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Random variable distribution question

  1. Jun 1, 2015 #1
    1. The problem statement, all variables and given/known data
    Random variable x is defined on interval (1,3) and it has probability mass function f(x) =A(x2 +1=
    a) Find PMF, g(y) for y=x2
    b)Expectation of y
    c)Variance of y
    d)Distribution function of y.
    e)most probable value of y

    3. The attempt at a solution
    As far as a), i integrated from 1 to 3 f(x) and found A=3/32
    So, for g(y) i inserted x=square root of y into f(x), plus due to it is parabola, it is multiplied by two. No problem there, and also, y is defined on [1,9]
    c) and d) is easy, just integrate

    d) Distribution function of y, how to find that?
    I know that distribution function is F(X):=P(X<=x)= integral from -inf to x f(t)dt
    How do i apply it on this very case?

    e) Just derivative

    Thanks for reading and spending time to answer me :)
     
  2. jcsd
  3. Jun 1, 2015 #2

    RUber

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    Homework Helper

    I am not sure why you multiplied by 2 in part a. What do you mean by it is parabola? x is only defined on the positive side, so there is no duplication in y=x^2.
    For d) you want to find the integral of f(y).
    ##1 = A\int_{1^2}^{3^2}(y+1) dx##
    the tricky part in the integral is properly changing dx into dy. Once you do that, you can check your answer to see that it still integrates to 1.
    Then you simply put it back together into the form you need:

    ##F(y) = \left\{ \begin{array}{l l} 0 &\text{for } y \leq 1\\
    A\int_{1}^{\sqrt{y}}(x^2+1) dx & \text{for} 1< y < 9 \\
    1 &\text{for } y \geq 9 \end{array} \right.##
     
  4. Jun 1, 2015 #3

    Ray Vickson

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    The random variable does not have a pmf; it has a pdf or a cdf. I assume you want the pdf in part (a) and the cdf in part (d).

    Please write exactly your formula for g(y); what you describe sounds wrong, but I cannot be sure until you show the details.

    You say (c) and (d) is easy: "just integrate"; then you go on to say you do not know how to do (d)! In fact, the safest way to get g(y) is to first get G(y) = P(Y ≤ y), then differentiate it to get g(y) = dG(y)/dy. (In other words, if I were doing it I would do (d) first, then do (a)!) In turn, to get G(y), look at what is the event {Y ≤ y} back in terms of X and x.
     
    Last edited: Jun 1, 2015
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