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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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.

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 Z_{n} 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.

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

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

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.

Let X_{1}, X_{2}, X_{3},.......... 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 $N(0,\ \sum)$.

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

The formula for the central limit theorem:

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 Z

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

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

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

Z

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.

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.

Let X

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

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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