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Discrete Random Variable - basic question in probability.

  1. Sep 12, 2009 #1
    1. The problem statement, all variables and given/known data

    2. Relevant equations

    Σ(n*2/5*(3/5)^(n-1)=5/2


    3. The attempt at a solution

    First I found the number of tosses needed to get heads, but I don't understand how to interpret this in the E[X] formula.
    I know that my
    p(x)=.40
    what is my x ? "tails for the first time" ? how to translate it in a numerical representation ?

    and the summation is from n=0 to n=2.5 ??


    Thanks,
    Roni.
     
  2. jcsd
  3. Sep 12, 2009 #2
    Seems like a geometric distribution.

    Let's define X = number of tosses needed until the first head appears.

    When x=1, it means that you get a head in your very first toss. So, p(x=1) = 0.4
    When x =2, it means that you get a tail in your first toss and a head in your second toss, so p(x=2) = (0.6)(0.4).
    When x = 3, ..., p(x=3) = (0.6)^2 (0.4)
    ...until...
    When x = n, ..., p (x=n) = (0.6)^(n-1) x (0.4)

    So, your E(X) = summation of x times p(x) [values of x is from 1 to infinity]
    = 1 x p(x=1) + 2 x p(x=2) + 3 x p(x=3) + ... + n x p(x=n)
    =1x0.4 + 2 x (0.6)(0.4) + 3 x (0.6)^2 x (0.4) + ... + n x (0.6)^(n-1) x (0.4)


    Factor out 0.4.

    Now, either the E(X) is a value or it might not be convergent. If it's the latter, than E(X) does not exist.

    Hope that helps. ;)
     
  4. Sep 12, 2009 #3
    Okay, I find the formula for calculating E(X) for a geometric distribution.

    E(X) = 1/p. (obtained by using the moment generating function)

    So in this case the answer is simply E(X) = 1/0.4 = 2.5 ;)
     
    Last edited: Sep 12, 2009
  5. Sep 12, 2009 #4
    Thank you for your help, it did help me understand.
    I just forgot to mention one thing. We should use calculus in our answers.
    I don't think we are supposed to use geometric distribution.

    However, geometric series might work here.
    I just don't know what to do with the 'n'.
    the sum of a geometric series is
    S=a/(1-r)
    the 'r' is 3/5
    but I can't figure out the 'a' since it's a constant number.

    Thanks.
     
    Last edited: Sep 12, 2009
  6. Sep 13, 2009 #5
    You're welcome. ;)
     
  7. Sep 13, 2009 #6
    I am still not able to solve the problem with the geometric series formula (a/(1-r)).
    Is it even possible ?


    Thanks
     
  8. Sep 14, 2009 #7
  9. Sep 14, 2009 #8
    I think you can use the moment generating function method to compute the sum. It's about differentiation and summation.

    http://answers.yahoo.com/question/index?qid=20080121114534AAXsi9J

    The mean is:

    E(X) =


    ∑ x * P(X = x)
    x = 0


    ∑ x * p * q ^ x
    x = 0

    change the index, the first terms of the sum is zero so we can start the index from 1


    ∑ x * p * q ^ (x - 1)
    x = 1

    factor out a p

    . . . ∞
    p * ∑ x * q ^ (x - 1)
    . . x = 1

    note that the arguement in the sum is a derivative:
    ∂ q^x / ∂q = x q ^ (x-1)

    . . . ∞
    p * ∑ ∂[q ^ (x - 1)] / ∂q
    . . x = 1

    it is not a trivial task to interchange a sumation and differentiation but it is valid in this case.

    . . . . . . . ∞
    p * ∂/∂q ∑ q ^ (x - 1)
    . . . . . . x = 1

    the sum is a geometric sum missing the first term and will evaluate to (1 / (1 - q) - 1)

    p * ∂/∂q * (1 / (1 - q) - 1)

    take the derivative

    p * (1 / (1 - q)²)

    write in terms of p

    p * ( 1/ (1 - (1-p)² )) = p / p² = 1/p

    qed.

    You can substitute in the value of p = 0.4 into the working shown above. ;)
     
  10. Sep 14, 2009 #9
    Yes, it is possible. Your expectation for this particular X is

    [tex]\sum_{n=1}^{\infty}n\left( \frac{2}{5} \right) \left( \frac{3}{5} \right)^{n-1}
    = \frac{2}{5} \sum_{n=1}^{\infty}n \left( \frac{3}{5} \right)^{n-1}
    [/tex]

    The sumation is the derivative of the geometric power series [itex]\sum x^n[/itex] evaluated at 3/5. Since that series sums to 1/(1-x) then our series is 1/(1-x)2.

    So

    [tex]\frac{2}{5} \sum_{n=1}^{\infty}n \left( \frac{3}{5} \right)^{n-1}
    =\frac{2}{5} \cdot \frac{1}{(1 - 3/5)^2} = \frac{2}{5} \cdot \frac{25}{4} = \frac{5}{2}
    [/tex].


    --Elucidus
     
  11. Sep 14, 2009 #10
    Thank you for your answer, you really helped me here.
    I was wondering if it's possible to solve it with Riemann Sums ?

    I know it's going to give me a good approximation.

    Thanks.
     
  12. Sep 14, 2009 #11
    Now that I am looking at your answer again... I don't think I understand how you are getting 1/(1-x)^2.
    I understand that 1/(1-x) would be perfect here if we didn't have the 'n'
    ....
    EDIT:

    Now, that I am looking at it again, I noticed that it's the derivative of 1/(1-x).

    Thanks.
     
    Last edited: Sep 14, 2009
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