Inverted gamma as a life distribution

M/croe/~tron. R¢//ab., Vol. 29,No. 4, pp. 619-626, 1989. Printed in Great Britain.

0026-2714/8953.00+ .00 © 1989 Pergamon Press plc

INVERTED GAMMA AS A LIFE DISTRIBUTION C.T.LIN, B.S.DURAN and T.O.LEWIS* Texas Tech University, Lubbock, Texas 79409, U.S.A. (Received for publication 11 November 1988)

Abstract

The inverted gamma distribution is studied as a prospective life distribution. The maximum likelihood estimation of the parameters of the inverted gamma is discussed, together with the estimation of t h e c o r r e s p o n d i n g reliability f u n c t i o n . A comparison with the lognormal and inverse Gaussian distribution is illustrated in terms of an example based on a maintenance data set. 1

Introduction

In reliability studies commonly used models in life testing include the gamma, log-normal and inverse Gaussian distributions. These models are usually chosen on the basis of what is understood about the failure mechanisms. If the failures are mainly due to aging or the wearing out process, then it is reasonable, in many applications, to choose one of the above mentioned distributions [1,4,5,6]. In this paper we consider the inverted gamma as a life distribution and compare it against the log-normal and inverse Gaussian distributions. From this study it is seen that the inverted gamma compares quite favorably with the log-normal and inverse Gaussian models and that it has advantages of being simple and easy to use. The probability density function (p.d.f) of the inverted gamma distribution with parameters a and ~ is

f(x;a,~)

=

1 F(a)fl~x,~+1 exp[-1/(x/~)], x > 0

=

o, otherwise,

(1)

where c~,~ > 0 [3]. It is observed that if X has an inverted gamma(a, ~) distribution then 1/X has a gamma(c~, ~) distribution. The mean of the inverted gamma distribution is 1 / [ ~ ( a - 1)], where c~ > 1, and its variance is 1 / [ ~ 2 ( a - 1)2(a-2)], where a > 2. If a = 1 we have the inverted exponential distribution which has no finite moments; however, in this study we will show by an example that it can be a good life distribution model. 2

Inverted

Exponential

As A Life Distribution

Model

The p.d.f, of an inverted exponential distribution with parameter ~ is

f(x;~)

=

~exp[-I/(x~)], ~ > o

=

O, otherwise,

*To whom correspondence should be addressed. 619

(2)

620

C.T. LIN et al. where 8 > 0. The reliability function, i.e. the probability of no failure before time t, is R ( t ) = 1 - F(t) = 1 - e -1/(,~), (3) where F ( t ) is the distribution function of X. The failure rate of a distribution at time t is defined as r(t) = f ( t ) / R ( t ) ,

t > O.

Thus the failure rate of an inverted exponential distribution with parameter 8 is r(t) = l/[St~(e'/('~) - 11, t > 0, (4) which is simple and can be easily computed for any parameter 8. Some properties of r(t) are the following: (i) {i~ r(t) = 0, (ii) l i m r ( t ) = O, d (iii) The maximum of r ( t ) can be computed by considering ~-~log r ( t ) = O. Thus, to determine the maximum of r(t), we consider logr(t) = - log8 - 21ogt - l o g [ e ll(ta) - 11,

(5)

which implies

r'(t)/r(t)= - 2 / t + el/(~)/[St2(el/('~) - 1)] = 0 or

28t[e l/ta) - 1] - e 1/(ta) = 0. By letting y = t/3 we obtain 2y[e 1/v

- -

1]

-

e 1/u = 0

or

e 1/' = 2 y l ( 2 y - 1).

(6)

The only solution to (6) is 0.6275005. Thus the only critical value of (5) is t = 0.6275005//~. Therefore, r(t) reaches its maximum at 0.6275005//L A graph of r(t) for various values of a and/~ is given in Figure 1. If xl, x2,..., x, are identical independent inverted exponential random variables with parameter ~, then the likelihood function is i x~-z) e x p [ - ~ i__~11/x,]. L(8; xl, x ~ , . . . , ~ , ) = ~1 ( fi=1 =

(7)

The maximum likelihood estimator (MLE) for 8 is ~ where rt

8

1 ~S 11~,.

