Download Theorem If X i ∼ Weibull(α i,β), for i = 1,2,...,n, and X 1,X2,...,Xn are

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Theorem If Xi ∼ Weibull(αi , β), for i = 1, 2, . . . , n, and X1 , X2 , . . . , Xn are mutually
independent random variables, then
"
min{X1 , X2 , . . . , Xn } ∼ Weibull 
n
X
(1/αi )
i=1
#−1

, β .
Proof The random variable Xi has probability density function
fXi (x) = (β/αi )xβ−1 e−x
β /α
x > 0,
i
for i = 1, 2, . . . , n. The associated survivor function of Xi is
SXi (x) = e−x
β /α
x>0
i
for i = 1, 2, . . . , n. Let Y = min {X1 , X2 , . . . , Xn }. The survivor function of Y is
SY (y) =
=
=
=
=
P (Y ≥ y)
P (min {X1 , X2 , . . . , Xn } ≥ y)
P (X1 ≥ y, X2 ≥ y, . . . , Xn ≥ y)
P (X1 ≥ y) P (X2 ≥ y) . . . P (Xn ≥ y)
SX1 (y)SX2 (y) . . . SXn (y)
β
β
1 )x
= e−(1/α
e−(1/α2 )x . . . e−(1/αn )x
Pn
β
= e− i=1 (1/αi )x
x > 0.
β
This survivor function can be recognized as that of a Weibull random variable with scale
P
parameter [ ni=1 (1/αi )]−1 and shape parameter β.
APPL illustration: The APPL statements to find the minimum of a Weibull(2, β) and
Weibull(3, β) are (note different parameterizations between APPL and the chart):
assume(beta > 0);
X1 := [[x -> exp(-x ^ beta / 2)], [0, infinity], ["Continuous", "SF"]];
X2 := [[x -> exp(-x ^ beta / 3)], [0, infinity], ["Continuous", "SF"]];
Minimum(X1, X2);
These statements yield a Weibull distribution for the minimum with a scale parameter
" n
X
i=1
(1/αi )
#−1
1 1
=
+
2 3
−1
=
6
5
and shape parameter β, which is consistent with the theorem.
1
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