Question

Suppose X, Y, and Z are random variables with the joint density function f(x, y, z) = Ce−(0.5x + 0.2y + 0.1z) if x ≥ 0, y ≥ 0, z ≥ 0, and f(x, y, z) = 0 otherwise. (a) Find the value of the constant C.

Answer 1

The value of the constant C is 0.01 .

Explanation:

Given:

Suppose X, Y, and Z are random variables with the joint density function,

`f(x,y,z) = \left \{ {{Ce^(-(0.5x + 0.2y + 0.1z)); x,y,z\geq0 } \atop {0}; Otherwise} \right.`

The value of constant C can be obtained as:

`\int_x( {\int_y( {\int_z {f(x,y,z)} \, dz }) \, dy }) \, dx = 1`

`\int\limits^\infty_0 ({\int\limits^\infty_0 ({\int\limits^\infty_0 {Ce^(-(0.5x + 0.2y + 0.1z)) } \, dz }) \, dy } )\, dx = 1`

`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`

`C\int\limits^\infty_0 {e^(-0.5x)(\int\limits^\infty_0{e^(-0.2y)([(-e^(-0.1z) )/(0.1) ]\limits^\infty__0 }) \, dy }) \, dx = 1`

`C\int\limits^\infty_0 {e^(-0.5x)(\int\limits^\infty_0 {e^(-0.2y)([(-e^(-0.1(\infty)) )/(0.1)+(e^(-0.1(0)) )/(0.1) ]) } \, dy }) \, dx = 1`

`C\int\limits^\infty_0 {e^(-0.5x)(\int\limits^\infty_0 {e^(-0.2y)[0+(1)/(0.1)] } \, dy }) \, dx =1`

`10C\int\limits^\infty_0 {e^(-0.5x)([(-e^(-0.2y) )/(0.2)]^\infty__0 }) \, dx = 1`

`10C\int\limits^\infty_0 {e^(-0.5x)([(-e^(-0.2(\infty)) )/(0.2)+(e^(-0.2(0)) )/(0.2)] } \, dx = 1`

`10C\int\limits^\infty_0 {e^(-0.5x)[0+(1)/(0.2)] } \, dx = 1`

`50C([(-e^(-0.5x) )/(0.5)]^\infty__0}) = 1`

`50C[(-e^(-0.5(\infty)) )/(0.5) + (-0.5(0))/(0.5)] =1`

`50C[0+(1)/(0.5) ] =1`

`100C = 1` ⇒
`C = (1)/(100)`

C = 0.01