Normal distribution or Gaussian

The normal distribution, Gauss distribution or Gaussian distribution, is discrete v.a. Z that measures the area included in the function represented by the Gaussian bell

Its probability function is:

\frac{1}{\sqrt{2 \cdot \pi}\cdot \sigma}\int^{+\infty}_Z e^m dx with m =-\frac{(x - \mu)^2}{2 \cdot \sigma^2}

E(\xi) = \mu

\sigma^2(\xi) = \sigma^2

\sigma(\xi) = \sigma

Properties of Normal

1. Is symmetric with respect to the axis x = \mu, P(\sigma > \mu) = P(\sigma < \mu) = \frac{1}{2}
2. When x \rightarrow \pm\infty we have an asymptote general y = 0
3. It has points of inflection at x = \mu = \sigma
4. Any v.a. built as a linear combination of normal v.a. also follows a normal distribution

Calculation a Normal

z \mu \sigma

\xi \approx N(\mu, \sigma)

E(\xi)
\sigma^2(\xi)
\sigma(\xi)

Normal typed

It \xi v.a. we will call typing another v.a. when:

Z = \frac{\xi - \mu \cdot \xi}{\sigma \cdot \xi}

If we type a z v.a. we have to:

If \xi \approx N(\xi, \sigma) \Rightarrow Z = \frac{\xi - \mu}{\sigma} \Rightarrow N(0, 1)

The probability function of the typed normal is:

P(Z > z) = \frac{1}{\sqrt{2 \cdot \pi}}\int^{+\infty}_Z e^m dx with m = -\frac{x^2}{2}, \forall x \in \mathbb{R}

E(\xi) = 0

\sigma^2(\xi) = 1

\sigma(\xi) = 1

Properties of the typed Normal

1. Is symmetric with respect to the axis x = 0, P(\sigma > 0) = P(\sigma < 0) = \frac{1}{2}
2. When x \rightarrow \pm\infty we have a horizontal asymptote
3. It has points of inflection at x = \pm 1

Calculation of a typed Normal

z

\xi \approx N(0, 1)

E(\xi)
\sigma^2(\xi)
\sigma(\xi)

Notes for Normal

It \xi_1, \cdots, \xi_n \approx v.to. (\mu_i, \sigma_i) with \xi = a_0 + a_1 \cdot \xi_1 + \cdots + a_n \cdot \xi_n, \forall i \in \{1, \cdots, n\}

1. E[\xi] = a_0 + a_1 \cdot \mu_1 + \cdots + a_n \cdot \mu_n
2. \sigma^2[\xi] = a_1^2 \cdot \sigma_1^2 + \cdots + a_n \cdot \sigma_n^2 + 2a_{1 2} \cdot Cov(\xi_1, \xi_2) + \cdots
3. If the \xi_i are independent or only incorreladas \sigma^2[\xi] = a_1^2 \cdot \sigma_1^2 + \cdots + a_n \cdot \sigma_n^2
4. If, in addition, \xi_i \approx N(\mu_i, \sigma_i), \forall i \in \{1, \cdots, n\} then:
   \begin{cases} \xi \approx N(\mu, \sigma)\text{ with }\mu = a_0 + a_1 \cdot \mu_1 + \cdots + a_n \cdot \mu_n \\ \sigma^2 = a_1^2 \cdot \sigma_1^2 + \cdots + a_n \cdot \sigma_n^2 + 2a_{1 2} \cdot Cov(\xi_1, \xi_2) + \cdots \end{cases}
5. If, in addition, \xi_i \approx N(\mu_i, \sigma_i), \forall i \in \{1, \cdots, n\} and independent then:
   \begin{cases}\mu = a_0 + a_1 \cdot \mu_1 + \cdots + a_n \cdot \mu_n \\ \sigma^2 = a_1^2 \cdot \sigma_1^2 + \cdots + a_n \cdot \sigma_n^2 \end{cases}

