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Homework Help: Proof for expectation

  1. Jun 15, 2010 #1
    hello!
    can any1 please help me with the following proofs? thanks


    let X and Y be random variables. prove the following:
    (a) if X = 1, then E(X) = 1

    (b) If X ≥ 0, then E(X) ≥ 0

    (c) If Y ≤ X, then E(Y) ≤ E(X)

    (d) |E(X)|≤ E(|X|)

    (e) E(X)= [tex]\sum[/tex]P(X≥n)
     
  2. jcsd
  3. Jun 15, 2010 #2

    EnumaElish

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    What is the definition of expectation? Have you attempted any of these proofs?
     
  4. Jun 15, 2010 #3
    expectation is the expected value or mean.

    I have tried the first one using probability density function. but am not sure of my answer. while the others I have no idea how to attempt them

    thank you
     
  5. Jun 15, 2010 #4

    EnumaElish

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    Do you mind copying your answer for (a)?
     
  6. Jun 15, 2010 #5
    E(X) = [-∞]\int[/∞] g(x).f(x) dx

    let g(x) = X

    E(X) = [-∞]\int[/∞] 1.f(x) dx

    = 1. [-∞]\int[/∞] f(x) dx

    = 1


    P.S: [-∞]\int[/∞] is the integral of -infinity to infinity
     
  7. Jun 15, 2010 #6

    EnumaElish

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    Just a notational remark,
    should be E(g(x)). Other than this it looks right.

    What about (b)? Any ideas?
     
    Last edited: Jun 15, 2010
  8. Jun 15, 2010 #7
    ok thanks.

    nope...no idea for the second part
     
  9. Jun 15, 2010 #8

    EnumaElish

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    What is the definition of integral that you've been using?
     
  10. Jun 15, 2010 #9
    its the probability density function for a continuous distribution

    the integral gives the total area under the pdf
     
  11. Jun 15, 2010 #10

    EnumaElish

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    What are the characteristics of a pdf? Which conditions must hold for a function to be a pdf?

    For example, can f(x) = -1 for x = 0 to 1 be a pdf?
     
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