數理統計與資料分析第三版習題 第3章 第19題 阿新 • • 發佈：2018-12-18 P(T1>T2)=∬t1>t2f(t1,t2)dt2dt1=∫0∞∫0t1αe−αt1∗βe−βt2dt2dt1=∫0∞αe−αt1∗β∗1−β[e−βy]0t1dt2=∫0∞(−αe−αt1e−βx+α−αt1)dt1=∫0∞(α−αt1)dt1−∫0∞αe−αt1e−βxdt1=1−αα+β=βα+β\begin{aligned} P(T_{1}>T_{2})&=\iint_{t_{1}>t_{2}} f(t_{1},t_{2})dt_{2}dt_{1}\\ &=\int_{0}^{\infty}\int_{0}^{t_{1}}αe^{-αt_{1}}*βe^{-βt_{2}}dt_{2}dt_{1}\\ &=\int_{0}^{\infty}αe^{-αt_{1}}*β*\frac{1}{-β}\left[e^{-βy}\right]_{0}^{t_{1}} dt_{2}\\ &=\int_{0}^{\infty}(-αe^{-αt_{1}}e^{-βx}+α^{-αt_{1}})dt_{1}\\ &=\int_{0}^{\infty}(α^{-αt_{1}})dt_{1}-\int_{0}^{\infty}αe^{-αt_{1}}e^{-βx}dt_{1}\\ &=1-\frac{α}{α+β}\\ &=\frac{β}{α+β} \end{aligned}