It \xi_S = \xi_1 + \cdots + \xi_n

1. E[\xi_S] = \mu_1 + \cdots + \mu_n
2. \sigma^2[\xi_S] = \sigma_1^2 + \cdots + \sigma_n^2 + 2 \cdot Cov(\xi_1, \xi_2) + \cdots
3. If the \xi_i are independent \sigma^2[\xi_S] = a_1^2 + \cdots + a_n^2
4. If, in addition, \xi_i \approx N(\mu_i, \sigma_i), \forall i \in \{1, \cdots, n\} then:
   \begin{cases} \xi_S \approx N(\mu, \sigma)\text{ with }\mu = \mu_1 + \cdots + \mu_n \\ \sigma^2 = a_1^2 \cdot \sigma_1^2 + \cdots + a_n \cdot \sigma_n^2 + 2a_{1 2} \cdot Cov(\xi_1, \xi_2) + \cdots \end{cases}
5. If, in addition, \xi_i \approx N(\mu_i, \sigma_i), \forall i \in \{1, \cdots, n\} and independent then:
   \begin{cases} \mu = \mu_1 + \cdots + \mu_n \\ \sigma^2 = \sigma_1^2 + \cdots + \sigma_n^2\end{cases}

It \xi_S v.a. independently and identically distributed with \xi_1, \cdots, \xi_n \approx v.a.i.i.d. (\mu, \sigma) and \xi_S = \xi_1 + \cdots + \xi_n \approx v.to.(n \cdot \mu, \sigma \cdot \sqrt{n})

1. E[\xi_S] = \overbrace{\mu + \cdots + \mu}^{n\;\rm times} = n \cdot \mu
2. \sigma^2[\xi_S] = \overbrace{\sigma^2 + \cdots + \sigma^2}^{n\;\rm times} = n \cdot \sigma^2
3. If the \xi_i are v.a.i.i.d. and normal:
   \xi_S \approx N(n \cdot \mu, \sqrt{n} \cdot \sigma)

Approaches

Approximation of the Binomial to the Normal

It B \approx B(n, p)

With B \approx number of successes in n equal and independent Bernoulli tests with probability of success p then:

B \approx B(n\cdot p, \sqrt{n \cdot p \cdot q})

Theorem of Moivre

It B \approx B(n, p) then:

\frac{B - m}{\sqrt{n \cdot p \cdot q}}\rightarrow N(0, 1)

So we have to:

E(B) = n \cdot p

\xi^2(B) = n \cdot p \cdot q

\xi(B) = +\sqrt{n \cdot p \cdot q}

It is considered a good approximation when n \cdot p \geq 5 and n \cdot q \geq 5 and then the Moivre Theorem is fulfilled with:

B \approx B(n \cdot p, \sqrt{n \cdot p \cdot q})

However, a discontinuity correction will have to be made to get the value sought, taking -0.5 if we look for the strictest or +0.5 in any other case

Examples:

P\{B < 4\} we will use P\{B < 3.5\}

P\{B \leq 4\} we will use P\{B \leq 4.5\}

P\{B > 4\} we will use P\{B > 4.5\}

P\{B \geq 4\} we will use P\{B \geq 4.5\}

The central limit theorem

\xi_1, \cdots, \xi_n \approx v.a.i.i.d. (\mu, \sigma) then:

\xi_T = \xi_1 + \cdots + \xi_n \approx v.to. (n \cdot \mu, \sqrt{n} \cdot \sigma) always happens

Theorem Lery-Lidenberg

It \{\xi_i\}, i \in N succession of v.a.i.i.d. then:

S_n = \xi_1 + \cdots + \xi_n \approx \frac{S_n - n \cdot \mu}{\sqrt{n} \cdot \sigma} \rightarrow N(0, 1)

It is considered a good approximation when n \geq 30 and then the Lery-Lidenberg Theorem is fulfilled by approx. normal any probability of the type:

\frac{S_n - n \cdot \mu}{\sqrt{n} \cdot \sigma}

Taylor series for Normal distribution

Approximation of Abramowitz and Stegun (1964) known as "Best Hastings Approach"

\tiny P(x) = 1 - \phi(x)(b_1 \cdot t + b_2 \cdot t^2 + b_3 \cdot t^3 + b_4 \cdot t^4 + b_5 \cdot t^5) + \epsilon(x)

\begin{cases} \phi(x) = \frac{1}{\sqrt{2 \cdot \pi}} \cdot e^{-\left(\frac{1}{2}\right) \cdot x^2} \\ t = \frac{1}{1 + (b_0 \cdot x)} \\ b_0 = 0.2316419 \\ b_1 = 0.319381530 \\ b_2 = -0.356563782 \\ b_3 = 1.781477937 \\ b_4 = -1.821255978 \\ b_5 = 1.330274429 \\ \|\epsilon(x)\| < 7.5 \cdot 10^{-8} \end{cases}

Replacing us we have: