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 Slides 6 TMA4275 LIFETIME ANALYSIS () 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 TMA4275 LIFETIME ANALYSIS () 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 Slides 6 TMA4275 LIFETIME ANALYSIS () 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 Slides 6 TMA4275 LIFETIME ANALYSIS () 4 / 12 EXAMPLE: NELSON-AALEN ESTIMATOR Bo Lindqvist Slides 6 TMA4275 LIFETIME ANALYSIS () 5 / 12 RESIDUAL LIFETIME 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 TMA4275 LIFETIME ANALYSIS ∞ 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 Slides 6 TMA4275 LIFETIME ANALYSIS () 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 Slides 6 TMA4275 LIFETIME ANALYSIS () 8 / 12 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 TMA4275 LIFETIME ANALYSIS () 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 Slides 6 TMA4275 LIFETIME ANALYSIS () 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 Slides 6 TMA4275 LIFETIME ANALYSIS () 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 TMA4275 LIFETIME ANALYSIS () 12 / 12

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