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hansbahia
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Homework Statement
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?
Homework Equations
fX(x) =(1/√2πσ)e^[-(x−µ)^2/2σ^2]=
∅(x)=(1/√2π)e^(-x^2/2) when µ = 0 and σ = 1
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 let's 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)?