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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

    mathman

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    The word variance seems to confuse you. By definition (almost) var(X)=E(X^2)-(E(X))^2.

    Also as you should know, E(A+B+C)=E(A)+E(B)+E(C)

    You should be able to work it out.
     
  4. Feb 25, 2008 #3
    So Var(x) = 3 = E(2^2-(E(4)+E(4x)+E(x^2))?

    Sorry I had a bad flu the past week so I missed my monday and wednesday lecture and our book is pretty lackluster
     
  5. Feb 25, 2008 #4
    Precisely.
     
  6. Feb 26, 2008 #5
    No, in your working, the brackets are in the wrong place, and you've ended up saying X = 2 in the first part, rather than E(X) = 2.

    Use the equations mathman said, rearrange your Var(X) equation to get E(X^2) = Var(X) + E(X)^2, and from there, its just plugging in what you already know.
     
  7. Feb 26, 2008 #6
    So I have 7=E(4)+E(4x)+E(x^2) but I'm confused what i do now. Do I plug in E(x)=2 so E(4x)=8 and E(x^2)=E(4)? if so where do you go from there, i'm just confused what i'm solving for
     
  8. Feb 26, 2008 #7
    Your not 'solving'

    I dont know where your getting this 7 from?

    E(4)+E(4X)+E(X^2), as you've stated, is E(4) + 4E(X) + E(X^2), which is E(4) + 4E(X) + Var(X) + E(X)^2, all of which you have values for. Again, you've said E(X^2) = E(4), here your saying X = 2, but X is a random variable, you cannot assume a value for X.
     
  9. Feb 26, 2008 #8
    E(X^2) = Var(X) + E(X)^2, with Var(X)=3 and E(X)^2=4 so 4+3 gives the 7.

    So do you not solve anything and it stays at 7=E(4)+4E(X)+E(X^2)?
     
  10. Feb 26, 2008 #9
    All your doing is working out the value of E[4+4X+X^2], and your using the fact that E(X^2) = 7 to help you, your not equating anything to 7.
     
  11. Feb 26, 2008 #10
    Your just trying to find the expected value of 4+4x+x^2, so all I have to say is E(4+4x+x6x^2)=E(4)+4E(x)+Var(X)+E(X^2)? Thats it? You dont say Var(X)=3 and E(X^2)=4 and plug that in?
     
    Last edited: Feb 26, 2008
  12. Feb 26, 2008 #11
    You've got to plug in the numerical values you know for E(4), E(X), Var(X) and E(X^2) to get a numerical answer.
     
  13. Feb 26, 2008 #12
    E(4)=4
    E(X)=2
    Var(X)=3
    E(X^2)=4

    E(4)+4E(X)+4E(X^2)=13?
     
  14. Feb 26, 2008 #13
    E(4+4X+X^2) = E(4) + 4E(X) + Var(X) + E(X)^2, as we discussed, I'm not sure where you've got your above equation from.
     
  15. Feb 26, 2008 #14
    Alright, I'm really confused. You said I know the values for E(4), E(X), Var(X) and E(X^2) and those need to be plugged in. You also said

    E(4+4x+X^2) = E(4)+4E(X)+E(X^2) = E(4)+4E(X)+Var(X)+E(X)^2

    Its given that E(X)=2, Var(X)=3. This gives E(X)^2=4 and I assumed E(4)=4? Plugging those in gives 4+(4*2)+3+4 which gives 19, is that the right approach? Forgot to multiply by 4 for that one term, typed the above one wrong, sorry. (I really appreciate the help)
     
    Last edited: Feb 26, 2008
  16. Feb 26, 2008 #15
    E(4) is 4, yes.

    So E(4)+4E(X)+Var(X)+E(X)^2 = 4 + 4*2 + 3 + 2^2 = 19
     
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