1.- Si la función de distribución conjunta de X y Y está dada por:
F
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otro caso
{\displaystyle F(x,y)={\begin{cases}\left(1-e^{-2x}\right)\left(1-e^{-3y}\right)&0<x<\infty ,0<y<\infty \\0&{\mbox{otro caso}}\end{cases}}}
Determina:
a) La densidad conjunta de X y Y .
Como se nos da F(x,y), es decir la distribución acumulada, de manera para obtenerla debimos integrar con respecto de x y y , así que para obtenerla, debemos derivar parcialmente con respecto de x y y :
f
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{\displaystyle f(x,y)}
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∂
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∂
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{\displaystyle \partial ^{2}\left[\left(1-e^{-2x}\right)\left(1-e^{-3y}\right)\right] \over \partial x\partial y}
∂
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∂
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{\displaystyle \partial \left[\left(2e^{-2x}\right)\left(1-e^{-3y}\right)\right] \over \partial y}
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6
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{\displaystyle =\left(2e^{-2x}\right)\left(3e^{-3y}\right)=6e^{-2x-3y}}
Es decir:
f
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6
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otro caso
{\displaystyle f(x,y)={\begin{cases}6e^{-2x-3y}&0<x<\infty ,0<y<\infty \\0&{\mbox{otro caso}}\end{cases}}}
P
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P
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6
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6
∫
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∫
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{\displaystyle \operatorname {P} (X<{\frac {1}{3}},Y>1)=\operatorname {P} (0<X<{\frac {1}{3}},1<Y<\infty )=\int _{0}^{\frac {1}{3}}\int _{1}^{\infty }6e^{-2x-3y}\,\operatorname {d} x\,\operatorname {d} y=6\int _{0}^{\frac {1}{3}}e^{-2x}\int _{1}^{\infty }e^{-3y}\,\operatorname {d} x\,\operatorname {d} y}
6
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{\displaystyle 6\int _{0}^{\frac {1}{3}}e^{-2x}\left[-{\frac {1}{3}}\,e^{-3y}\right]{\Bigg |}_{1}^{\infty }\operatorname {d} y=6\int _{0}^{\frac {1}{3}}e^{-2x}\left[{\frac {1}{3}}\,e^{-3}\right]\operatorname {d} y=6\left({\frac {1}{3}}\,e^{-3}\right)\int _{0}^{\frac {1}{3}}e^{-2x}\operatorname {d} y=2e^{-3}\left[-{\frac {1}{2}}e^{-2x}\right]{\Bigg |}_{0}^{\frac {1}{3}}}
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11
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≈
0.0242
{\displaystyle =2e^{-3}\left(-{\frac {1}{2}}e^{-{\frac {2}{3}}}+{\frac {1}{2}}\right)=e^{-{\frac {11}{3}}}+e^{-3}\approx 0.0242}
c)
h
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6
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6
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{\displaystyle \operatorname {h} (y)=\int _{0}^{\infty }6e^{-2x-3y}\,\operatorname {d} x=6e^{-3y}\int _{0}^{\infty }e^{-2x}\,\operatorname {d} x=6e^{-3y}\left[-{\frac {1}{2}}e^{-2x}\right]{\Bigg |}_{0}^{\infty }=6e^{-3y}\left({\frac {1}{2}}\right)=3e^{-3y}}
d)
M
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E
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6
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{\displaystyle \operatorname {M_{y}} (t)=\operatorname {E} \left[e^{xt}\right]=\int _{0}^{\infty }\int _{0}^{\infty }e^{xt}6e^{-2x-3y}\,\operatorname {d} x\,\operatorname {d} y=6\int _{0}^{\infty }e^{-3y}\int _{0}^{\infty }e^{(t-2)x}\,\operatorname {d} x\,\operatorname {d} y}
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6
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{\displaystyle =6\int _{0}^{\infty }e^{-3y}\left[{\frac {e^{(t-2)x}}{t-2}}\right]{\Bigg |}_{0}^{\infty }\,\operatorname {d} y=6\int _{0}^{\infty }e^{-3y}\left(-{\frac {1}{t-2}}\right)\,\operatorname {d} y=-{\frac {6}{t-2}}\int _{0}^{\infty }e^{-3y}\,\operatorname {d} y=-{\frac {6}{t-2}}\left[-{\frac {1}{3}}e^{-3y}\right]{\Bigg |}_{0}^{\infty }}
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{\displaystyle =-{\frac {6}{t-2}}\left({\frac {1}{3}}\right)=-{\frac {2}{t-2}}}
![{\displaystyle \partial ^{2}\left[\left(1-e^{-2x}\right)\left(1-e^{-3y}\right)\right] \over \partial x\partial y}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ccf18883687d3d786cbfd2c751b9662baf28a56c)
![{\displaystyle \partial \left[\left(2e^{-2x}\right)\left(1-e^{-3y}\right)\right] \over \partial y}](https://wikimedia.org/api/rest_v1/media/math/render/svg/26309db2bdef5354c0cee4f77439d26b70830e12)
![{\displaystyle 6\int _{0}^{\frac {1}{3}}e^{-2x}\left[-{\frac {1}{3}}\,e^{-3y}\right]{\Bigg |}_{1}^{\infty }\operatorname {d} y=6\int _{0}^{\frac {1}{3}}e^{-2x}\left[{\frac {1}{3}}\,e^{-3}\right]\operatorname {d} y=6\left({\frac {1}{3}}\,e^{-3}\right)\int _{0}^{\frac {1}{3}}e^{-2x}\operatorname {d} y=2e^{-3}\left[-{\frac {1}{2}}e^{-2x}\right]{\Bigg |}_{0}^{\frac {1}{3}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/67235551cb57d556a645b852763bffc5845e02ae)
![{\displaystyle \operatorname {h} (y)=\int _{0}^{\infty }6e^{-2x-3y}\,\operatorname {d} x=6e^{-3y}\int _{0}^{\infty }e^{-2x}\,\operatorname {d} x=6e^{-3y}\left[-{\frac {1}{2}}e^{-2x}\right]{\Bigg |}_{0}^{\infty }=6e^{-3y}\left({\frac {1}{2}}\right)=3e^{-3y}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/cc68a6a9112ca7e8dc035567b669a6ecd2c9df04)
![{\displaystyle \operatorname {M_{y}} (t)=\operatorname {E} \left[e^{xt}\right]=\int _{0}^{\infty }\int _{0}^{\infty }e^{xt}6e^{-2x-3y}\,\operatorname {d} x\,\operatorname {d} y=6\int _{0}^{\infty }e^{-3y}\int _{0}^{\infty }e^{(t-2)x}\,\operatorname {d} x\,\operatorname {d} y}](https://wikimedia.org/api/rest_v1/media/math/render/svg/919732936c94c2f3091cb938152efae4d9ac4671)
![{\displaystyle =6\int _{0}^{\infty }e^{-3y}\left[{\frac {e^{(t-2)x}}{t-2}}\right]{\Bigg |}_{0}^{\infty }\,\operatorname {d} y=6\int _{0}^{\infty }e^{-3y}\left(-{\frac {1}{t-2}}\right)\,\operatorname {d} y=-{\frac {6}{t-2}}\int _{0}^{\infty }e^{-3y}\,\operatorname {d} y=-{\frac {6}{t-2}}\left[-{\frac {1}{3}}e^{-3y}\right]{\Bigg |}_{0}^{\infty }}](https://wikimedia.org/api/rest_v1/media/math/render/svg/5e803f237a238269641eecae75f96b07a70cc26f)
