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Central Limit Theorem

The central limit theorem states that for any population with mean $\mu$ and standard deviation $\sigma$, the distribution of sample mean for sample size N have mean $\mu$ and standard deviation $\frac{\sigma}{\sqrt n}$.

We can summarize the properties of the Central Limit Theorem for sample means with following statements:

  • Sampling is form any distribution with mean and standard deviation.
  • Provided that n is large (n $\geq$ 30), as a rule of thumb), the sampling distribution of the sample mean $\bar X$ will be approximately normally distributed with a mean $\mu$ and a standard deviation $\sigma$ = $\frac{\sigma}{\sqrt{n}}$.
  • If the sampling distribution is normal, the sampling distribution of the sample means will be an exact normal distribution for any sample size.

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Central Limit Theorem Definition

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The central limit theorem states that whenever a random sample of size n is taken from any distribution with mean $\mu$ and variance $\sigma^2$, then the sample mean $\bar x$ will be approximately normally distributed with mean $\mu$ and variance $\sigma^2$. The larger the value of the sample size n, the better the approximation to the normal.

For example, it allows us (if the sample size is fairly large) to use hypothesis tests which assume normality even if our data appear non-normal. This is because the tests use the sample mean $\bar X$, which the Central Limit Theorem tells us will be approximately normally distributed.

Central Limit Theorem Formula

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The formula for the central limit theorem:
Central Limit Theorem for Sample Means Z-Score

Z = $\frac{\bar x - \mu}{\frac{\sigma}{\sqrt{n}}}$

Let f be a density with mean $\mu$ and finite variance $\sigma^2$. Let $\bar X_n$ be the sample mean of a random sample of size n from f.

Let the random variable Zn be defined by

$Z_n$ = $\frac{\bar X_n - \mu}{\frac{\sigma}{\sqrt{n}}}$

Then, the distribution of $Z_n$ approaches the standard normal distribution as n approaches to infinity.

Central Limit Theorem Proof

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Suppose that $x_1, x_2, x_3,......,x_n$ are independent and identically distributed with mean $\mu$ and finite variance $\sigma^2$ . Define the random variable $Z_n$ as follows

Zn = $\frac{\bar X_n - \mu}{\frac{\sigma}{\sqrt{n}}}$, where $\bar x_n$ = $\frac{1}{n} \sum_{i = 1}^n$ $x_i$.

Then the distribution function of $Z_n$ converges to the standard normal distribution function as n increases without bound.

Define a random variable $U_i$ by

$U_i$ = $\frac{x_i - \mu}{\sigma}$

$E(U_i)$ = $0$ and $V(U_i)$ = $1$ 

Thus, the moment generating function can be written as

$m_u(t)$ = 1 + $\frac{t^2}{2}$ + $\frac{t^3}{3!}$ $E(U_i^3) + ........$

also $Z_n$ = $\sqrt{n}$ $(\frac{\bar X - \mu}{\sigma})$ 

= $\frac{1}{\sqrt{n}}$ $\sum U$.

Since the random variables $x_i$ are independent, so are the random variables $U_i$ are independent, so are the random variables.

$\Rightarrow$ $m_u(t)$ = $(1\ +$ $\frac{t^2}{2n}$ $+$ $\frac{t^3}{3! n^{\frac{3}{2}}}$$E(U_i^3)\ +\ ...........)^n$

or $ln\ m_u(t)$ = $n\ ln$$( 1\ +$ $\frac{t^2}{2n}$ $+$ $\frac{t^3}{3! n^{\frac{3}{2}}}$$E(U_i^3)\ +\ ...........)$

We know that Taylor series expansion for

$ln(1\ +\ x)$ = $x\ -$ $\frac{x^2}{2}\ +\ \frac{x^3}{3}\ -\ \frac{x^4}{4}\ +$ $........$

If $x$ = $\frac{t^2}{2n}\ +\ \frac{t^3}{3! n^{\frac{3}{2}}}$$E(U_i^3)$

then $ln(m_u(t))$ = $n\ ln\ (1\ +\ x)$ = $n(x\ -$ $\frac{x^2}{2}\ +\ \frac{x^3}{3}$ $-\ ........)$

If we multiply through by the initial $n$, all terms except the first have some positive power of n in the denominator. Consequently, as $n\ \rightarrow\ \infty$ , all terms but the first go to zero.

$\Rightarrow\ \lim_{n->\infty}\ ln(m_u(t))$ = $\frac{t^2}{2}$ and $\lim_{n->\infty}\ (m_u(t))$ = $\exp$$(\frac{t^2}{2})$

$\Rightarrow$ The moment-generating function for a standard normal random variable. The essential implication here is that probability statements about $Z_n$ can be approximated by probability statements about the standard normal random variable if $n$ is large.

Central Limit Theorem Standard Deviation

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The central limit theorem states that for any population with mean $\mu$ and standard deviation $\sigma$, the distribution of sample mean for sample size N have mean $\mu$ and standard deviation $\frac{\sigma}{\sqrt n}$. To determine the standard error of the mean, the standard deviation for the population and divide by the square root of the sample size.

Z = $\frac{\bar X - \mu}{\sigma_{\bar X}}$

where, $\sigma_{\bar X}$ = $\frac{\sigma}{\sqrt{N}}$

$\bar X$ = sample mean

$\mu$ = mean of sampling distribution

$\sigma_{\bar X}$ = standard deviation of the sampling distribution or standard error of the mean.

Multivariate Central Limit Theorem

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Let X1, X2, X3,.......... be a sequence of independently and identically distributed p-dimensional random variables each with mean $\mu$ and covariance matrix.

$\bar X_n$ = $\frac{1}{n}\ \sum_{i = 1}^n$ $X_i$, n $\geq$ 1

Then as n-> $\infty$, the asymptotic distribution of

$\sqrt n(\bar X_n - \mu)$ = $n$$^{\frac{-1}{2}\sum_{i = 1}^n(X_i - \mu)}$ is $N(0,\ \sum)$.

Martingale Central Limit Theorem

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The martingale central limit theorem generalizes the central limit theorem, that is, the sum of many independent identically-distributed random variables, when scaled appropriately, converges in distribution to a standard normal distribution for random variables to martingales. For a sequence of random variables $Z_1, Z_2, ........$ with variance one, the central limit theorem for large n is approximately standard normal. Consider a mean zero martingale with $x_0$ = 0 and write $Z_i = X_i - X_{i-1}$, i $\geq$ 1.

Then, $X_n$ = $X_n - X_0 = \sum_{i=1}^n(X_i - X_{i-1}) = \sum_{i=1}^n Z_i$
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