| Autor/a | Fecha de Publicación |
|---|---|
| Jenniffer Alemán | 14 de Julio de 2024 |
Solución:
Se tiene que:
\[ Y \rightsquigarrow Weibull(\theta, \lambda) \]
Función de densidad de probabilidad de la Distribución Weibull
\[ f(Y; \lambda , \theta) = \frac{\lambda}{\theta^{}} y^{(\lambda -1)} e^{- \left (\frac{y}{\theta} \right)^{\lambda}} \] Si: \[\lambda = 2\]
Entonces:
\[ f(Y; \lambda , \theta) = \frac{2}{\theta^2} y^{(2 -1)} e^{- \left (\frac{y}{\theta} \right)^{2}} \] \[ f(Y; \lambda , \theta) = \frac{2}{\theta^2} y e^{- \left (\frac{y}{\theta} \right)^{2}} \]
Considerando que: \(e^{ln(x)} = x\) , entonces se tiene:
\[\begin{align} f(Y; \lambda , \theta) &= e^{ln \left [\frac{2}{\theta^2} y e^{- \left (\frac{y}{\theta} \right)^{2}} \right]} \\ f(Y; \lambda , \theta) &= e^{ln \left (\frac{2}{\theta^2} \right) ln(y) ln \left (e^{- \left (\frac{y}{\theta} \right)^{2}} \right) } \end{align}\]Donde:
\[\begin{align} ln \left (\frac{2}{\theta^2} \right) &= ln(2) - 2ln(\theta) \\ ln \left (e^{- \left (\frac{y}{\theta} \right)^{2}} \right) &= - \left (\frac{y}{\theta} \right)^{2} \end{align}\]\(\Rightarrow\)
\[ f(Y; \lambda , \theta) = e^{ ln(2) - 2ln(\theta) + ln(y) - \left (\frac{y}{\theta} \right)^{2}} \]
Dado que:
\[ \int f(Y; \lambda , \theta) dy = 1 \]
Siendo: \(f(Y; \lambda , \theta)\) una función de densidad de probabilidad de una variable aleatoria continua.
Utilizando lo anterior, se tiene:
\[ \int f(Y; \lambda , \theta) dy = \int e^{ ln(2) - 2ln(\theta) + ln(y) - \left (\frac{y}{\theta} \right)^{2}} dy \]
Teniendo en cuenta máxima verosimilitud:
\[\begin{align} \frac{d}{d \theta} \int f(Y; \lambda , \theta) dy &= \frac{d}{d \theta}\int e^{ ln(2) - 2ln(\theta) + ln(y) - \left (\frac{y}{\theta} \right)^{2}} dy \\ \frac{d}{d \theta} (1) &= \frac{d}{d \theta}\int e^{ ln(2) - 2ln(\theta) + ln(y) - \left (\frac{y}{\theta} \right)^{2}} dy \\ 0 &= \int \frac{d}{d \theta} \left [ e^{ ln(2) - 2ln(\theta) + ln(y) - \left (\frac{y}{\theta} \right)^{2}} \right] dy \end{align}\]Resolviendo:
\[ 0 = \int e^{ ln(2) - 2ln(\theta) + ln(y) - \left (\frac{y}{\theta} \right)^{2}} * \frac{d}{d \theta} \left [ ln(2) - 2ln(\theta) + ln(y) - \left (\frac{y}{\theta} \right)^{2} \right] dy \]
Resolviendo las derivadas por separado:
\[\begin{align} \frac{d}{d \theta} ln(2) &= 0 \\ \frac{d}{d \theta} ln(y) &= 0 \\ \frac{d}{d \theta} 2ln(\theta) &= \frac{2}{\theta} \\ \frac{d}{d \theta} \left ( \frac{y}{\theta} \right )^{2} &= - \frac{2y^{2}}{\theta ^{3}} \end{align}\]Sustituyendo:
\[\begin{align} 0 &= \int e^{ ln(2) - 2ln(\theta) + ln(y) - \left (\frac{y}{\theta} \right)^{2}} * \frac{d}{d \theta} \left [ ln(2) - 2ln(\theta) + ln(y) - \left (\frac{y}{\theta} \right)^{2} \right] dy \\ 0 &= \int f(Y; \lambda , \theta) * \left ( - \frac{2}{\theta} + \frac{2y^{2}}{\theta ^{3}} \right ) dy \\ 0 &= \int \left ( f(Y; \lambda , \theta) * - \frac{2}{\theta} \right ) + \left ( f(Y; \lambda , \theta) * \frac{2y^{2}}{\theta ^{3}} \right ) \\ 0 &= - \frac{2}{\theta} \int f(Y; \lambda , \theta) + \left ( f(Y; \lambda , \theta) * \frac{2y^{2}}{\theta ^{3}} \right ) dy \\ 0 &= - \frac{2}{\theta} \int f(Y; \lambda , \theta) dy + \int \left ( f(Y; \lambda , \theta) * \frac{2y^{2}}{\theta ^{3}} \right ) dy \end{align}\]Recordando:
