Cómo calcular la probabilidad conjunta: $\Pr \left( \tfrac{g_1}{g_3} \geq \theta_1, \tfrac{g_2}{g_3} \geq \theta_2, g_3 > \theta_3 \right)$ ?

Question

Pregunta : ¿Cómo se calcula lo siguiente? $$\Pr \left( \dfrac{g_1}{g_3} \geq \theta_1, \dfrac{g_2}{g_3} \geq \theta_2, g_3 > \theta_3 \right),$$ donde $g_i, i \in \{1, 2, 3\}$ es una variable aleatoria distribuida exponencialmente con la pdf dada por $$f_{g_i}(x) = \dfrac{1}{\Omega_i} \exp\left( \dfrac{-x}{\Omega_i}\right),$$ y $\theta_i, \Omega_i \in \mathbb R^+ \forall i$ . Además, todos los $g_i$ se supone que son mutuamente independientes.

Intento : Para el caso en que tengamos que calcular lo siguiente $$\Pr\left( \dfrac{g_1}{g_3} \geq \theta_1, g_3 > \theta_3 \right) = \dfrac{1}{\Omega_1 \Omega_3} \int_{y = \theta_1 \theta_3}^{y = \infty} \int_{x = \theta_3}^{x = y/\theta_1} f_{g_3}(x) f_{g_1}(y) \ \mathrm dx \ \mathrm dy.$$ Sin embargo, estoy atascado debido a la tercera variable $g_2$ . Se agradece cualquier pista.

Answer 1

Supongo que esos $G_i$ son independientes. Entonces, condicionado a $G_3$ simplificará los cálculos:

$$ \begin{align} & \Pr\left\{\frac {G_1} {G_3} \geq \theta_1, \frac {G_2} {G_3} \geq \theta_2, G_3 > \theta_3\right\} \\ = & \int_{\theta_3}^{+\infty} \Pr\left\{\frac {G_1} {G_3} \geq \theta_1, \frac {G_2} {G_3} \geq \theta_2, G_3 > \theta_3 ~\Bigg|~ G_3 = g \right\} \frac {1} {\omega_3} \exp\left\{- \frac {g} {\omega_3} \right\} dg \\ = & \int_{\theta_3}^{+\infty} \Pr\{G_1 \geq g\theta_1, G_2 \geq g\theta_2\} \frac {1} {\omega_3} \exp\left\{- \frac {g} {\omega_3} \right\} dg \\ = & \int_{\theta_3}^{+\infty} \Pr\{G_1 \geq g\theta_1\}\Pr\{G_2 \geq g\theta_2 \} \frac {1} {\omega_3} \exp\left\{- \frac {g} {\omega_3} \right\} dg \\ = & \int_{\theta_3}^{+\infty} \exp\left\{- \frac {g\theta_1} {\omega_1} \right\} \exp\left\{- \frac {g\theta_2} {\omega_2} \right\} \frac {1} {\omega_3} \exp\left\{- \frac {g} {\omega_3} \right\} dg \\ = & \frac {1} {\omega_3} \int_{\theta_3}^{+\infty} \exp\left\{- \left(\frac {\theta_1} {\omega_1} + \frac {\theta_2} {\omega_2} + \frac {1} {\omega_3} \right)g\right\} dg \\ = & - \frac {1} {\omega_3} \left(\frac {\theta_1} {\omega_1} + \frac {\theta_2} {\omega_2} + \frac {1} {\omega_3} \right)^{-1} \left. \exp\left\{- \left(\frac {\theta_1} {\omega_1} + \frac {\theta_2} {\omega_2} + \frac {1} {\omega_3} \right)g\right\} \right|_{\theta_3}^{+\infty} \\ =& \left(\frac {\theta_1\omega_3} {\omega_1} + \frac {\theta_2\omega_3} {\omega_2} + 1 \right)^{-1} \exp\left\{- \left(\frac {\theta_1} {\omega_1} + \frac {\theta_2} {\omega_2} + \frac {1} {\omega_3} \right)\theta_3 \right\} \end{align} $$