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Find probability density function from Central Limit theorem

  1. Oct 25, 2012 #1
    1. The problem statement, all variables and given/known data
    How can I derive the probability density function by using the Central Limit theorem?
    For an example, let's say that we have a random variable Xi corresponding to the base at
    the ith position; to make even simpler, let's say all probabilities are equal. If we have four variable, P(Xi = A) =P (Xi = B) = P(Xi = C) = P(Xi = D) the probability density function would be
    fX(x) =(1/√2πσ)e^[-(x−µ)^2/2σ^2]= (1/√2π)e^(-x^2/2) since µ = 0 and σ = 1.

    Plus we would have a probability of 1/4 for all i.
    Now consider random strands of length l, where l is very large. How can I use Use Central Limit Theorem to find the probability density function corresponding to finding the length N in A(P(Xi = A)) occurrence times?


    2. Relevant equations
    fX(x) =(1/√2πσ)e^[-(x−µ)^2/2σ^2]=
    ∅(x)=(1/√2π)e^(-x^2/2) when µ = 0 and σ = 1



    3. The attempt at a solution

    The theorem says : "Let X1, X2, . . . , Xn , . . . be a sequence of
    independent discrete random variables, and let Sn = X1 + X2 +
    · · · + Xn. For each n, denote the mean and variance of Xn by µn
    and σ(^2)n, respectively. Define the mean and variance of Sn to be mnand s^(2)n, respectively, and assume that sn → ∞. If there exists a constant A, such that |Xn| ≤ A for all n, then for a < b,

    limn→∞P (a < Sn − mn/sn< b)=1/√2π∫(from a to b)e^−x(^2)/2dx"

    Well lets say XA be the random variable corresponding to the number of oc-
    curences of the base A in the strand. My guessed would be that the mean length for every occurrence of XA, results in a mean length of 4 bases. Because bases are equally likely, then Xa occurs in roughly 1/4 of the positions.
    Therefore variance would be 16 and all I would do is plug it in

    fX(x) =(1/√2πσ)e^[-(x−µ)^2/2σ^2]

    However how do I use the Central Limit theorem to find p(x)?
     
  2. jcsd
  3. Oct 26, 2012 #2
    I got it. Since Sn=X1+X2+X3, all I have to do is to calculate the mean from Sn and the variance from Sn

    Sn=1/4(P{A})+1/4+(P{G}).....
    Sn=(1/4)^2....

    than just plug in the mean and variance into the probability density function
     
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