\[ \int f(Y; \lambda , \theta) dy = 1 \]
Entonces:
\[\begin{align} 0 &= - \frac{2}{\theta} (1) + \int \left ( f(Y; \lambda , \theta) * \frac{2y^{2}}{\theta ^{3}} \right ) dy \\ 0 &= - \frac{2}{\theta} + \int \left ( f(Y; \lambda , \theta) * \frac{2y^{2}}{\theta ^{3}} \right ) dy \\ 0 &= - \frac{2}{\theta} + \int \left ( f(Y; \lambda , \theta) * \frac{2}{\theta ^{3}} y^{2} \right ) dy \\ 0 &= - \frac{2}{\theta} + \int \left ( \frac{2}{\theta ^{3}} y^{2} f(Y; \lambda , \theta) \right ) dy \\ 0 &= - \frac{2}{\theta} + \frac{2}{\theta ^{3}} \int y^{2} f(Y; \lambda , \theta) dy \end{align} \]Recordando:
\[ f(Y; \lambda , \theta) = \frac{2}{\theta^2} y e^{- \left (\frac{y}{\theta} \right)^{2}} \] Entonces:
\[ 0 = - \frac{2}{\theta} + \frac{2}{\theta ^{3}} \int y^{2} f(Y; \lambda , \theta) dy \\ 0 = - \frac{2}{\theta} + \frac{2}{\theta ^{3}} \int y^{2} \frac{2}{\theta^2} y e^{- \left (\frac{y}{\theta} \right)^{2}}dy \]
Simplificando:
\[\begin{align} 0 &= - \frac{2}{\theta} + \frac{2}{\theta ^{3}} \left ( \frac{2}{\theta^2} \int y^{2} y e^{- \left (\frac{y}{\theta} \right)^{2}}dy \right) \\ 0 &= - \frac{2}{\theta} + \frac{2}{\theta ^{3}} \left ( \frac{2}{\theta^2} \int y^{3} e^{- \left (\frac{y}{\theta} \right)^{2}}dy \right) \\ 0 &= - \frac{2}{\theta} + \frac{2}{\theta ^{3}} * \frac{2}{\theta^2}\left ( \int y^{3} e^{- \left (\frac{y}{\theta} \right)^{2}}dy \right) \end{align}\]Simplificando fracciones:
\[ \frac{2}{\theta ^{3}} * \frac{2}{\theta^2} = \frac{4}{\theta ^{5}} \]
Entonces:
\[ 0 = - \frac{2}{\theta} + \frac{4}{\theta ^{5}} \left ( \int y^{3} e^{- \left (\frac{y}{\theta} \right)^{2}}dy \right) \]
Si se realiza el siguiente cambio de variables:
\[ u= \frac{y}{\theta} \Rightarrow y=u\theta \\ dy = \theta du \]
Al sustituir se tiene que:
\[\begin{align} 0 &= - \frac{2}{\theta} + \frac{4}{\theta ^{5}} \left ( \int (u \theta)^{3} e^{-(u)^{2}} \theta du \right) \\ 0 &= - \frac{2}{\theta} + \frac{4}{\theta ^{5}} \left ( \int u^{3} \theta ^{3} e^{-(u)^{2}}\theta du \right) \\ 0 &= - \frac{2}{\theta} + \frac{4}{\theta ^{5}} \left ( \int u^{3} \theta ^{3} \theta e^{-(u)^{2}} du \right) \\ 0 &= - \frac{2}{\theta} + \frac{4}{\theta ^{5}} \left ( \int u^{3} \theta ^{4} e^{-(u)^{2}} du \right) \\ 0 &= - \frac{2}{\theta} + \frac{4}{\theta ^{5}} \left ( \theta ^{4} \int u^{3} e^{-(u)^{2}} du \right) \\ 0 &= - \frac{2}{\theta} + \frac{4}{\theta ^{5}} \theta ^{4} \left ( \int u^{3} e^{-(u)^{2}} du \right) \\ 0 &= - \frac{2}{\theta} + \frac{4}{\theta} \left ( \int u^{3} e^{-(u)^{2}} du \right) \end{align}\]Si se realiza otro cambio de variables:
\[ v = u^{2} \Rightarrow \sqrt{v} = u \\ dv = 2 u du \Rightarrow du = \frac{dv}{2u} \Rightarrow du = \frac{dv}{2 \sqrt{v}} \\ u= \frac{y}{\theta} \Rightarrow y=u\theta \\ dy = \theta du \]
Recordando:
\[ 0 = - \frac{2}{\theta} + \frac{4}{\theta} \left ( \int u^{3} e^{-(u)^{2}} du \right) \]
Sustituyendo en la integral:
