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Math Help - Probability help please

  1. #1
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    Probability help please

    Suppose we choose a random point in the interval (-2, 1) and denote the distance to 0 by X. Prove that X is a continuous random variable.
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by clockingly View Post
    Suppose we choose a random point in the interval (-2, 1) and denote the distance to 0 by X. Prove that X is a continuous random variable.
    Define what you mean by a random point - no such thing without an
    explict method for generating it.

    Then define what you mean by a continuous random variable.

    RonL
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  3. #3
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    Quote Originally Posted by clockingly View Post
    Suppose we choose a random point in the interval (-2, 1) and denote the distance to 0 by X. Prove that X is a continuous random variable.
    If you begin with a random variable Y that is uniformly distributed on the interval (-2,1), then it is reasonable to think of Y as a randomly chosen point. Then by definition X=|Y|, Ys distance from zero. Now one characterization of continuous random variable is: the range of the variable includes an interval of real numbers, bounded or unbounded. The range of X is [0,2).
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    Thanks, CaptainBlack and Plato for responding!

    I was wondering - would I be correct by saying that the density of X is
    1/(-2-1) = -1/3

    and the mean of X is

    (1+-2)/2 = 1/2

    ?
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  5. #5
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    Quote Originally Posted by clockingly View Post
    I was wondering - would I be correct by saying that the density of X is
    Assuming that I am correct about Y being uniformly distributed on the interval (-2,1), and X=|Y| then the density function is a bit hard to see.
    Consider this: f(x) = \left\{ {\begin{array}{lc}<br />
   {\frac{2}{3}} & {0 \le x \le 1}  \\<br />
   {\frac{1}{3}} & {1 < x \le 2}  \\<br />
\end{array}} \right.

    You may want to check out the integral of f(x) over [0,2] to see how it all works.
    In particular, you want use the integral of xf(x) to find the mean.
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