### TMA4275 LIFETIME ANALYSIS Slides 6: Nelson-Aalen

```TMA4275 LIFETIME ANALYSIS
Slides 6: Nelson-Aalen estimator and TTT plot
Bo Lindqvist
Department of Mathematical Sciences
Norwegian University of Science and Technology
Trondheim
http://www.math.ntnu.no/∼bo/
[email protected]
NTNU, Spring 2014
Bo Lindqvist
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1 / 12
WHY IS AN ESTIMATE OF Z (t) USEFUL?
Note first that Z 0 (t) = z(t). Thus,
T is IFR ⇔ z(t) is increasing ⇔ Z (t) is convex
T is DFR ⇔ z(t) is decreasing ⇔ Z (t) is concave
Thus a plot of an estimate Zˆ (t) can give us information on whether the
distribution of T is IFR (increasing failure rate) or DFR (decreasing failure
rate).
Bo Lindqvist
Slides 6
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2 / 12
ESTIMATING Z (t) BY THE KM-ESTIMATOR
Recall that R(t) = e −Z (t) , so
Z (t) = − ln R(t)
ˆ KM (t) is the KM-estimator for R(t), then we can define,
Thus - if R
ˆ KM (t)
ZˆKM (t) = − ln R
Y ni − di
= − ln
ni
T(i) ≤t
=−
X
ln 1 −
T(i) ≤t
≈
di ni
X di
ni
T(i) ≤t
where we used that for small x is
− ln(1 − x) ≈ x
Bo Lindqvist
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3 / 12
THE NELSON-AALEN ESTIMATOR FOR Z (t)
The Nelson-Aalen estimator (NA-estimator) is simply defined by
ZˆNA (t) =
X di
ni
T(i) ≤t
It can then be shown that its variance can be estimated by
Var\
(ZˆNA (t)) =
X di
ni2
T(i) ≤t
Note: The Nelson-Aalen estimator is not included in MINITAB (only
“hazard plot” which is in fact not a correct). For this course has been
made a MINITAB Macro (see MINITAB Macros on the Software
webpage).
In the following we shall have a closer look at how the Nelson-Aalen
estimator can be motivated from properties of the exponential distribution.
Bo Lindqvist
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EXAMPLE: NELSON-AALEN ESTIMATOR
Bo Lindqvist
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5 / 12
Suppose an item with lifetime T is still alive at time s. The probability of
surviving an additional t time is then
R(t | s) ≡ P(T > s + t | T > s)
P(T > s + t ∩ T > s)
=
P(T > s)
R(s + t)
=
R(s)
This is called the conditional survival function of the item at time t, or the
distribution of the residual life. The following is its expectation, called
Mean Residual Life:
Z
MRL(t) =
=
Bo Lindqvist
Slides 6
∞
R(t | s)dt =
0
Z ∞
1
R(t)dt
R(s) s
Z
0
∞
R(s + t)
dt
R(s)
()
6 / 12
PROPERTIES OF THE EXPONENTIAL DISTRIBUTION:
1. The memoryless property
Write T ∼ expon(λ) if f (t) = λe −λt ; R(t) = P(T > t) = e −λt , t > 0.
For T ∼ expon(λ) we therefore have
R(t | s) = P(T > s + t | T > s) =
R(s + t)
e −λ(s+t)
=
= e −λt = R(t).
R(s)
e −λs
This is called the memoryless property of the exponential distribution.
For any age s, the remaining life has same distribution as for a new item.
Bo Lindqvist
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7 / 12
PROPERTIES OF THE EXPONENTIAL DISTRIBUTION
2. Let T ∼ expon(λ) and let W = aT . Then W ∼ expon(λ/a).
Proof:
P(W > w ) = P(aT > w ) = P(T >
λ
w
) = e −( a )w
a
3. Let Ti for i = 1, . . . , n be independent, with TiP
∼ expon(λi ).
Let W = min(T1 , . . . , Tn ).. Then W ∼ expon( ni=1 λi ).
Proof:
P(W > w ) = P(min(T1 , · · · , Tn ) > w )
= P(T1 > w , T2 > w , · · · , Tn > w )
= P(T1 > w )P(T2 > w ) · · · P(Tn > w )
= e −(λ1 +···+λn )w ,
so W ∼ expon(λ1 + · · · + λn )
Bo Lindqvist
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PROPERTIES OF THE EXPONENTIAL DISTRIBUTION
4. In particular if T1 , . . . , Tn are independent each with
distribution expon(λ), then
W = min(T1 , . . . , Tn ) ∼ expon(nλ)
So a series system of n components with lifetimes that are independent
and exponentially distributed with hazard rate λ, has a lifetime which is
exponenital with hazard rate nλ and hence
MTTF =
Bo Lindqvist
Slides 6
1
Component MTTF
=
nλ
n
()
9 / 12
PROPERTIES OF THE EXPONENTIAL DISTRIBUTION
5. Let T1 , . . . , Tn be independent each with distribution
expon(λ). Let the ordering of these be
T(1) < T(2) < · · · < T(n)
Then
nT(1)
(n − 1)(T(2) − T(1) )
(n − 2)(T(3) − T(2) )
..
.
(n − i + 1)(T(i) − T(i−1) )
..
.
(T(n) − T(n−1) )
are independent and identically distributed as expon(λ).
Bo Lindqvist
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10 / 12
PROPERTIES OF THE EXPONENTIAL DISTRIBUTION
5b. Let T1 , . . . , Tn be independent each with distribution
expon(λ). Let the ordering of these be
T(1) < T(2) < · · · < T(n)
Then
T(1) ∼ expon(nλ)
T(2) − T(1) ∼ expon((n − 1)λ)
T(3) − T(2) ∼ expon((n − 2)λ)
..
.
T(i) − T(i−1) ∼ expon((n − i + 1)λ)
..
.
T(n) − T(n−1) ∼ expon(λ)
are independent with the displayed exponential distributions.
Bo Lindqvist
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11 / 12
PROOF OF THE EQUIVALENT CLAIMS OF 5 AND 5b
To go from 5b to 5, we use property 2 of the exponential distribution.
Thus we prove only 5b here.
Assume that n units are put on test at time 0. Potential lifetimes of these
are T1 , . . . , Tn , and hence T(1) = min(T1 , . . . , Tn ), so by property 4 above
we already have T(1) ∼ expon(nλ).
After time T(1) there are n − 1 unfailed units. At time s = T(1) each of
these has by property 1 a remaining lifetime which is expon(λ). It follows
from this that we from time T(1) and onwards have the same situation as
at time 0, only that there are now n − 1 instead of n units on test.
Therefore the time to next failure, T(2) − T(1) , is distributed as the
minimum of n − 1 expon(λ) variables and hence is expon((n − 1)λ). That
T(2) − T(1) is independent of T(1) follows from property 1 which says that,
for the exponential distribution, the distribution of the remaining lifetime
is the same whatever be the age of the item.
This reasoning can be continued at time T(2) in an obvious fashion, and
we finish by concluding that T(n) − T(n−1) is expon(λ).
Bo Lindqvist
Slides 6