Respuesta :
Answer:
The value of the constant C is 0.01 .
Step-by-step explanation:
Given:
Suppose X, Y, and Z are random variables with the joint density function,
[tex]f(x,y,z) = \left \{ {{Ce^{-(0.5x + 0.2y + 0.1z)}; x,y,z\geq0 } \atop {0}; Otherwise} \right.[/tex]
The value of constant C can be obtained as:
[tex]\int_x( {\int_y( {\int_z {f(x,y,z)} \, dz }) \, dy }) \, dx = 1[/tex]
[tex]\int\limits^\infty_0 ({\int\limits^\infty_0 ({\int\limits^\infty_0 {Ce^{-(0.5x + 0.2y + 0.1z)} } \, dz }) \, dy } )\, dx = 1[/tex]
[tex]C\int\limits^\infty_0 {e^{-0.5x}(\int\limits^\infty_0 {e^{-0.2y }(\int\limits^\infty_0 {e^{-0.1z} } \, dz }) \, dy }) \, dx = 1[/tex]
[tex]C\int\limits^\infty_0 {e^{-0.5x}(\int\limits^\infty_0{e^{-0.2y}([\frac{-e^{-0.1z} }{0.1} ]\limits^\infty__0 }) \, dy }) \, dx = 1[/tex]
[tex]C\int\limits^\infty_0 {e^{-0.5x}(\int\limits^\infty_0 {e^{-0.2y}([\frac{-e^{-0.1(\infty)} }{0.1}+\frac{e^{-0.1(0)} }{0.1} ]) } \, dy }) \, dx = 1[/tex]
[tex]C\int\limits^\infty_0 {e^{-0.5x}(\int\limits^\infty_0 {e^{-0.2y}[0+\frac{1}{0.1}] } \, dy }) \, dx =1[/tex]
[tex]10C\int\limits^\infty_0 {e^{-0.5x}([\frac{-e^{-0.2y} }{0.2}]^\infty__0 }) \, dx = 1[/tex]
[tex]10C\int\limits^\infty_0 {e^{-0.5x}([\frac{-e^{-0.2(\infty)} }{0.2}+\frac{e^{-0.2(0)} }{0.2}] } \, dx = 1[/tex]
[tex]10C\int\limits^\infty_0 {e^{-0.5x}[0+\frac{1}{0.2}] } \, dx = 1[/tex]
[tex]50C([\frac{-e^{-0.5x} }{0.5}]^\infty__0}) = 1[/tex]
[tex]50C[\frac{-e^{-0.5(\infty)} }{0.5} + \frac{-0.5(0)}{0.5}] =1[/tex]
[tex]50C[0+\frac{1}{0.5} ] =1[/tex]
[tex]100C = 1[/tex] ⇒ [tex]C = \frac{1}{100}[/tex]
C = 0.01
