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Math Help - Functions of a Discrete Random Variable

  1. #1
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    Functions of a Discrete Random Variable

    Hi there,

    I just wanna verify my solution to this:

    Let X be a Discrete Uniform random variable over the interval [-n, n], n \ge 1. What is the distribution of |X|?

    The PMF of the Discrete Uniform RV is given by: p_X(x) = \frac{1}{n-m+1}, x \in [m, n]

    Let Y be the RV such that Y = |X|

    Thus p_Y(y) = \sum_{\{x | |x| = y\}}p_X(x)

    Note that in this case x \in [-n, n], thus p_X(x) = \frac{1}{2n+1}

    p_Y(0) = \sum_{\{x | |x| = 0\}}p_X(x) = \sum_{\{x | x = 0\}}p_X(x) = \frac{1}{2n+1} since 0 \in [-n, n] as n \ge 1

    p_Y(1) = \sum_{\{x | |x| = 1\}}p_X(x) = \sum_{\{x | x = -1, 1\}}p_X(x) = p_X(-1) + p_X(1) = \frac{2}{2n+1}

    p_Y(2) = \sum_{\{x | |x| = 1\}}p_X(x) = \sum_{\{x | x = -2, 2\}}p_X(x) = p_X(-2) + p_X(2) = \frac{2}{2n+1}

    p_Y(n) = \sum_{\{x | |x| = 1\}}p_X(x) = \sum_{\{x | x = -n, n\}}p_X(x) = p_X(-n) + p_X(n) = \frac{2}{2n+1}

    Thus the PMF of Y = |X| is given by:

    p_Y(y) = \frac{1}{2n+1} if y = 0 and p_Y(y) = \frac{2}{2n+1} if y = 1, 2, \cdots, n

    Thanks
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  2. #2
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    Re: Functions of a Discrete Random Variable

    Correct indeed
    Thanks from usagi_killer
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