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Probability Question

  1. Mar 10, 2013 #1
    1. The problem statement, all variables and given/known data

    Random variable X takes values a,a+Δ,a+2Δ,...,a+nΔ with equal probabilities. (a) What is Ex? (b) What is Var X?

    2. Relevant equations



    3. The attempt at a solution

    X = [a,a+Δ,a+2Δ,...,a+nΔ]
    X = a+(X'-1)Δ
    where X' is some random variable
    X' = [1,2,...,n+1]
    I test to make sure this is true
    X = [a+(1-1)Δ,a+(2-1)Δ,a+(3-1)Δ...,a+(n+1-1)Δ]=[a,a+Δ,a+2Δ...,a+nΔ]
    So it's true.

    E[X] = E[a+(X'-1)Δ] = a+(E[X']-1)Δ

    E[X']=[itex]\sum^{n+1}_{X'=1}[/itex]X'P(X')

    Since X has equal probabilities then X' has equal probabilities so it's just
    P(X') = [itex]\frac{1}{number of entities}[/itex]
    number of entities in this case is n+1 so
    E[X']=[itex]\sum^{n+1}_{X'=1}[/itex]X'([itex]\frac{1}{n+1})[/itex]
    So I can pull out the constant from the summation
    =[itex]\frac{1}{n+1}\sum^{n+1}_{X'=1}X'=\frac{1}{n+1}(1+2+...+n+1)[/itex]
    I know that the quantity 1+2+...+n+1 can be expressed as [itex]1+\frac{(n+1)n}{2}[/itex] so...
    [itex]\frac{1}{n+1}(1+\frac{(n+1)n}{2}) = \frac{1}{n+1}+\frac{n}{2}[/itex]

    however the answer key which my professor made up says that it should be [itex]\frac{n}{2}+1[/itex]. However I don't see how my answer is wrong.

    Thanks for any help that anyone can provide me in solving this problem. My professor didn't show in the answer key how hey got this and just wrote it without showing any work. I think he may be wrong.
     
  2. jcsd
  3. Mar 10, 2013 #2

    Ray Vickson

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    [tex]\sum_{i=0}^n i = \sum_{i=1}^n i = \frac{1}{2}n(n+1).[/tex]
     
  4. Mar 10, 2013 #3
    But I want to sum to n+1 and not n. How does this effect the answer? I haven't taken calculus 2 in several semesters. I remember some change of base thing.

    [itex]\sum^{n+1}_{X'=1}X'=\sum^{n}_{X'=0}(X'-1)[/itex]

    Is this correct?
     
  5. Mar 10, 2013 #4
    So now that I think of it using the information you provided

    [itex]\frac{1}{n+1}\sum^{n+1}_{X'=1}X'=\frac{1}{n+1}\sum [1 + 2 + 3 + ... + n+1] = \frac{1}{n+1}(1+[1+ 2+ 3 + ... + n]) = \frac{1}{n+1}(1+\frac{1}{2}n(n+1))=\frac{1}{n+1}+.5 n[/itex]

    I still get the same answer
     
  6. Mar 10, 2013 #5

    Ray Vickson

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    It is still wrong. Start again, and do not bother to look at X'; just go with the original X.
     
  7. Mar 10, 2013 #6
    What am I doing wrong though? I can't seem to find what it is exactly that I am doing wrong.
     
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