(s)

---- r t i - - 1

Since 1 / x i , i = 1, 2 , . . . , n, are independent and identically distributed exrt

ponential random variables with parameter/~, n/~ = ~ " ~ l / x i has a gamma(n, 8) iml

distribution. Thus E(/~) =/3, that is, f} is an unbiased estimator for fL To find a 100(1 - a)% lower confidence limit (LCL) for the reliability function R(t), we regard R ( t ) = Rt(8) as a function of fL For every t > O, R , ( 8 ) = 1 - e -l/(tB) is a decreasing function of 8 since P~(/3) = -e-~(t-~2 ) < 0. Letting/~ denote the 100(1 - a)% upper confidence limit (UCL) for 8, we have 1-a

=

P[~
=

PIR,(~) > R,(~)].

Inverted gamma as a life distribution

621

8H

L..

--t

,,= =5,13=1 ot=l , p=l •

2

01J 0

1

2

3

~=1,13=5

4

5

6

7

8

9

10

Repair time Figure 1 - Failure rate of inverted gamma(a.fl)

Therefore, r , ( ~ ) = 1 - e-'/'~

(9)

is a 100(1-c~)% LCL for R(~). To determine ~ consider the fact that n~ has a gamma(n, ~) distribution and that 2n~/~ has a chi-square distribution, denoted by X~(2n), with 2n degrees of freedom. If X~(2n) denotes the c~-th quantile of X2(2n), then Z-

,~

=

P[2.#'/~ > X~(2n)]

= Pp.~/x~(2.) >_~1. Hence a 100(1 -- a)% UCL for B is

= 2.~/x~(2~).

(10)

The MLE/~(~) of R($) of the inverted exponential distribution is

~t) = I

-

e -v'#,

(ii)

where fi is given in (8). The estimator/~(t) is a biased estimator for R(~). This can be determined as follows. The probability density function of/~ is

/(/~;/~)--r ( n~-:~("-'-~ )#.p , /~>o,/~>o. e

02)

C.T. LINet al.

622 We have

~:[R(t)] = E [ 1 - e-*#a] =

1 - E[e -l#a]

=

1-

-Tv

Let V = n~/~, so that the last expression can be written as 1

1-~(-~_

~o

_ _

_~.

y~ le (~+~,,)dy.

To evaluate the integral we use a result due to Watson [7], _

_

1~

a

r._7.l.

f0 y "e (~v+,)dy = 2 ( ~ ) 2 K._l(2v/a--b) where K~(.) is the modified Bessel function of the second kind of order v. Also, K~(.) = K_~(.) for v = 0 , 1 , 2 , . . . . Substituting - v = n - 1, a = 1 and b = Tt into the equation above we obtain

E[/~(t)l = 1

(n -1)! (

)-~IQ(2

)

(13)

which shows t h a t / ~ ( t ) is a biased estimator of R(0. To determine an estimator of R(t) which is a uniformly minimum variance unbiased estimator (UMVUE) we proceed as follows. Let 1, if XI_> t

T(X1) =

O, otherwise,

which is a unbiased estimator of R(t). Observe that 8, given by (8), is a complete sufficient estimator of R(t). Sufficiency of ~ follows by using (7) and (12) to write n

L ( ~ , , ~ ..... ~da;~)

=

':' a . _ , e _ ~ ~

r(.)~-,"

r(,~)

1

n

,~"II~/~-' i=1

which is independent of ~. The completeness of ~ follows from Lehmann and Scheffe [2]. By the Rao-Blackwell Theorem the UMVUE of R(t) is given by

~l(t) = E[T(X1)[~].

(14)

To derive the conditional p.d.f Xl given ~ partition the n-dimensional random vector (X1, X 2 , . . . , Xn) into components X1 and ( X 2 , . . . , X , ) . Let

]I1 = 1/X,

(15)

and

?_

1_ ~ I / X , . n

I

(16)

i=2

Since I"1 is distributed as a gamma(l, ~) independent of (n - 1)I2 which is distributed as a gamma(n - 1,/~), the joint p.d.f, of Y1 and I~ is

Inverted gamma as a life distribution

1

Y(v,,~)

=

,

y,,

(n -

7expi-Tlr(

.

