Proof of central limit theorem

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Proof of central limit theorem

Notably, there are two proofs of the central limit theorem.

1. Using the Moment Generating function
2. Using cumulants

1. Using the Moment Generating function

The Moment Generating function is defined as follows in a random variable X;

$M_X\left(t\right)=E\left[e^{tX}\right]$

We then expand the Taylor series of $e^{tX}$ and have $M_x\left(t\right)=\sum _{n\stackrel{}{\to 0}}^{\infty }\frac{E\left[X^n\right]}{n!} t^n$

The Moment generating function is thus just the exponential generating function for the moments of X. To be precise ${M_x}^{\left(n\right)}\left(0\right)=E\left[X^n\right]$

For instance, Finding the CLT of a Bernoulli distribution using the MGF;

Recall on how we calculated the MGF of bernoulli distribution. Check on the proof HERE

The MGF is: $(1-p)+p{e}^t=1+(e^t-1)p$

The Binomial function is usually the sum of several Bernoulli variables and we have;

$(1-p)+p{e}^t=1+(e^t-1)p$

$=(1+{(e^t-1)p)}^n$

Considering the Asymptotic behavior of Binomial Distribution we have;

$={(1+\frac{(e^t-1)\lambda }{n})}^n\underset{}{\to } e^{\lambda (e}$^t-1)

2. Using Cumulants

First, we define the cumulant generating function of a random variable X:

$K_X\left(t\right)=\log M_X\left(t\right)$

We know that $M_X\left(t\right)=1+0\left(t\right) which follows a Taylor Series \log (1+x)$

$\log (1+x)=x-\frac{x^2}{2}+...................$

We then expand the $K_X\left(t\right)$ as a power series which is done as following:

$K_X\left(t\right)=\left(E\right[X]t+\frac{E\left[X^2\right]}{2} t^2 + .....) - \frac{{\left(E\right[X]t +.....)}^2}{2} +.........$

$=\left(E\right[X]t+\frac{E\left[X^2\right]-E{\left[X\right]}^2}{2} t^2 +.............$

Notably, the first two coefficients of $K_X\left(t\right)$ are the expectation and the variance. Thus, they are referred to as cumulants.

Thus the nth cumulant will be defined as $K_n\left(X\right)=K_x^{\left(n\right)}\left(0\right)$

Suppose $X_i$.....are independent random variables with mean zero. Thus we have:

$K_{\frac{X_1+........+X_n}{\sqrt{n}}} \left(t\right)=K_{X_I} \left(\frac{t}{\sqrt{n}}\right) + ...........+ K_{X_n} (\frac{t}{\sqrt{n}}$

Rephrasing the terms in terms of the cumulants we have;

$K_m\left[\frac{X_1+.......+X_n}{\sqrt{n}}\right] = \frac{K_{m }\left[X_1\right]+.......+K_m\left[X_n\right]}{n^{{\displaystyle \frac{m}{2}}}}$

Plausibly, the first cumulant will be a zero since it is the mean while the second cumulant is the variance. If the cumulants are bounded by a given constant C, then for all the values of m>2 we shall have;

$K_m\left[\frac{X_1+.......+X_n}{\sqrt{n}}\right] = \frac{nC}{n^{{\displaystyle \frac{m}{2}}}}\stackrel{}{\to 0}$

Thus the cumulative generating function will be $\frac{{\delta }^2}{2}{t}^2$ which follows a normal distribution.

That is $N(0, {\delta }^2)$

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