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Statistics: independently distributed mean and variance

  1. Jul 9, 2014 #1
    1. The problem statement, all variables and given/known data

    Math and verbal SAT scores are each N(500, 10000)

    1)If the math and verbal SAT scores were independently distributed, which is not the case, then what would be the distribution of the overall SAT scores? Find its mean and variance.


    2. Relevant equations



    3. The attempt at a solution

    So originally they are not independent variables, yet the share the same mean and variance. So, if we now assume they are independent, the mean and variance for the two would still be the same, therefore the answer is still N(500,10000). Is this correct?
     
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  3. Jul 9, 2014 #2

    Ray Vickson

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    How is the overall SAT score determined from the math and verbal scores? If it is their sum or their arithmetic average, then NO, you are incorrect.
     
  4. Jul 9, 2014 #3
    "How is the overall SAT score determined from the math and verbal scores? If it is their sum or their arithmetic average, then NO, you are incorrect. "

    Im not sure I follow... I do not know how the OVERALL scores are determined. I gave the problem word for word as it was given.

    Are you saying I should treat it as say...

    X = Math where X~N(500,10000), Y = Verbal where Y~N(500,10000)

    and then I find E(X+Y) and V(X+Y)???

    E(X+Y) = E(X)+E(Y) = 500+500=1000

    V(X+Y) = V(X) + V(Y) +2cov(X,Y) where X&Y are independent so cov(X,Y) = 0
    .... = 10000+10000=20000

    ???
     
  5. Jul 9, 2014 #4

    Orodruin

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    Something like that, yes. Assuming that the total score is the sum, otherwise you need to do the corresponding thing to whatever other formula is used.
     
  6. Jul 9, 2014 #5
    Ok, thanks...

    The question continues...

    2) Next assume that the correlation coefficient between the math and verbal scores is .75, Find the mean and variance of the resulting distribution

    so... I got .75 = cov(X,Y)/(sigma_x * sigma_y) = [E(XY) - E(X)E(Y)]/(sigma_x * sigma_y)

    where x & y are from the same distributions so X = Y...

    = [E(X^2) - (E(X))^2]/(sigma_x^2) = V(X)/sigma_x^2 = V(X)/V(X)=1 = incorrect???
     
  7. Jul 9, 2014 #6

    Orodruin

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    That ##E(X) = E(Y)## does not imply that ##E(XY) = E(X^2)## (consider the case where they were independent variables). Instead, I suggest solving for the covariance and inserting it instead of 0 in your original expression in post #3.
     
  8. Jul 9, 2014 #7
    So cov(X,Y) = E(XY) - E(X)E(Y)

    I looked through my book and searched the web but I do not see how to solve E(XY). I know E(XY) = x*y*p(x,y) but how do I do it when all I know is that X & Y ~ N(500,10000)????
     
    Last edited: Jul 9, 2014
  9. Jul 9, 2014 #8

    Orodruin

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    The point is that you do not need to compute E(XY). You have:

    as well as

    V(X+Y) = V(X) + V(Y) + 2 cov(X,Y)
     
  10. Jul 9, 2014 #9


    I must be missing something... My thought was I solve V(X+Y) for cov(X,Y) and then plug it into my .75 = ... equation.

    cov(X,Y) = [V(X+Y) - V(X) - V(Y)] = V(X+Y) - 20000

    .75 = [V(X+Y) - 20000]/(sigma_x*sigma_y) = V(X+Y) - 20000/(100*100)

    therefore V(X+Y) = .75(10000)+20000 = 27500 I feel like im off track???

    27500 = V(X) + V(Y) + 2 cov(X,Y) = 20000 + 2cov(X,Y)

    cov(X,Y) = (27500-20000)/2 = 3750

    I don't think I should be using the the "old" variance to find the new mean and variance... how does this end of giving me a new mean and variance
     
  11. Jul 9, 2014 #10

    Orodruin

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    Why? You just computed the variance of X+Y which was the aim ... (although missing a factor of two when solving for the covariance)
    The more straightforward way would have been to solve for the covariance first:
    $$
    0.75 = \frac{{\rm cov}(X,Y)}{\sqrt{V(X)V(Y)}}= \frac{{\rm cov}(X,Y)}{V(X)} \quad \Longrightarrow \quad {\rm cov}(X,Y) = 0.75 V(X) = 7500
    $$
    It follows that
    $$
    V(X+Y) = V(X) + V(Y) + 2{\rm cov}(X,Y) = 20000 + 15000 = 35000
    $$
     
  12. Jul 9, 2014 #11
    Ohhh okay, I don't think I really understood what the question was asking. So the question is asking for the TOTAL or COMBINED score distribution given that correlation coefficient is .75, which is why I solved for V(X+Y).

    Soo then what about the E(X+Y), is that simply E(X) + E(Y) = 1000, because the correlation coefficient doesn't really involve the expected values. Probably not the best way of explaining it but am I on the right track.
     
  13. Jul 10, 2014 #12

    Orodruin

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    The expectation value is linear so E(X+Y) is always equal to E(X)+E(Y) regardless of how X and Y are distributed. The reason you are not allowed to do E(XY) = E(X^2) is that, although they are distributed in the same way, X and Y are different stochastic variables. In any given outcome, X and Y may take on different values, while in X^2 the value of X is always fully correlated with the value of X.
     
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