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Math Help - Some random variable confusion!

  1. #1
    epv
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    Cool Any help here would be stupendous!!!

    Hello,

    I was a bit confused by this problem that appeared in my prob and stat textbook:

    "Suppose that X had a uniform distribution on the interval [0,5] and that the random variable Y is defined by Y = 0 if X <= 1, Y = 5 if X >= 3, and
    Y = X otherwise. Sketch the d.f. of Y"

    I wanted to solve this problem by first finding the mixed probability function of Y, f_Y. To do this I first found the probability function of X, f_X:

    f_X(x) = { 0 if x < 0; 1/5 if 0 <= x <= 5; 0 if x > 5 } by the fact that X has uniform distribution on [0,5].

    Then to find f_Y, the mixed probability function of Y:

    Pr(Y = 0) = Pr(X <= 1) = 1/5

    Also, it's given that Y = 5 if X >= 3 which means that

    Pr(Y = 5) = Pr(X >= 3) = 2/5

    The only solution I could think of was that after summing the places of non-zero probability mass at Y = 0 and Y = 5 to get 1/5 + 2/5 = 3/5, that then it forces Pr(0 < Y < 5) = 2/5. Then maybe would I interpret "Y = X otherwise" to mean that simply Y distrubutes the probability 2/5 uniformly over the interval (0,5)? In that case I would have:

    f_Y(y) = { 0 if y < 0; 1/5 if y = 0; 2/25 if 0 < y < 5; 2/5 if y = 5; 0 if y > 5 }

    And then the Pr(Y = 0) + Pr(0 < Y < 5) + Pr(Y = 5) = 1/5 + 10/25 + 2/5 =
    25/25 = 1 as needed. But I'm not sure that that's what is meant by "Y = X otherwise" Oh well, I'd really appreciate any help on determining the correct formulas for the mixed probability function of Y. Thanks in advance for any help,

    Last edited by epv; July 19th 2005 at 10:36 AM.
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  2. #2
    hpe
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    Quote Originally Posted by epv
    Hello,

    I was a bit confused by this problem that appeared in my prob and stat textbook:

    "Suppose that X had a uniform distribution on the interval [0,5] and that the random variable Y is defined by Y = 0 if X <= 1, Y = 5 if X >= 3, and
    Y = X otherwise. Sketch the d.f. of Y"
    Try this:

    <br />
F_Y(x) = \begin{cases} 0 \quad (x < 0}) <br />
\\ \frac15 \quad(0 \le x < 1) \\ <br />
\frac{x}{5} \quad (1 \le x < 3) <br />
\\ \frac35 \quad(3 \le x < 5)\\<br />
1 \quad (x \ge 5)<br />
\end{cases}<br />
    This expresses the fact that 1/5 of the probability is concentrated at x=0, 2/5 is spread out uniformly over [1,3], and 2/5 is concentrated at x=5.
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  3. #3
    epv
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    Thanks

    Thanks a bunch. Ah OK yeah I see where this is coming from now.

    E
    Last edited by epv; July 20th 2005 at 05:26 PM. Reason: self-realization
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