\[\begin{align} \int u^{3} e^{-(u)^{2}} du &= \int (\sqrt{v})^{3} e^{ - (\sqrt{v})^{2}} \frac{dv}{2 \sqrt{v} } \\ &= \int v^{\frac{3}{2}} e^{ -v} \frac{dv}{2v^{\frac{1}{2}} } \\ &= \int \frac{v^{\frac{3}{2}}}{2v^{\frac{1}{2}} } e^{-v} dv \\ &= \int \frac{v}{2} e^{-v} dv \\ &= \frac{1}{2} \int v e^{-v} dv \\ \end{align}\]Agregando a la ecuación con la que se estaba trabajando:
\[ 0 = - \frac{2}{\theta} + \frac{4}{\theta} \left [ \frac{1}{2} \int v e^{-v} dv \right] \]
Trabajando con la integral:
\[ \int v e^{-v} dv \]
Si se realiza el siguiente cambio de variables:
\[ w = v \Rightarrow dw = dv \\ z = -e^{-v} \Rightarrow dz = e^{-v} dv \]
Al reemplazar en la integral se obtiene:
\[\begin{align} \int v e^{-v} dv &= \int w dz \\ &= wz - \int z dw \\ &= -v e^{-v} - \int -e^{-v} dv \\ &= -ve^{-v} + \int e^{-v} dv \\ \int e^{-v} dv &= -e^{-v} \\ \int v e^{-v} dv &= -v e^{-v} -e^{-v} \\ &= -e^{-v} (v+1) \end{align}\]Recordando:
\[ 0 = - \frac{2}{\theta} + \frac{4}{\theta} \left [ \frac{1}{2} \int v e^{-v} dv \right] \]
Entonces:
\[ 0 = - \frac{2}{\theta} + \frac{4}{\theta} \left [ \frac{1}{2} ( -e^{-v} (v+1)) \right] \\ 0 = - \frac{2}{\theta} + \frac{4}{\theta} \left [ - \frac{1}{2} ( e^{-v} (v+1)) \right] \\ \]
Como \(v = u^{2}\) , entonces:
\[ e^{-v} (v+1) = e^{-u^{2}} (u^{2} +1) \]
Reemplazando:
\[\begin{align} 0 &= - \frac{2}{\theta} + \frac{4}{\theta} \left [ - \frac{1}{2} ( e^{-v} (v+1)) \right] \\ 0 &= - \frac{2}{\theta} + \frac{4}{\theta} \left [ - \frac{1}{2} ( e^{-u^{2}} (u^{2} +1)) \right] \end{align}\]Recordando que \(u=\frac{y}{\theta}\) , entonces se tiene:
\[\begin{align} 0 &= - \frac{2}{\theta} + \frac{4}{\theta} \left [ - \frac{1}{2} ( e^{-u^{2}} (u^{2} +1)) \right] \\ 0 &= - \frac{2}{\theta} + \frac{4}{\theta} \left [ - \frac{1}{2} \left( e^{- \left (\frac{y}{\theta} \right)^{2}} \left (\left( \frac{y}{\theta} \right)^{2} +1 \right) \right) \right] \\ 0 &= - \frac{2}{\theta} + \frac{4}{\theta}* \left ( - \frac{1}{2} \right) \left[ e^{- \left (\frac{y}{\theta} \right)^{2}} \left (\left( \frac{y}{\theta} \right)^{2} +1 \right) \right] \\ 0 &= - \frac{2}{\theta} - \frac{4}{2 \theta} \left[ e^{- \left (\frac{y}{\theta} \right)^{2}} \left (\left( \frac{y}{\theta} \right)^{2} +1 \right) \right] \\ 0 &= - \frac{2}{\theta} - \frac{2}{\theta} \left[ e^{- \left (\frac{y}{\theta} \right)^{2}} \left (\left( \frac{y}{\theta} \right)^{2} +1 \right) \right] \end{align}\]Distribuyendo \(\frac{2}{\theta}\) , queda:
\[\begin{align} 0 &= - \frac{2}{\theta} - \frac{2}{\theta} \left[ e^{- \left (\frac{y}{\theta} \right)^{2}} \left (\left( \frac{y}{\theta} \right)^{2} +1 \right) \right] \\ 0 &= - \frac{2}{\theta} - \frac{2}{\theta} e^{- \left (\frac{y}{\theta} \right)^{2}} \left (\left( \frac{y}{\theta} \right)^{2} +1 \right) \\ 0 &= - \frac{2}{\theta} \left [ 1 + e^{- \left (\frac{y}{\theta} \right)^{2}} \left (\left( \frac{y}{\theta} \right)^{2} +1 \right) \right ] \end{align}\]El estimador de máxima verosimilitud de \(\theta\) se obtiene resolviendo la ecuación anterior. Aunque no es posible obtener una solución análıticamente, se puede calcular \(\theta\) utilizando métodos numéricos como el método de Newton-Raphson.
