The joint distribution for the lenght of life of two different types of components operating in a
system is given by
0 , elsewhere
The relative efficiency of the two types of components is measured by U= Y2 / Y1.
a.) Let V = Y1. Find the joint density for U and V.
b.) Find the marginal density for U.
c.) Find E(V|U = 1) The joint distribution for the lenght of life of two different types of
components operating in a system is given by f( y1 , y2) = q1/8 y 1 e -(y 1 + y 2)/2 y1 >0 , y2
>0 , elsewhere The relative efficiency of the two types of components is measured by U= Y2 /
Y1. Let V = Y1. Find the joint density for U and V. Find the marginal density for U. Find
E(V|U = 1)
Solution
a) Let U = Y?/Y? and V = Y? <==> Y? = V and Y? = UV.
So, the boundary lines transform as follows:
y? = 0 ==> v = 0
y? = 0 ==> u = 0.
(Moreover any point in the first quadrant of (y?, y?) is mapped to a point in the first quadrant of
(u, v).)
Next, the Jacobian ?(y?, y?)/?(u, v) equals
|0 1|
|v u| = -v.
So, g(u, v) = f(y?, y?) * |?(y?, y?)/?(u, v)|
................= (1/8)y? e^(-(y? + y?)/2) * |-v|
................= (1/8)v e^(-(v + uv)/2) * v
................= (1/8)v^2 e^(-v(1+u)/2) for u, v ? 0 (and 0 otherwise)
--------------
b) Integrating through the domain of V,
g_u(u) = ?(v = 0 to ?) (1/8)v^2 e^(-v(1+u)/2) dv
..........= ?(t = 0 to ?) (1/8) (2t/(1+u))^2 e^(-t) * 2 dt/(1+u), letting t = v(1+u)/2
..........= (1+u)^(-3) * ?(t = 0 to ?) t^2 e^(-t) dt
..........= (1+u)^(-3) * 2, via gamma function or repeated integration by parts
..........= 2/(1+u)^3 for u ? 0 (and 0 otherwise).
--------------
c) g_v(v | U = 1) = [(1/8)v^2 e^(-v(1+u)/2)] / [2/(1+u)^3] {at u = 1}
........................= (1/2)v^2 e^(-v) for v ? 0 (and 0 otherwise).
So, E[V | U = 1]
= ?(v = 0 to ?) v * (1/2)v^2 e^(-v) dv
= (1/2) * ?(v = 0 to ?) v^3 e^(-v) dv
= (1/2) * 3!
= 3..
The joint distribution for the lenght of life of two different type.pdf
1. The joint distribution for the lenght of life of two different types of components operating in a
system is given by
0 , elsewhere
The relative efficiency of the two types of components is measured by U= Y2 / Y1.
a.) Let V = Y1. Find the joint density for U and V.
b.) Find the marginal density for U.
c.) Find E(V|U = 1) The joint distribution for the lenght of life of two different types of
components operating in a system is given by f( y1 , y2) = q1/8 y 1 e -(y 1 + y 2)/2 y1 >0 , y2
>0 , elsewhere The relative efficiency of the two types of components is measured by U= Y2 /
Y1. Let V = Y1. Find the joint density for U and V. Find the marginal density for U. Find
E(V|U = 1)
Solution
a) Let U = Y?/Y? and V = Y? <==> Y? = V and Y? = UV.
So, the boundary lines transform as follows:
y? = 0 ==> v = 0
y? = 0 ==> u = 0.
(Moreover any point in the first quadrant of (y?, y?) is mapped to a point in the first quadrant of
(u, v).)
Next, the Jacobian ?(y?, y?)/?(u, v) equals
|0 1|
|v u| = -v.
So, g(u, v) = f(y?, y?) * |?(y?, y?)/?(u, v)|
................= (1/8)y? e^(-(y? + y?)/2) * |-v|
................= (1/8)v e^(-(v + uv)/2) * v
................= (1/8)v^2 e^(-v(1+u)/2) for u, v ? 0 (and 0 otherwise)
--------------
2. b) Integrating through the domain of V,
g_u(u) = ?(v = 0 to ?) (1/8)v^2 e^(-v(1+u)/2) dv
..........= ?(t = 0 to ?) (1/8) (2t/(1+u))^2 e^(-t) * 2 dt/(1+u), letting t = v(1+u)/2
..........= (1+u)^(-3) * ?(t = 0 to ?) t^2 e^(-t) dt
..........= (1+u)^(-3) * 2, via gamma function or repeated integration by parts
..........= 2/(1+u)^3 for u ? 0 (and 0 otherwise).
--------------
c) g_v(v | U = 1) = [(1/8)v^2 e^(-v(1+u)/2)] / [2/(1+u)^3] {at u = 1}
........................= (1/2)v^2 e^(-v) for v ? 0 (and 0 otherwise).
So, E[V | U = 1]
= ?(v = 0 to ?) v * (1/2)v^2 e^(-v) dv
= (1/2) * ?(v = 0 to ?) v^3 e^(-v) dv
= (1/2) * 3!
= 3.
![b) Integrating through the domain of V,
g_u(u) = ?(v = 0 to ?) (1/8)v^2 e^(-v(1+u)/2) dv
..........= ?(t = 0 to ?) (1/8) (2t/(1+u))^2 e^(-t) * 2 dt/(1+u), letting t = v(1+u)/2
..........= (1+u)^(-3) * ?(t = 0 to ?) t^2 e^(-t) dt
..........= (1+u)^(-3) * 2, via gamma function or repeated integration by parts
..........= 2/(1+u)^3 for u ? 0 (and 0 otherwise).
--------------
c) g_v(v | U = 1) = [(1/8)v^2 e^(-v(1+u)/2)] / [2/(1+u)^3] {at u = 1}
........................= (1/2)v^2 e^(-v) for v ? 0 (and 0 otherwise).
So, E[V | U = 1]
= ?(v = 0 to ?) v * (1/2)v^2 e^(-v) dv
= (1/2) * ?(v = 0 to ?) v^3 e^(-v) dv
= (1/2) * 3!
= 3.](data:image/gif;base64,R0lGODlhAQABAIAAAAAAAP///yH5BAEAAAAALAAAAAABAAEAAAIBRAA7)