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Basic Expected Value Problem (probability)

  1. Feb 25, 2008 #1
    E[X]=2
    Var(X)=3
    Find E[4+4x+x^2]

    I'm just confused what its asking. The expected value of this function is 2 so the average of it is 2 and the variance is how much it varies which is 3? Every example I have for expected values is related to an example such as cards, not just a polynomial
     
  2. jcsd
  3. Feb 25, 2008 #2

    Dick

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    Var(X)=E(X^2-E(X)^2). Just solve that for E(X^2). Then you can find E of the quadratic.
     
  4. Feb 25, 2008 #3
    Is the E(X^2-E(X)^2) = E(2^2-(4+4x+X^2)^2)?
     
  5. Feb 25, 2008 #4

    Dick

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    No.... Var(X)=3=E(X^2)-E(X)^2. E(X)=2. What's E(X^2)?? E(4+4X+X^2)=E(4)+E(4X)+E(X^2). Right? Etc. Use the linearity properties of 'E'.
     
    Last edited: Feb 25, 2008
  6. Feb 25, 2008 #5
    I think I'm mixing up the terms E(X^2) and e(X)^2. Which one is E[X]=2?

    So E(X^2)=E(4)^2+E(4x)^2+(x^2)^2?
     
  7. Feb 25, 2008 #6

    Dick

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    E(X)^2=4, since E(X)=2.
     
  8. Feb 25, 2008 #7
    Right but whats the difference between E(X^2) and E(X)^2?

    Is E(X^2)=E(4^2)+E(4x^2)+E((x^2)^2) with x=2?
     
  9. Feb 25, 2008 #8

    Dick

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    No! E(X^2) is not the same as E(X)^2. They aren't directly related to each other. The only way you can find E(X^2) from the information you are given is to use Var(X)=3.
     
  10. Feb 25, 2008 #9
    Var(X)+E(X)^2=E(X^2)
    3+4=7=E(x^2)

    Then use the fact that E(4+4X+X^2)=E(4)+E(4X)+E(X^2).
     
  11. Feb 25, 2008 #10

    Dick

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    Yes, that's it.
     
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