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# 17.3: Apéndice C- Datos sobre algunas distribuciones comunes

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## Distribuciones discretas

Función de indicador$$X = I_E$$$$P(X = 1) = P(E) = p$$$$P(X = 0) = q = 1 - p$$

$$E[X] = p$$$$\text{Var} [X] = pq$$$$M_X (s) = q + pe^s$$$$g_X (s) = q + ps$$

Variable aleatoria simple$$X = \sum_{i = 1}^{n} t_i I_{A_i}$$ (una forma primitiva)$$P(A_i) = p_i$$

$$E[X] = \sum_{i = 1}^{n} t_ip_i$$$$\text{Var} [X] = \sum_{i = 1}^{n} t_i^2 p_i q_i - 2 \sum_{i < j} t_i t_j p_i p_j$$$$M_X(s) = \sum_{i = 1}^{n} p_i e^{st_i}$$

Binomial$$(n, p)$$$$X = \sum_{i = 1}^{n} I_{E_i}$$ con$$\{I_{E_i} : 1 \le i \le n\}$$ iid$$P(E_i) = p$$

$$P(X = k) = C(n, k) p^k q^{n - k}$$

$$E[X] = np$$$$\text{Var} [X] = npq$$$$M_X (s) = (q + pe^s)^n$$$$g_X (s) = (q + ps)^n$$

MATLAB:$$P(X = k) = \text{ibinom} (n, p, k)$$$$P(X \ge k) = \text{cbinom} (n, p, k)$$

Geométrico ($$p$$)$$P(X = k) = pq^k$$$$\forall k \ge 0$$

$$E[X] = q/p$$$$\text{Var} [X] = q/p^2$$$$M_X (s) = dfrac{p}{1 - qe^s}$$$$g_X (s) = \dfrac{p}{1- qs}$$

Si$$Y - 1$$ ~ geométrica$$(p)$$, así que$$P(Y = k) = pq^{k - 1}$$$$\forall k \ge 1$$, entonces

$$E[Y] = 1/p$$$$\text{Var} [X] = q/p^2$$$$M_Y (s) = \dfrac{pe^s}{1 - qe^s}$$$$g_Y (s) = \dfrac{ps}{1 - qs}$$

Binomial negativo$$(m, p)$$,$$X$$ es el número de fracasos antes del éxito$$m$$ th.

$$P(X = k) = C(m + k - 1, m - 1) p^m q^k$$$$\forall k \ge 0$$

$$E[X] = mq/p$$$$\text{Var} [X] = mq/p^2$$$$M_X (s) = (\dfrac{p}{1 - qe^s})^m$$$$g_X (s) = (\dfrac{p}{1 - qs})^m$$

Para$$Y_m = X_m + m$$, el número del juicio en el que se$$m$$ produce el éxito. $$P(Y = k) = C(k - 1, m - 1) p^m q^{k - m}$$$$\forall k \ge m$$.

$$E[Y] = m/p$$$$\text{Var} [Y] = mq/p^2$$$$M_Y(s) = (\dfrac{pe^s}{1 - qe^s})^m$$$$g_Y (s) = (\dfrac{ps}{1 - qs})^m$$

MATLAB:$$P(Y = k) = \text{nbinom} (m, p, k)$$

Poisson$$(\mu)$$. $$P(X = k) = e^{-\mu} \dfrac{\mu^k}{k!}$$$$\forall k \ge 0$$

$$E[X] = \mu$$$$\text{Var}[X] = \mu$$$$M_X (s) = e^{\mu (e^s - 1)}$$$$g_X (s) = e^{\mu (s - 1)}$$

MATLAB:$$P(X = k) = \text{ipoisson} (m, k)$$$$P(X \ge k) = \text{cpoisson} (m, k)$$

## Distribuciones absolutamente continuas

Uniforme$$(a, b)$$$$f_x (t) = \dfrac{1}{b - a}$$$$a < t < b$$ (cero en otra parte)

$$E[X] = \dfrac{b + a}{2}$$$$\text{Var} [X] = \dfrac{(b - a)^2}{12}$$$$M_X (s) = \dfrac{e^{sb} - e^{sa}}{s(b - a)}$$

