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Coefficient of correletion problem

  1. Jul 23, 2011 #1
    a vase contains one red ball two white balls and three black balls. n balls are take out of the vase and (each ball returned to it afterwards). let B denote the number of black balls taken out and R denote the number of red balls taken out. what is the coefficient of correlation between R and B?

    well i know that coeffieicnt of correlation = [itex]\frac{cov(R,B)}{\sigma R*\sigma B}[/itex]
    and that R and B are simply bernouli trials with x and n-x trials respectivly.

    my problem is calculating the covariance = E[RB] -E[R]E (E is the mean).
    E[R] and E are straightforward but E[RB] is a bit trickier for me.

    i thought of using the definition of the mean with a multinomial vector for the shared probability for R and B and summing over 0<x<n, but is there an easier way?
  2. jcsd
  3. Jul 23, 2011 #2
    It would be easy if it weren't for the white balls... As it is, I don't see any way of doing it short of writing out the sum.
  4. Jul 23, 2011 #3
    The number of red balls is R = sum_{i=1}^n I[ball i is red] where I is the indicator function so to calculate E[RB] you just need to expand the product.
  5. Jul 23, 2011 #4
    bpet, i am sorry, but can you elaborate?

    and i thought of using the smoothing theorom,
    Due to the fact that it is easier to think in terms of conditional probability in this problem.



    B is distributed binomially with n trials and probability P(B).

    the problem here is that i get a different answer for E[E[RB|R]]
    what am i doing wrong?
  6. Jul 23, 2011 #5
    sry for the double post but believe i understood the part about expanding the product :).

    tell me if this is correct

    [itex]R = R1+R2 +R3 ..... + Rn[/itex]

    [itex] R = 1 , p=1/6 [/itex]
    [itex] R = 0 , p=5/6 [/itex]

    and likewise for B

    [itex]E[RB] = E[(R1+R2+...+Rn)(B1+B2+...+Bn)= (n^2-n)*P(B)*P(R)][/itex]
    Ri*Bi is always 0 becuase one being 1 implies the other being 0.

    [itex]E[RB]-E[R]E = (n^2-n)*P(B)*P(R)-n^2*P(B)*P(R) =- n*P(R)*P(B) = -n/12[/itex]
    Last edited: Jul 23, 2011
  7. Jul 23, 2011 #6

    If you change the RHS to E[B*(n-B)*P(R|B)] does that work?
  8. Jul 24, 2011 #7
    i don't see how P(R|B) changes anything becuase (n-B)*P(R) is the mean of R|B (where B=s , 0<s<n).
    P(R) is the probability that in a given "trial" (pulling a ball out) you get red, which is independent from the amount of black balls you took out.
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