Central limit theorem - finding cumulants

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SUMMARY

The discussion centers on the application of the Central Limit Theorem (CLT) to find the cumulants of the standardized variable y, defined as y = (Σx_i - N⟨x⟩) / √N. The key results established include that the cumulants of y are ⟨y⟩_c = 0, ⟨y²⟩_c = ⟨x²⟩_c, and for m > 2, ⟨y^m⟩_c = ⟨x^m⟩_c N^(1 - m/2). The discussion also details the use of generating functions, specifically the cumulant generating function ln(ŧp(k)) = Σ(-ik)^n/n!⟨x^n⟩_c, to derive these results.

PREREQUISITES
  • Understanding of the Central Limit Theorem (CLT)
  • Familiarity with cumulants and their properties
  • Knowledge of generating functions in probability theory
  • Proficiency in mathematical notation and series expansions
NEXT STEPS
  • Study the derivation of cumulants in probability distributions
  • Learn about the properties of generating functions in statistical mechanics
  • Explore advanced topics in asymptotic analysis related to the CLT
  • Investigate applications of cumulants in statistical inference
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Students and researchers in statistics, probability theory, and mathematical physics who are interested in understanding the implications of the Central Limit Theorem and cumulants in data analysis and modeling.

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Homework Statement


Given:
y=\frac{\sum_{i}x_i-N\left \langle x \right \rangle}{\sqrt{N}}

Show that the cumulants of y are:
<br /> \begin{matrix}<br /> \left \langle y \right \rangle_c=0&amp; &amp; \left \langle y^2 \right \rangle_c=\left \langle x^2 \right \rangle_c &amp; &amp; \left \langle y^m \right \rangle_c=\left \langle x^m \right \rangle_c N^{1-m/2}\begin{matrix}<br /> &amp; for &amp; m&gt;2<br /> \end{matrix}<br /> \end{matrix}<br />

Homework Equations


Generating function:
\tilde{p}\left ( k \right )=\sum_{n=0}^{\infty }\frac{\left ( -ik \right )^n}{n!}\left \langle x^n \right \rangle
Cumulant generating function:
ln \left ( \tilde{p}\left ( k \right ) \right )=\sum_{n=0}^{\infty }\frac{\left ( -ik \right )^n}{n!}\left \langle x^n \right \rangle_c

For independent
X=\left ( x_1,x_2,...,x_N \right )
if
y=a_0+a_1x_1+a_2x_2+...+a_Nx_N
then:
\tilde{p}_y\left ( k \right )=e^{-ika_0}\prod_{i}\tilde{p}_i\left ( a_ik \right )

The Attempt at a Solution


In my case:
y=\frac{\sum_{i}x_i-N\left \langle x \right \rangle}{\sqrt{N}}=-\frac{N\left \langle x \right \rangle}{\sqrt{N}}+\frac{1}{\sqrt{N}}x_1+...+\frac{1}{\sqrt{N}}x_N
that is:
\begin{matrix} a_0=-\frac{N\left \langle x \right \rangle}{\sqrt{N}} &amp; &amp; a_1=...=a_N=\frac{1}{\sqrt{N}}<br /> \end{matrix}
Substituting into
\tilde{p}_y\left ( k \right )=e^{-ika_0}\prod_{i}\tilde{p}_i\left ( a_ik \right )
gives:
\tilde{p}_y\left ( k \right )=e^{\frac{ikN\left \langle x \right \rangle}{\sqrt{N}}}\prod_{i}\tilde{p}_i\left ( \frac{1}{\sqrt{N}}k \right )
Cumulant generating function:
ln \left ( \tilde{p}_y\left ( k \right ) \right )=\frac{ikN\left \langle x \right \rangle}{\sqrt{N}}+\sum_{i}ln\left ( \tilde{p}_i\left ( \frac{1}{\sqrt{N}}k \right ) \right )
Next step I tried is to represent ln function as series:
\sum_{n=0}^{\infty }\frac{\left ( -ik \right )^n}{n!}\left \langle y^n \right \rangle_c = \frac{ikN\left \langle x \right \rangle}{\sqrt{N}}+\sum_{i}^{N}\sum_{n=0}^{\infty }\frac{\left ( -i\frac{1}{\sqrt{N}}k \right )^n}{n!}\left \langle x^n \right \rangle_c=\frac{ikN\left \langle x \right \rangle}{\sqrt{N}}+N\sum_{n=0}^{\infty }\frac{\left ( -i\frac{1}{\sqrt{N}}k \right )^n}{n!}\left \langle x^n \right \rangle_c
And, finally, here I'm stuck. Can someone help me how to continue from here?
Thanks in advance.
 
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