623

1~,"-I .-.,-i expr (n ~.1)~] 1)/~._,/ l

(" - 1)"-' "-~ expi-"Jl'7

(17)

where Yl > 0 and ~ > 0. From (15) and (16), we have/~ = ~[Y1 1 + ( n - 1)17"] and by using a simple transformation we find the joint p.d.f, of Y1 and/~ to be 72

/(v,,~) - r(~ - i)Z -(~i - v')"-~ exp[-~]'

o < v, <

~.

The conditional p.d.f, of Y1 given j is given by Yl ~,-~ S@,ID)=~ n - 1 ( 1 _ ~, , o<~,< From (14), /~(Q is the conditional probability of equivalent to Y1 < 1/t given/~. We thus obtain

n - 1 L:t/'(1

~(t)_

i

- (1

-

_>

t given j , which is

y,/nj)"-~d~,

---J

=

X,

n/~.

l/n~ll)

"-|

1

, ~ ~ llnj, , 0 < t < llnj,

(18)

which is the UMVUE estimator of R(t).

3

Illustration

The following maintenance data were reported on active repair times (hours) for an airborne communication transceiver (Von Alven [3], page 156). Repair Time: 0.2, 1.0, 2.5, 7.0,

0.3, 1.0, 2.7, 7.5,

0.5, 1.0, 3.0, 8.8.,

0.5, 1.0, 3.0, 9.0,

0.5, 1.1, 3.3, 10.3,

0.5, 1.3, 3.3, 22.0,

0.6, 1.5, 4.0, 24.5.

0.6, 0.7, 1.5, 1.5, 4.0, 4.5,

0.7, 1.5, 4.7,

0.7, 2.0, 5.0,

0.8, 2.0, 5.4,

Suppose we model the data with the inverted exponential distribution. From the data we estimate f? to be 0.88 and consequently we find the inverted exponential distribution with/~ = 0.88 provides a good fit to the above repair time data. These data have been modeled with the log-normal distribution [4] and the inverse Gauss,an distribution [1]. To compare the three models the calculated value of the Kolmororov-Smirnov (K-S) test statistic is 0.0760 for the inverted exponential, 0.0526 for the inverse Gauss, an, and 0.0807 for the log-normal, and these values are smaller than their corresponding values expected at the five percent significance level. The five percent significance level for the K-S test statistic with sample size n = 46 is 0.200. The MLE ~(t) of r(t) of the inverted exponential distribution is

÷(t) = 1/[/~t~(e '/'~` - 1] and a corrsponding graph along with the graphs of the MLE's of the inverse Gauss,an and log-normal for the above data set are given in Figure 2. It can be seen from the graphs of Figure 2 that for small value of t, say less than or equal 1, the inverted exponential distribution models the empirical failure rate better than either the inverse Gauss,an or the log-normal. However, for t large, the inverse Gauss,an seems to hold the advantage. A graph of the MLE of R(t) and a graph of the UMVUE of R(t) along with a corresponding graph of the 95% LCL of R(t) are given in Figure 3. MR 29:4-K

0.8, 2.2, 5.4,

C.T. LINet al.

624 0.8

0.7-

0.6

~'~ 0.5

0.4 Log normal E

Inverse Gaussian

m 0.3

Inved~l exponential

0,2

0.1

0.01~

0

=

I

=

J

i

i

i

l

I

I

2

4

6

8

10

12

14

16

18

20

Repair time Figure 2 - MLE of failure rate

1.0

0.8

~- 0,6 J3

"~ 0.4

R(t)

n-

-UMVUE of R(t) ~

/ ~

"

~

~

95% LCL for R(t) 0.0 0.0

I

I

I

I

2.5

5.0

7.5

10.0

l

12.5

Repair time Figure 3 - UMVUE, MLE, and 95% LCL of R(t)

