)}{P(X<1)}

    on the other hand: p(X<1)=p(X=0) cuz  X \rightsquigarrow [0,2] \Longrightarrow F(x|X<1)=\sum\limits_{0}^{2}\frac{p(X=0|x) *p(x)}{P(X=0)} = x^2


    (ii) in the case of F(x|X>1)=x-1 :
    F(x|X>1)=\sum\limits_{0}^{2} p(x|X>1) …. I apply the Baye’s theorem: F(x|X>1)=\sum\limits_{0}^{2} \frac{p(X>1|x) *p(x)}{P(X>1)}

    on the other hand: p(X>1)=p(X=2) cuz X \rightsquigarrow [0,2] \Longrightarrow F(x|X>1)=\sum\limits_{0}^{2} \frac{p(X=2|x) *p(x)}{P(X=2)} = x-1


    (iii) p(X=0)=\frac 3 4 - p(X=2)

    p(X=2)=\frac 3 4 - p(X=0)

    Note: notice the difference between X and x (uppercase and lowercase)

    then I think I'm wrong from the foundation, we would appreciate if someone guide me in this problem, and another thing, is   P (X = 1) = P (1)?
    Last edited by killertapia; January 4th 2010 at 10:13 PM. Reason: switch to LaTex
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  2. #2
    MHF Contributor matheagle's Avatar
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    is this zero again?
    I have no idea what F (x | X <1) means.
    is that P(X<x|X<1)?
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  3. #3
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    Smile

    Quote Originally Posted by matheagle View Post
    is this zero again?
    I have no idea what F (x | X <1) means.
    is that P(X<x|X<1)?
    ready, so I edited in code "LaTex" I hope it is better understood
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  4. #4
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    Quote Originally Posted by killertapia View Post
    ready, so I edited in code "LaTex" I hope they will understand better
    I'm strugglng to understand what you're trying to say here.

    It appears that X is a discrete random variable that can take on the value 0, 1 and 2 with non-zero probability.

    You know Pr(X = 1) = 1/4.
    From E(X) = 1 it follows that Pr(X = 2) is 3/8.
    From Pr(X = 0) + Pr(X = 1) + Pr(X = 2) = 1 it follows that Pr(X = 0) = 3/8.

    So the distribution is now completely defined. I have no idea what F is meant to be but whatever it is, F(1) can surely be calculated from what I have said above.
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