Distribuciones de Probabilidad

Distribución Binomial

Modela el número de éxitos en $n$ ensayos independientes, cada uno con probabilidad de éxito $p$.

$$X \sim \text{Bin}(n, p)$$

$$P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}, \quad k = 0, 1, \ldots, n$$

Parámetros:

• $E[X] = np$
• $\text{Var}(X) = np(1-p)$

Cuándo usarla

• Número fijo de ensayos $n$
• Cada ensayo es independiente
• Solo dos resultados posibles (éxito / fracaso)
• $p$ constante en todos los ensayos

from scipy.stats import binom

n, p = 10, 0.3
X = binom(n, p)

print(X.pmf(3))      # P(X = 3)
print(X.cdf(4))      # P(X <= 4)
print(X.mean())      # np = 3.0
print(X.std())       # sqrt(np(1-p))

Distribución de Poisson

Modela el número de eventos que ocurren en un intervalo fijo de tiempo o espacio, con tasa promedio $\lambda$.

$$X \sim \text{Pois}(\lambda)$$

$$P(X = k) = \frac{e^{-\lambda} \lambda^k}{k!}, \quad k = 0, 1, 2, \ldots$$

Parámetros:

• $E[X] = \lambda$
• $\text{Var}(X) = \lambda$

La media y la varianza son iguales — una forma rápida de verificar si Poisson es adecuada.

Aproximación Binomial → Poisson

Cuando $n$ es grande y $p$ es pequeña, $\text{Bin}(n, p) \approx \text{Pois}(\lambda)$ con $\lambda = np$.

from scipy.stats import poisson

lam = 4.5
X = poisson(lam)

print(X.pmf(3))      # P(X = 3)
print(X.cdf(6))      # P(X <= 6)
print(X.mean())      # 4.5

Distribución Exponencial

Modela el tiempo entre eventos de un proceso de Poisson. Si los eventos ocurren a tasa $\lambda$, el tiempo entre eventos es $\text{Exp}(\lambda)$.

$$f(x) = \lambda e^{-\lambda x}, \quad x \geq 0$$

$$F(x) = 1 - e^{-\lambda x}$$

Parámetros:

• $E[X] = \frac{1}{\lambda}$
• $\text{Var}(X) = \frac{1}{\lambda^2}$

Propiedad de falta de memoria

$$P(X > s + t \mid X > s) = P(X > t)$$

El tiempo restante no depende del tiempo ya transcurrido — la distribución exponencial es la única continua con esta propiedad.

from scipy.stats import expon

# tasa lambda = 2 (media = 1/2)
X = expon(scale=1/2)

print(X.pdf(1))      # densidad en x=1
print(X.cdf(1))      # P(X <= 1)
print(X.ppf(0.9))    # percentil 90

Resumen comparativo

Distribución	Soporte	$E[X]$	$\text{Var}(X)$	Uso típico
Binomial$(n,p)$	$\{0,\ldots,n\}$	$np$	$np(1-p)$	Conteo de éxitos en $n$ ensayos
Poisson$(\lambda)$	$\{0,1,2,\ldots\}$	$\lambda$	$\lambda$	Eventos raros en tiempo/espacio
Exponencial$(\lambda)$	$[0,\infty)$	$1/\lambda$	$1/\lambda^2$	Tiempo entre eventos

Binomial Distribution

Models the number of successes in $n$ independent trials, each with success probability $p$.

$$X \sim \text{Bin}(n, p)$$

$$P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}, \quad k = 0, 1, \ldots, n$$

Parameters:

• $E[X] = np$
• $\text{Var}(X) = np(1-p)$

When to use it

• Fixed number of trials $n$
• Each trial is independent
• Only two outcomes (success / failure)
• Constant $p$ across all trials

from scipy.stats import binom

n, p = 10, 0.3
X = binom(n, p)

print(X.pmf(3))      # P(X = 3)
print(X.cdf(4))      # P(X <= 4)
print(X.mean())      # np = 3.0
print(X.std())       # sqrt(np(1-p))

Poisson Distribution

Models the number of events occurring in a fixed interval of time or space, with average rate $\lambda$.

$$X \sim \text{Pois}(\lambda)$$

$$P(X = k) = \frac{e^{-\lambda} \lambda^k}{k!}, \quad k = 0, 1, 2, \ldots$$

Parameters:

• $E[X] = \lambda$
• $\text{Var}(X) = \lambda$

Mean and variance are equal — a quick check for whether Poisson is appropriate.

Binomial → Poisson approximation

When $n$ is large and $p$ is small, $\text{Bin}(n, p) \approx \text{Pois}(\lambda)$ with $\lambda = np$.

from scipy.stats import poisson

lam = 4.5
X = poisson(lam)

print(X.pmf(3))      # P(X = 3)
print(X.cdf(6))      # P(X <= 6)
print(X.mean())      # 4.5

Exponential Distribution

Models the time between events in a Poisson process. If events occur at rate $\lambda$, the time between events is $\text{Exp}(\lambda)$.

$$f(x) = \lambda e^{-\lambda x}, \quad x \geq 0$$

$$F(x) = 1 - e^{-\lambda x}$$

Parameters:

• $E[X] = \frac{1}{\lambda}$
• $\text{Var}(X) = \frac{1}{\lambda^2}$

Memoryless property

$$P(X > s + t \mid X > s) = P(X > t)$$

The remaining time does not depend on time already elapsed — the exponential is the only continuous distribution with this property.

from scipy.stats import expon

# rate lambda = 2 (mean = 1/2)
X = expon(scale=1/2)

print(X.pdf(1))      # density at x=1
print(X.cdf(1))      # P(X <= 1)
print(X.ppf(0.9))    # 90th percentile

Comparative Summary

Distribution	Support	$E[X]$	$\text{Var}(X)$	Typical use
Binomial$(n,p)$	$\{0,\ldots,n\}$	$np$	$np(1-p)$	Count of successes in $n$ trials
Poisson$(\lambda)$	$\{0,1,2,\ldots\}$	$\lambda$	$\lambda$	Rare events in time/space
Exponential$(\lambda)$	$[0,\infty)$	$1/\lambda$	$1/\lambda^2$	Time between events