Se tiene la distribución de Weibull:
\[ f(Y; \lambda , \theta) = \frac{\lambda}{\theta^{}} y^{(\lambda -1)} e^{- \left (\frac{y}{\theta} \right)^{\lambda}} \] Para que esta distribución pertenezca a la familia exponencial se debe considerar el parámetro \(\lambda\) como una constante, en este caso se tiene que \(\lambda = 2\) entonces se puede trabajar como:
Recordado:
\[ f(Y; \lambda , \theta) = e^{ ln(2) - 2ln(\theta) + ln(y) - \left (\frac{y}{\theta} \right)^{2}} \\ f(Y; \lambda , \theta) = e^{ ln(2) - 2ln(\theta) + ln(y) - y^{2} \left (\frac{1}{\theta^{2}} \right)} \] Reordenando de forma adecuada:
\[ f(Y; \lambda , \theta) = e^{ y^{2} \left ( - \frac{1}{\theta^{2}} \right) + ln(2) - 2ln(\theta) + ln(y)} \]
De modo que:
\[\begin{align} a(y) &= y^{2} \\ b(\theta) &= - \frac{1}{\theta} \\ c(\theta) &= ln(2) - 2ln(\theta) \\ d(y) &= ln(y) \end{align}\]Por lo que la distribución de Weibull cuando \(\lambda =2\) si pertenece a las familias exponenciales.
Si las \(Y_{i}\) son variables aleatorias independientes, provenientes de Weibull con los mismos parámetros, entonces:
Solución:
Recordando:
\[ f(Y; \lambda , \theta) = e^{ y^{2} \left ( - \frac{1}{\theta^{2}} \right) + ln(2) - 2ln(\theta) + ln(y)} \]
Entonces la forma de la distribución de la probabilidad conjunta es:
\[ f(Y_{i}; \lambda , \theta_{i}) = exp \left [ \sum_{i=1}^{n}y_{i} \left ( - \frac{1}{\theta_{i}^{2}} \right) + \sum_{i=1}^{n} [ln(2)- 2ln(\theta_{i}] + \sum_{i=1}^{n} ln(y_{i}) \right] \]
De nuevo se parte desde la distribución de Weibull:
\[ Y \rightsquigarrow Weibull(\theta, \lambda) \]
Donde:
\[ f(Y; \lambda , \theta) = e^{ y^{2} \left ( - \frac{1}{\theta^{2}} \right) + ln(2) - 2ln(\theta) + ln(y)} \] La función de máxima verosimilitud esta definida como:
\[ L(\theta \mid \mathbf{x}) = \prod_{i=1}^{n} f(x_i \mid \theta) \]
La función de log-verosimilitud es:
\[ \ell(\theta \mid \mathbf{x}) = \log L(\theta \mid \mathbf{x}) = \sum_{i=1}^{n} \log f(x_i \mid \theta) \]
Aplicando esto a la distribución de Weibull, se tiene:
\[ L(\theta, \lambda \mid \mathbf{y}) = \prod_{i=1}^{n} \frac{\theta}{\lambda} \left( \frac{y_i}{\lambda} \right)^{\theta-1} e^{-\left( \frac{y_i}{\lambda} \right)^\theta} \]
\[ \ell(\theta, \lambda \mid \mathbf{y}) = \log L(\theta, \lambda \mid \mathbf{y}) = \sum_{i=1}^{n} \left[ \log \left( \frac{\theta}{\lambda} \right) + (\theta-1) \log \left( \frac{y_i}{\lambda} \right) - \left( \frac{y_i}{\lambda} \right)^\theta \right] \]
\[\begin{align} \end{align}\]