Triangular simétrico$$(-a, a)$$$$f_X (t) = \begin{cases} (a + t)/a^2 & -a \le t < 0 \\ (a - t)/a^2 & 0 \le t \le a \end{cases}$$

$$E[X] = 0$$$$\text{Var} [X] = \dfrac{a^2}{6}$$$$M_X (s) = \dfrac{e^{as} + e^{-as} - 2}{a^2 s^2} = \dfrac{e^{as} - 1}{as} \cdot \dfrac{1 - e^{-as}}{as}$$

Exponencial$$(\lambda)$$$$f_X(t) = \lambda e^{-\lambda t}$$$$t \ge 0$$

$$E[X] = \dfrac{1}{\lambda}$$$$\text{Var} [X] = \dfrac{1}{\lambda^2}$$$$M_X (s) = \dfrac{\lambda}{\lambda - s}$$

Gamma$$(\alpha, \lambda)$$$$f_X(t) = \dfrac{\lambda^{\alpha} t^{\alpha - 1} e^{-\lambda t}}{\Gamma (\alpha)}$$$$t \ge 0$$

$$E[X] = \dfrac{\alpha}{\lambda}$$$$\text{Var} [X] = \dfrac{\alpha}{\lambda^2}$$$$M_X (s) = (\dfrac{\lambda}{\lambda - s})^{\alpha}$$

MATLAB:$$P(X \le t) = \text{gammadbn} (\alpha, \lambda, t)$$

Normal$$N(\mu, \sigma^2)f_X (t) = \dfrac{1}{\sigma \sqrt{2\pi}} \text{exp} (-\dfrac{1}{2} (\dfrac{t - \mu}{\sigma})^2)$$

$$E[X] = \mu$$$$\text{Var} [X] \sigma^2$$$$M_X (s) = \text{exp} (\dfrac{\sigma^2 s^2}{2} + \mu s)$$

MATLAB:$$P(X \le t) = \text{gaussian} (\mu, \sigma^2, t)$$

Beta$$(r, s)$$

$$f_X (t) = \dfrac{\Gamma (r + s)}{\Gamma (r) \Gamma (s)} t^{r -1} (1 - t)^{s - 1}$$$$0 < t < 1$$,$$r > 0$$,$$s > 0$$

$$E[X] = \dfrac{r}{r + s}$$$$\text{Var} [X] = \dfrac{rs}{(r + s)^2 (r + s + 1)}$$

MATLAB:$$f_X (t) = \text{beta} (r, s, t)$$$$P(X \le t) = \text{betadbn} (r, s, t)$$

Weibull ($$\alpha, \lambda, \nu$$)

$$F_X (t) = 1 - e^{-\lambda (t - \nu)^{\alpha}}$$,$$\alpha > 0, \lambda >0, \nu \ge 0, t \ge \nu$$

$$E[X] = \dfrac{1}{\lambda^{1/\alpha}} \Gamma (1 + 1/\alpha) + \nu$$$$\text{Var} [X] = \dfrac{1}{\lambda^{2/\alpha}} [\Gamma (1 + 2/\lambda) - \Gamma^2 (1 + 1/\lambda)]$$

MATLAB: ($$\nu = 0$$solo)

$$f_X (t) = \text{weibull} (a, l, t)$$$$P(X \le t) = \text{weibull} (a, l, t)$$

## Relación entre distribuciones gamma y Poisson

• Si$$X$$ ~ gamma$$(n, \lambda)$$, entonces$$P(X \le t) = P(Y \ge n)$$ donde$$Y$$ ~ Poisson$$(\lambda t)$$.
• Si$$Y$$ ~ Poisson$$(\lambda t)$$, entonces$$P(Y \ge n) = P(X \le t)$$ donde$$X$$ ~ gamma$$(n, \lambda)$$.

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