Find probability density function from Central Limit theorem

hansbahia
Messages
54
Reaction score
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)?
 
Physics news on Phys.org
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
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
Back
Top