Find probability density function from Central Limit theorem

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SUMMARY

The discussion focuses on deriving the probability density function (PDF) using the Central Limit Theorem (CLT) for a random variable with equal probabilities across four outcomes (A, B, C, D). The PDF is expressed as fX(x) = (1/√2π)e^(-x^2/2) with mean (µ) = 0 and standard deviation (σ) = 1. The participants clarify that to find the PDF corresponding to the occurrence of a specific base in a random strand, one must calculate the mean and variance of the sum of occurrences (Sn) and substitute these values into the PDF formula.

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  • Understanding of the Central Limit Theorem (CLT)
  • Familiarity with probability density functions (PDFs)
  • Knowledge of mean and variance calculations
  • Basic concepts of random variables and their distributions
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  • Learn how to calculate mean and variance for discrete random variables
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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)?
 
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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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