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\begin{document}
\subsection*{Funci\'on distribuci\'on acumulada de X}
$$ F_{X}(x) = P(X \leq x) = \displaystyle \int_{-\infty}^{x} f_{X}(u)\, du $$
$$ E(X) = \mu_{X} = \displaystyle \int_{-\infty}^{\infty} x . f_{X}(x)\, dx $$
$$ V(X) = \sigma^{2}_{X} = \displaystyle \int_{-\infty}^{\infty} (x - \mu_{X})^2 . f_{X}(x)\, dx $$
\subsection*{Distribuci\'on uniforme continua}
Si la f.d.p. de X es $ f_{X}(x) = \frac{1}{b - a}$ entonces:
$$ \mu_{X} = \frac{a + b}{2} $$
$$ \sigma^{2}_{X} = \frac{(b - a)^2}{12} $$
\subsection*{Distribuci\'on normal}
$$ f_{X}(x; \mu; \sigma) = \frac{1}{2 . \pi . \sigma} . e ^ \frac{-(x - \mu)^2}{2 . \sigma ^ 2} $$
$$ E(X) = \mu $$
$$ V(X) = \sigma^2 $$
$$ Z (\mu = 0, \sigma = 1) = \frac{X - \mu_{X}}{\sigma_{X}}$$
\subsection*{Aproximaci\'on normal a v.a. X binomial de par\'ametros n, p }
$$ Z = \frac{X - n . p}{n . p . (1 - p)}$$ 
Z aproxima a X, tanto mejor según crecen $n.p$ y $n.p.(1-p)$
\subsection*{Aproximaci\'on normal a v.a. X $\sim$ Poisson, parámetro $\lambda$  }
$$ Z = \frac{X - \lambda}{\sqrt{\lambda}}$$
\subsection*{Aproximaci\'on normal a v.a. X binomial de par\'ametros n, p }
$$ Z = \frac{X - n . p}{n . p . (1 - p)}$$ 
Z aproxima a X, tanto mejor según crecen $n.p$ y $n.p.(1-p)$
\subsection*{Distribuci\'on exponencial X asociada a Poisson de parámetro $\lambda$}
La f.d.p. de X es $f_{X}(x; \lambda) = \lambda . e^{-\lambda.x}$, para $0 \leq x \le \infty$ 
$$ E(X) = \frac{1}{\lambda} $$
$$ V(X) = \frac{1}{\lambda^2} $$
\subsection*{Distribuci\'on Erlang X de par\'ametro r, asociada a Poisson de parámetro $\lambda$}
La f.d.p. de X es $f_{X}(x; \lambda; r) = \frac{\lambda^{r}.x^{r - 1}.e^{-\lambda.x}}{(r - 1)!}$ 
$$ E(X) = \frac{r}{\lambda} $$
$$ V(X) = \frac{r}{\lambda^2} $$
\subsection*{Funci\'on gamma}
$$\Gamma (r) = \displaystyle \int_{0}^{\infty} x^{r - 1}.e^{-x}\, dx $$
$$ E(X) = \frac{r}{\lambda} $$
$$ V(X) = \frac{r}{\lambda^2} $$
X con f.d.p. $f_{X}(x; \lambda\; r) = \frac{\lambda^r.x^{r - 1}.e^{-\lambda.x}}{\Gamma(r)}$
tiene una distribuci\'on \textbf{gamma} con par\'ametros $\lambda$ y r.
$$ E(X) = \mu_{x} = \frac{r}{\lambda} $$
$$ V(X) = \sigma^2_{x}\frac{r} = {\lambda^2} $$
\subsection*{Distribuci\'on de Weibull}
Sea X con f.d.p. $f_{X}(x; \delta ; \beta) = \frac{\beta}{\delta}.(\frac{x}{\delta})^{\beta - 1}. e^{-(\frac{x}{\delta})^\beta}$
\\La f.d. acumulada de X es $F_{X}(x) = 1 - e^{-(\frac{x}{\delta})^\beta}$
$$E(X) = \delta . \Gamma(1 + \frac{1}{\beta})$$
$$V(X) = \delta^2 . \Gamma(1 + \frac{2}{\beta}) - \delta^2.[ \Gamma(1+\frac{1}{\beta})]^2$$

\end{document}
