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Math Help - Random Variable.

  1. #1
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    Random Variable.

    I don't know where to start

    please help me, thanks

    Last edited by mr fantastic; November 8th 2011 at 12:12 PM. Reason: Title.
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  2. #2
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    Re: Random Variable VERY TRICKY

    Quote Originally Posted by Helpmeplease1 View Post
    Let X be the larger of the two numbers shown.
    I assume that the two numbers are on the faces of the dice.

    Let E=\{1,\dots,6\} and \mathcal{E}=2^E; then E is the domain of X. The definition says that X:\Omega\to E, X(x,y)=\max(x,y) is an (E,\mathcal{E})-valued random variable if X^{-1}(B)\in\mathcal{F}_2 for all B\in\mathcal{E}. That is, for any subset B of values that X may have, the set of outcomes from \Omega that produce an answer in B is a legitimate event, i.e., belongs to the set of events \mathcal{F}_2. For this example, this fact is obvious because \mathcal{F}_2 includes all possible subsets of \Omega.

    In contrast, X is not a random variable with respect to (\Omega,\mathcal{F}_1,\mathbb{P}) and (E,\mathcal{E}). E.g., consider X^{-1}(\{2\}).
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    Re: Random Variable VERY TRICKY

    Quote Originally Posted by emakarov View Post
    I assume that the two numbers are on the faces of the dice.

    Let E=\{1,\dots,6\} and \mathcal{E}=2^E; then E is the domain of X. The definition says that X:\Omega\to E, X(x,y)=\max(x,y) is an (E,\mathcal{E})-valued random variable if X^{-1}(B)\in\mathcal{F}_2 for all B\in\mathcal{E}. That is, for any subset B of values that X may have, the set of outcomes from \Omega that produce an answer in B is a legitimate event, i.e., belongs to the set of events \mathcal{F}_2. For this example, this fact is obvious because \mathcal{F}_2 includes all possible subsets of \Omega.

    In contrast, X is not a random variable with respect to (\Omega,\mathcal{F}_1,\mathbb{P}) and (E,\mathcal{E}). E.g., consider X^{-1}(\{2\}).
    Thanks so the inverse of {2} = (2,2) and you take the smallest one instead which is 2. And 2 isn't there, is this correct?
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  4. #4
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    Re: Random Variable VERY TRICKY

    No, here X=\max and \max\nolimits^{-1}(\{2\})=\{(1,2),(2,1),(2,2)\} because the maximum of all three pairs is 2. Now, \{(1,2),(2,1),(2,2)\}\notin\mathcal{F}_1.
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  5. #5
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    Re: Random Variable VERY TRICKY

    If I wanted to describe the distribution function Fx, I take it I do a step function?
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  6. #6
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    Re: Random Variable VERY TRICKY

    Quote Originally Posted by Helpmeplease1 View Post
    If I wanted to describe the distribution function Fx, I take it I do a step function?
    Anyone, i am also struggling to do the step function

    I know the chance of 1 = 1/36
    2 = 3/36
    3 = 5/26
    4 = 7/36
    5 = 9/36
    6 = 11/36
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  7. #7
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    Re: Random Variable VERY TRICKY

    Yes, so the cumulative distribution function F_X is from {1, ..., 6} to [0, 1]; F_X(n)=\mathop{\mbox{Pr}}(X\le n). So,

    F_X(1)=\mathop{\mbox{Pr}}(X=1)=1/36,

    F_X(2)=\mathop{\mbox{Pr}}(X=1\mbox{ or }X=2)=1/36+3/36,

    F_X(3)=\mathop{\mbox{Pr}}(X=1\mbox{ or }X=2\mbox{ or }X=3)=1/36+3/36+5/36,

    etc.
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  8. #8
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    Re: Random Variable VERY TRICKY

    Quote Originally Posted by emakarov View Post
    Yes, so the cumulative distribution function F_X is from {1, ..., 6} to [0, 1]; F_X(n)=\mathop{\mbox{Pr}}(X\le n). So,

    F_X(1)=\mathop{\mbox{Pr}}(X=1)=1/36,

    F_X(2)=\mathop{\mbox{Pr}}(X=1\mbox{ or }X=2)=1/36+3/36,

    F_X(3)=\mathop{\mbox{Pr}}(X=1\mbox{ or }X=2\mbox{ or }X=3)=1/36+3/36+5/36,

    etc.
    So to plot a step function

    would i draw a horizontal line between 0 and 1 on the x axis and 1 on the y -axis
    then a horizontal line between 1 and 2 on the x axis and 4 on the y -axis?

    or should i just put a dot instead of a line?
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  9. #9
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    Re: Random Variable VERY TRICKY

    Quote Originally Posted by emakarov
    Yes, so the cumulative distribution function F_X is from {1, ..., 6} to [0, 1]
    Correction: looking at the Wikipedia page, it seems that CDF is always defined on real numbers, even for discrete random variables. So,

    F_X(x)=0 for x<1,

    F_X(x)=1/36 for 1\le x<2,

    F_X(x)=1/36+3/36 for 2\le x<3,

    ...

    F_X(x)=1 for x\ge 6.

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