Download Order variables and extreme variables

Order variables and extreme variables
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(Note written by Stian Lydersen, NTNU, and Jan Terje Kvaløy, HiS)
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The distribution of order variables
Let X1 , X2 , . . . , Xn be independent identically distributed random variables with cumulative distribution function FX (x) = P (X ≤ x). We sort
the variables in increasing order and denote them X(1) , X(2) , . . . , X(n) , where
X(1) ≤ X(2) ≤ · · · ≤ X(n) . This is called order variables. In particular we
have
X(1) = min(X1 , X2 , . . . , Xn )
X(n) = max(X1 , X2 , . . . , Xn )
and these are called extreme variables. The median is defined by:
X̃ =
X((n+1)/2)
1
(X
+ X(n/2+1) )
(n/2)
2
if n is odd
if n is even
The range is defined by: X(n) − X(1) .
FV (v) = P (V ≤ v)
P (max(X1 , X2 , . . . , Xn ) ≤ v)
P ((X1 ≤ v) ∩ (X2 ≤ v) ∩ . . . ∩ (Xn ≤ v))
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indep..
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P (X1 ≤ v) · P (X2 ≤ v) · · · P (Xn ≤ v)
[FX (v)]n
1 − P (X1 > u) · P (X2 > u) · · · P (Xn > u)
1 − [1 − FX (u)]n
If X is having a continuous distribution:
fU (u) =
d
FU (u) = n[1 − FX (u)]n−1 fX (u)
du
The distribution of X(k) becomes
FX(k) (x) = P (k or more Xi s are ≤ x) =
n
j=k
n
[FX (x)]j [1 − FX (x)]n−j
j
since the number of Xi s ≤ x is having a binomial distribution with parameters n and p = P (Xi ≤ x) = FX (x). The probability density function is
found by taking the derivative with respect to x and after some calculation
it can be written:
fX(k) (x) = n
The distribution of V = X(n) = max(X1 , X2 , . . . , Xn ) can be found by
using the fact that if the largest of X1 , . . . , Xn is less than or equal to v then
all of X1 , . . . , Xn must be less than or equal to v:
1 − P (min(X1 , X2 , . . . , Xn ) > u)
1 − P ((X1 > u) ∩ (X2 > u) ∩ . . . ∩ (Xn > u))
n−1
[FX (x)]k−1 [1 − FX (x)]n−k fX (x)
k−1
Example: Series systems and parallel systems
Let us look at the lifetime of a system made up of components with
independent lifetimes. First we will look at a system which is functioning if
and only if all of the components are functioning. This can be illustrated by
a system with the components in series:
If X is having a continuous distribution:
fV (v) =
d
FV (v) = n[FX (v)]n−1 fX (v)
dv
Similarly, the distribution of U = X(1) = min(X1 , X2 , . . . , Xn ) becomes:
FU (u) = P (U ≤ u)
=
P (min(X1 , X2 , . . . , Xn ) ≤ u)
1
X1
q
X2
q
q
Xn
Let Xi be the lifetime of component number i. The system is functioning
until the first component fails. The lifetime of the system is then U =
min(X1 , X2 , . . . , Xn ).
2
In systems or subsystems where high reliability is demanded components
are often in parallel:
Exercises
X1
X2
q
Exercise 1. Consider a parallel system made up of 2 independent components. The lifetime of each component is having an exponential distribution
with parameter λ, i.e.
FX (x) =
q
q
Xn
The system is functioning as long as at least one of the components are
functioning. The lifetime of the system is then V = max(X1 , X2 , . . . , Xn ).
Series system: Let X1 , X2 , . . . , Xn be independent exponentially distributed variables having an exponential distribution with parameter λ such
that fX (x) = λe−λx for x ≥ 0, λ > 0. The distribution of U = min(X1 , X2 , . . . , Xn )
becomes:
FU (u) = P (U ≤ u)
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indep.
P (min(X1 , X2 , . . . , Xn ) ≤ u)
1 − P (min(X1 , X2 , . . . , Xn ) > u)
1 − P ((X1 > u) ∩ (X2 > u) ∩ . . . ∩ (Xn > u))
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1 − P (X1 > u) · P (X2 > u) · · · P (Xn > u))
1 − [1 − FX (u)]n
=
1 − [1 −
=
1 − [1 −
=
=
1 − [1 − (1 − e−λu )]n
1 − e−nλu
,for x > 0
u
0
u
0
1 − e−λx for x ≥ 0
0
otherwise
Let V be the lifetime of the system. Find the distribution of V and E(V ).
Exercise 2. Consider a series system made up of n independent components. The lifetime of each component is having a distribution with cumulative distribution function:
FX (x) =
1 − e−αx
0
for x ≥ 0
otherwise
where α > 0 and β > 0. This is a Weibull distribution with parameters
α > 0 and β > 0.
Find the cumulative distribution function for the lifetime of the system.
Which probability distribution is this?
Calculate the expected lifetime of a component and the expected lifetime
of the whole system when α = 0.1 and β = 0.5.
fX (x)dx]n
λe−λx dx]n
Thus the lifetime of this series system is having an exponential distribution
1
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with parameter nλ. The expected lifetime of the system becomes E(U) = nλ
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β
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