Find probability density function from Central Limit theorem

In summary: X(x) = (1/√2π)e^(-x^2/2).In summary, the conversation discusses how to derive the probability density function using the Central Limit theorem for a random variable corresponding to the number of occurrences of a specific base in a strand of DNA. The mean and variance of the random variable are calculated from the sum of independent discrete random variables, and then plugged into the probability density function to find the desired result.
  • #1
hansbahia
55
0

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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  • #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
 

1. What is the Central Limit Theorem?

The Central Limit Theorem is a fundamental concept in statistics that states that when independent random variables are added, their sum will tend towards a normal distribution, regardless of the underlying distribution of the individual variables.

2. How is the Central Limit Theorem used to find a probability density function?

The Central Limit Theorem is used to find a probability density function by taking a large sample size from a population and calculating the mean of each sample. The distribution of these sample means will approximate a normal distribution, which can then be used to estimate the probability density function of the population.

3. Can the Central Limit Theorem be applied to any type of distribution?

Yes, the Central Limit Theorem can be applied to any type of distribution as long as the sample size is sufficiently large. However, the larger the sample size, the closer the resulting distribution will be to a normal distribution.

4. What is the significance of the Central Limit Theorem in statistics?

The Central Limit Theorem is significant in statistics because it allows for the use of normal distribution assumptions in many practical situations. It also provides a way to approximate the probability distribution of a population, even if the underlying distribution is unknown. This makes it a powerful tool in hypothesis testing and confidence interval estimation.

5. Are there any limitations to the Central Limit Theorem?

Yes, there are limitations to the Central Limit Theorem. One of the main limitations is that it only applies to independent random variables. It also assumes that the sample size is sufficiently large. If these assumptions are not met, the resulting distribution may not approximate a normal distribution. Additionally, the Central Limit Theorem may not work well for distributions with heavy tails or outliers.

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