There are cases when we are interested in the maximum or minimum of a random sample.
Maximum:
For instance, the maximum can be used to model:
◦
The lifetime of a system of
n
independent components connected in parallel,
◦
The completion time of a project of
n
independent subprojects, which can be completed
simultaneously.
Setup of problem:
You are given a pdf of n independent random variables
X
1
, X
2
, ..., X
n
.
Question: Find the pdf of the maximum of
X
1
, X
2
, ..., X
n
Steps:
Let
Y
= max(
X
1
, X
2
, ..., X
n
).
To find the pdf of
Y
we start by finding its cdf.
F
Y
(
y
) =
P
(
Y
≤
y
)
=
P
(
X
1
≤
y, X
2
≤
y, ..., X
n
≤
y
)
=
P
(
X
1
≤
y
)
P
(
X
2
≤
y
)
...P
(
X
n
≤
y
) (because
X
i
’s are independent)
=
F
X
1
(
y
)
F
X
2
(
y
)
...F
X
n
(
y
)
Furthermore, if all the
X
i
s have the same pdf,
F
Y
(
y
) = [
F
X
(
y
)]
n
11
We found the cdf of
Y
, now how do we get the pdf of
Y
?
f
Y
(
y
) =
F
0
Y
(
y
)
=
n
[
F
X
(
y
)]
n

1
f
X
(
y
)
We will work on ex 4.8 and 4.9 from the course text.
Example
8
.
A system consists of five components connected in parallel.
The lifetime (in
thousands of hours) of each component is an exponential random variable with mean
μ
= 3.
(a) Calculate the median and standard deviation for each component
(b) Calculate the probability that the system fails before 3500 hours. Compare this with
the probability that a component fails before 3500 hours.
(c) Calculate the median life, mean life and standard deviation for the system.
12
Minimum:
The minimum can be used to model:
◦
The lifetime of a system of
n
independent components connected in series,
◦
The completion time of a project pursued by
n
independent competing teams
Setup of problem:
You are given the pdf of n independent random variables
X
1
, X
2
, ..., X
n
.
Question: Find the pdf of the minimum of
X
1
, X
2
, ..., X
n
Steps:
Let
Y
= min(
X
1
, X
2
, ..., X
n
).
Again, to find the pdf of
Y
we start by finding its cdf.
F
Y
(
y
) =
P
(
Y
≤
y
)
= 1

P
(
Y > y
)
= 1

P
(
X
1
> y, X
2
> y, ..., X
n
> y
)
= 1

[
P
(
X
1
> y
)
P
(
X
2
> y
)
...P
(
X
n
> y
)] (because
X
i
’s are independent)
= 1

[1

F
X
1
(
y
)][1

F
X
2
(
y
)]
...
[1

F
X
n
(
y
)]
Furthermore, if all the
X
i
s have the same pdf,
F
Y
(
y
) = 1

[1

F
X
(
y
)]
n
How do we get the pdf of
Y
?
f
Y
(
y
) =
F
0
Y
(
y
)
=

n
[1

F
X
(
y
)]
n

1
(

f
X
(
y
))
=
n
[1

F
X
(
y
)]
n

1
f
X
(
y
)
Example
9
.
A system consists of five components connected in series.
The lifetime (in
thousands of hours) of each component is an exponential random variable with mean
μ
= 3.
(a) Calculate the probability that the system fails before 3500 hours. Compare this with
the probability that a component fails before 3500 hours.
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 Summer '11
 YewWei
 Statistics, Probability theory, σ, λ, 3500 hours