1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 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


    than just plug in the mean and variance into the probability density function
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook