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Thread: mathematical statistics

  1. #1
    May 2015

    mathematical statistics

    please help me~!!mathematical statistics-1.png
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  2. #2
    MHF Contributor

    Aug 2006

    Re: mathematical statistics

    Quote Originally Posted by qnwk007 View Post
    please help me~!!Click image for larger version. 

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  3. #3
    MHF Contributor
    Sep 2012
    Washington DC USA

    Re: mathematical statistics

    I'm not great with this stuff, but I believe that what it's saying is this:
    The r.v. X takes on values in {1, 2, 3, 4, ... } with probabilites given by the formula $\displaystyle p_{X}(x)$. So for example, the probability that X = 5 is 1/32.
    When X takes on a value, that will mean that Z takes on a value. For example, if X came up 5, then Z would be 9 ($\displaystyle (5-2)^2$).
    Which values could Z take on? Could Z be negative? (No - it's the square of a number, so never negative.) Could it ever be anything other than an integer? (No - it's the square of something that's always an integer.).
    The probabilities for the things that Z can never be are always 0 - but what that set exactly is depends on exactly how the problem is phrased. Prob(Z = -1) = 0.
    Now get to the more meaty cases... and ask which non-negative integers Z might end up being.
    We already know that Z could end up being 9 (like when X = 5). Could Z end up being 8? No. 7? No. 6, or 5? No. 4? YES - when X = 4, since ($\displaystyle 4 = (4-2)^2$).
    Realize that Z could end up being any of the integer-squares, {0, 1, 4, 9, 16, 25, ... }, and that Z has 0 probability of taking on any value other than those in that set (Prob(Z = 5) = 0.)
    Now the question becomes figuring out the probabilities for each of those possible values that Z could be. That's what the pmf of Z is!
    So, for instance, what's the probability that Z = 0? Well that only happens when $\displaystyle (X-2)^2 = 0$, so when $\displaystyle X = 2$. And $\displaystyle X = 2$ happens with probability 1/4.
    Thus Prob(Z = 0) = 1/4.
    What's the probability that Z = 16? Well that only happens when $\displaystyle (X-2)^2 = 16$, so when $\displaystyle X = 6 or X = -2$. But $\displaystyle X = -2$ never happens (or has probability 0), and so Z = 16 only when X = 6... and X = 6 happens with probability 1/64.
    Thus Prob(Z = 16) = 1/64.
    This is how you figure out the pmf of Z... you find all these probabilities. They basically all work like the examples I did, with one exception. Computing Prob(Z = 1) requires just a little more attention.
    Hope that helps.
    Thanks from qnwk007
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