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Math Help - Distribution function of a random variable X

  1. #1
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    Distribution function of a random variable X

    The distribution function of X is given by

    F(b) = \\

    0 if b < 0 \\*

    \frac{b}{4} if 0 \leq b < 1\\

    \frac{1}{2} +\frac{b-1}{4} if 1 \leq b < 2\\

    \frac{11}{12} if 2 \leq b < 3\\

    1 if 3 \leq b\\

    I have to find
    a) P({X= i}), i = 1,2,3
    b) P({\frac{1}{2} < X < \frac{3}{2}})

    For a) I figured possible values for the random variable X were 0,1,2,3

    Doing some subtraction I got:

    P(X=0) = \frac{b}{4}

    P(X=1) = \frac{1}{4}

    P(X=2) = \frac{11}{12} - (\frac{1}{2} + \frac{b+1}{4})

    P(X=3) = 1 - \frac{11}{12} = \frac{1}{12}

    I am confused. What is b? Is it a possible value that the random variable can take on? Also, is part a) the same thing as saying "find the probability mass function"?

    And for part b), I am not sure where to begin with this one. Any help is appreciated.

    Thank you very much.

    -Jame

    (Also I apologize if the table is hard to read, that's the nicest I could get it with tex)
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  2. #2
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    None of those should contain a variable.
    P(X=1)=\frac{1}{4}.

    It is P(X=k)=F(k)-F(k-) where \displaystyle F(k-)=<br />
\lim _{b \to k^ -  } F(b)
    In other word, find the amount of the 'jump' at k.
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  3. #3
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    F(b) \equiv P(X \le b). b is just a dummy variable. Your answers should not have any b's in them.

    Also, no, X can assume values other than 0, 1, 2, 3. It can actually attain any value in (0, 2] \cup \{3\}.
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  4. #4
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    What do you mean by k-? I've never seen that notation before.

    Thank you for answering.
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  5. #5
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    Quote Originally Posted by Jame View Post
    What do you mean by k-? I've never seen that notation before.
    If you read reply #2 very carefully it was indeed explained there.
    It is very common notation. F(k-) denotes the left hand limit at k, \displaystyle\lim _{b \to k^ -  } F(b).

    The cumulative distribution function is continuous from the right.
    So we have F(k)=F(k+) where F(k+)=\displaystyle\lim _{b \to k^ +  } F(b).

    This is convenient notation for CDF's.
    P(X\le k)=F(k)~, P(X<k)=F(k-)~, P(j\le X \le k)=F(k)-F(j-).
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  6. #6
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    Oh yes, I see what you mean. We are concerned with the probability at the "jump". Sorry I was kinda thrown off by the b at first.

    The correct probabilities are

    P(X=1) = \frac{1}{4}
    P(X=2) = \frac{1}{6}
    P(X=3) = \frac{1}{12}

    Thanks again for the clarification, I see how this works now.

    Cheers
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