I

15.0

Inverted gamma as a life distribution

625

The MLE ]?/(t) of R(t) for the inverted exponential compares favorably with the empirical reliability. To illustrate this, if t = 1, the empirical reliability is 0.6304 while ~ ( t = 1) = 0.6791. If t = 2, then the empirical reliability is 0.4565 while R(t -- 2) = 0.4334 and if t = 22 the empirical reliability is 0.0217 while ]~(t = 22) = 0.0503. Also, the graph of the inverted exponential/~(t) compares favorably with the corresponding graphs of the inverse Gaussian and the log-normal ]~(t)'s. If there is an advantage, over all, of the inverted exponential to the other two models it is in the simplicity of calculation. In summary the inverted exponential when used as a life distribution model is certainly comparable to the inverse Gaussian or the log-normal and has the advantage of being very simple with only one parameter to be estimated. Also, one can calculate the 100(1 -c~)% LCL for R(t) with ease. 4

Inverted

Gamma

As A Life Distribution

Model

If we use the inverted g a m m a distribution (I) as life distribution, then the reliabilityfunction is

R(t) = P[T > t l = P[

< 71.

Since l I T is distributed as a gamma(a,/~) then n(0

1

=

['#

-

r(a) r,/~,(,0,

1

(10)

where r v ( a ) is the incomplete gamma function. The failure rate is

,.(t)= [#~t°+' e'/o'r,/,,(-)l -'.

(20)

The graph of r(t) for different values of a and /3 is shown in Figure 1. In each case the failure rate increases from 0 to its maximum and then decreases to 0 as t ~ oo. For a random sample X l , X 2 , . . . , x , from an inverted gamma distribution with parameters a and r , the log likelihood function is log L(a,/3; x,.,x2, • • •, x , ) n

=

-

+ i) Elog

n

l/xi.

,-

(21)

i-~ l

The MLE's of a and/~ are solutions of the following system,

0

L(~,~;xl,... ,x,)

= - n ¢ ( a ) - n l o g B - Y'~logxl -- 0 i=l

OlogL - ~ .(a,/3;xl,...,x,) =

n~ 1 ----~--F ~ ~ I/x, = O,

(22)

where ¢(.) is the derivative of the g a m m a function and is called the digamma function. A closed form for a and/3 does not exist, however, the equation in (22) are equivalent to the system

n¢(~) + n log[! O~ r t .__

1(1 ~

8-- g

1

~1. i=1

i

)1 + ~ log ~, = 0, i=1

(23)

C.T. LIN et al.

626

The first equation involves only one unknown c~, so it can be solved numerically. As an illustration we model the data given in Section III with the inverted gamma distribution. The MLE's obtained as a solution of (23) are & = 1.078 and/~ = 0.816. The calculated K-S test statistic is .0918 for the inverted gamma distribution with parameters c~ = 1.078 and ~ = 0.816. This value is greater than .0760 for the corresponding value for the inverted exponential distribution. The MLE/~(t) of R(t) for the inverted gamma is superior to that of the inverted exponential for large values of T(T > 7.0), equally comparable for small values of T(T < 1.0), and inferior for values of T(1.0 < T < 7.0) in the middle range. From the above illustration we see that the inverted gamma life distribution model does not provide an overall improvement to the inverted exponential life distributed model, b-hrthermore, it loses the desirable feature of simplicity.

References [1] Chhikara, R. S. and Folks, J. L., (1977). "The Inverse Gaussian Distribution As A Life Model," Teehnometries, 19, 461-468. [2] Lehmann, E. L. and Scheffe, H., (1950, 1955). "Completeness, similar regions and unbiased estimation," Sankhya I0, 305-340; 10, 219-236.

[3]

Raiffa, H. and Schlaifer, R. (1961), Applied Statistical Decision Theory, Graduate School of Business Administration. Harvard University, Boston.

[4]

Sherif, Y. S. and Smith, M. L. (1980). "First-Passage Time Distribution of Brownlan Motion As A Reliability Model," IEEE Transaction of Reliability, Vol. R-29, No. 5, Dec. 1980, 425-426.

[5] Sinha, S. K. and Kale, B. K. (1980). Life Testing and Reliability Estimation, John Wiley ~z Sons. [6] Von Alven, W. H. (ed.) (1964). Reliabilty Engineering by ARINC, Prentice-Hali, Inc., New Jersey. [7] Watson, G.N. (1952). Treatise On The Theory Of Bessel Functions (2nd ed.), Cambridge University Press.