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Homework Help: Prob and stat expected value of x

  1. Jan 2, 2009 #1
    1. The problem statement, all variables and given/known data
    The distribution function of a random variable X is given by

    F(x) = {
    0 if x < -3
    3/8 if -3 [tex]\leq[/tex] x < 0
    1/2 if 0 [tex]\leq[/tex] x < 3
    3/4 if 3 [tex]\leq[/tex] x < 4
    1 if x [tex]\geq[/tex] 4


    Calculate E(X), E(X[tex]^{2}[/tex] - 2|X|), E(X|X|).


    2. Relevant equations



    3. The attempt at a solution
    I actually have no idea how to start this. None of the chapter examples seem to give me a clue on what to do. I have calcuated E(X) for probability mass function equations, but not distributive functions like the above. Can someone give me a strong hint on how to approach this for E(X) and I can take it from there? Thank you for any help.

    And sorry for the formatting, I could not get it to look right in latex, but I think it should be understandable.
     
  2. jcsd
  3. Jan 2, 2009 #2

    radou

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    Well, what is the definition of E(X)? Further on, what is the relation between F(X) and f(x) (i.e. the probability density function f of the random variable X)?
     
  4. Jan 2, 2009 #3

    radou

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    Okay, another hint. Since you need to find f(x) to calculate E(X), is there a way to calculate f(x) from F(x)? [Hint 2: f(xi) = F(xi) - F(xi-), where F(xi-) is the left limit of F when x --> xi form the left.]
     
  5. Jan 2, 2009 #4
    Well, E(X) is the expected value of X. E(X) = [tex]\sumxp(x)[/tex] where p(x) is the probability mass function. My main issue with the problem is that I am confused by the lack of an equation...

    okay, so I started typing that and now I am wondering if it is as simple as this:
    -3 * 3/8 + 0 * 1/8 + 3 * 2/8 + 4 * 2/8 = 5/8

    I only looked at the numbers where it jumped on a graph and I looked at actual chances of each happening. In other words I subtracted f(x2) - f(x1) to get the chances of f(x2). I know the answer is 5/8, am I doing this correct or did I just get a coincidentally equal number?
     
  6. Jan 2, 2009 #5

    radou

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    Yes, my result is 5/8 too. So, the random variable is given with [tex]X = \[ \left( \begin{array}{cccc}-3 & 0 & 3 & 4 \\ 3/8 & 1/8 & 1/4 & 1/4 \end{array} \right)\] [/tex]. If you "calculate" F(x) from X, you'll get your given function F.
     
  7. Jan 2, 2009 #6
    so then what I really mean to say would be:
    -3 * (3/8 - 0) + 0 * (1/2 - 3/8) + 3 * (3/4 - 1/2) + 4 * (1 - 3/4) = 5/8

    technically the same thing, but clearer in showing what I was doing

    and for E(X - 2|X|) would I then plug in -3 and replace the -3 above with the solution? such as (-3[tex]^{2}[/tex] - 2|-3|) = 3
    so the equation above would start as 3 * (3/8 - 0) + ... ?
    *edit*dropped the square, fixed now
     
  8. Jan 2, 2009 #7
    answered my own question by working it out to find that I got the correct answer. Thanks for the hints radou, it is making sense now!
     
  9. Jan 2, 2009 #8

    radou

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    No problemo! It's nice to hear when someone works the answers out alone. :wink:
     
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