3 The Weibull distribution

The Weibull distribution was first used to describe the behaviour of light bulbs as they undergo the ageing process. Other applications are widespread and include the description of structural failure, ball-bearing failure and the failure of a variety of electronic components.

Key Point 2

If the random variable X is defined by a probability density function of the form

f ( x ) = α β ( α x ) β − 1 e − ( α x ) β

then X is said to follow a Weibull distribution.

The hazard function or conditional failure rate function H ( t ) , which gives the probability that a system or component fails after it has been in use for a given time, is constant for the exponential distribution but for a Weibull distribution is proportional to x β − 1 . This implies that β = 1 gives a constant hazard, β < 1 gives a reducing hazard and β > 1 gives an increasing hazard. Note that α is simply a scale factor while the case β = 1 reduces the Weibull distribution to

f ( x ) = α e − α x

which you may recognize as one form of the exponential distribution.

Figure 3 below shows the Weibull distribution for various values of β . For simplicity, the graphs assume that α = 1 . Essentially the plots are of the function f ( x ) = β x β − 1 e − x β

Figure 3

{The Weibull distribution with beta =1, 2, 4 and 8}

In applications of the Weibull distribution, it is often useful to know the cumulative distribution function (cdf). Although it is not derived directly here, the cdf F ( x ) of the Weibull distribution, whose probability density function is f ( x ) = α β ( α x ) β − 1 e − ( α x ) β , is given by the function

F ( x ) = 1 − e − ( α x ) β

The relationship between f ( x ) and F ( x ) may be seen by remembering that

f ( x ) = d d x F ( x )

Differentiating the cdf F ( x ) = 1 − e − ( α x ) β gives the result

f ( x ) = d d x F ( x ) = ( α x ) β β e − ( α x ) β x = α β β x β − 1 e − ( α x ) β = α β ( α x ) β − 1 e − ( α x ) β

3.1 Mean and variance of the Weibull distribution

The mean and variance of the Weibull distribution are quite complicated expressions and involve the use of the Gamma function, Γ ( x ) . The outline explanations given below defines the Gamma function and shows that basically in terms of the application here, the function can be expressed as a simple factorial.

It can be shown that if the random variable X follows a Weibull distribution defined by the probability density function

f ( x ) = α β ( α x ) β − 1 e − ( α x ) β

then the mean and variance of the distribution are given by the expressions

E ( X ) = 1 α Γ ( 1 + 1 β ) and V ( X ) = 1 α 2 Γ ( 1 + 2 β ) − Γ ( 1 + 1 β ) 2

where Γ ( r ) is the gamma function defined by

Γ ( r ) = ∫ 0 ∞ x r − 1 e − x d x

Using integration by parts gives a fairly straightforward identity satisfied by the gamma function is:

Γ ( r ) = ( r − 1 ) Γ ( r − 1 )

Using this relationship tells us that

Γ ( r ) = ( r − 1 ) ( r − 2 ) ( r − 3 ) … Γ ( 1 ) for integral r

but

Γ ( 1 ) = ∫ 0 ∞ x 1 − 1 e − x d x = ∫ 0 ∞ e − x d x = − e − x 0 ∞ = 1

so that

Γ ( r ) = ( r − 1 ) ( r − 2 ) ( r − 3 ) … ( 3 ) ( 2 ) ( 1 ) = ( r − 1 ) ! for integral r

Essentially, this means that when we calculate the mean and variance of the Weibull distribution using the expressions

E ( X ) = 1 α Γ ( 1 + 1 β ) and V ( X ) = 1 α Γ ( 1 + 2 β ) − Γ ( 1 + 1 β ) 2

we are evaluating factorials and the calculation is not as difficult as you might imagine at first sight.

Note that to apply the work as presented here, the expressions

( 1 + 1 β ) and ( 1 + 2 β )

must both take positive integral values (i.e. 1 β must be an integer) in order for the factorial calculation to make sense. If the above expressions do not assume positive integral values, you will need to know much more about the values and behaviour of the Gamma function in order to do calculations. In practice, we would use a computer of course. As you might expect, the Gamma function is tabulated to assist in such calculations but in the examples and exercises set in this Workbook, the values of 1 β will always be integral.

Example 4

The main drive shaft of a pumping engine runs in two bearings whose lifetime follows a Weibull distribution with random variable X and parameters α = 0.0002 and β = 0.5 .

  1. Find the expected time that a single bearing runs before failure.
  2. Find the probability that any one bearing lasts at least 8000 hours.
  3. Find the probability that both bearings last at least 6000 hours.
Solution
  1. We know that E ( X ) = 1 α Γ ( 1 + 1 β ) = 5000 × Γ ( 1 + 2 ) = 5000 × 2 = 10000 hours.
  2. We require P ( X > 8000 ) , this is given by the calculation: P ( X > 8000 ) = 1 − F ( 8000 ) = 1 − ( 1 − e − ( 0.0002 × 8000 ) 0.5 ) = e − ( 0.0002 × 8000 ) 0.5 = e − 1.265 = 0.282
  3. Assuming that the bearings wear independently, the probability that both bearings last at least 6000 hours is given by { P ( X > 6000 ) } 2 . But P ( X > 6000 ) is given by the calculation

    P ( X > 6000 ) = 1 − F ( 6000 ) = 1 − ( 1 − e − ( 0.0002 × 6000 ) 0.5 ) = 0.335

    so that the probability that both bearings last at least 6000 hours is given by

    { P ( X > 6000 ) } 2 = 0.33 5 2 = 0.112

Task!

A shaft runs in four roller bearings the lifetime of each of which follows a Weibull distribution with parameters α = 0.0001 and β = 1 ∕ 3 .

  1. Find the mean life of a bearing.
  2. Find the variance of the life of a bearing.
  3. Find the probability that all four bearings last at least 50,000 hours. State clearly any assumptions you make when determining this probability.
  1. We know that E ( X ) = 1 α Γ ( 1 + 1 β ) = 10000 × Γ ( 1 + 3 ) = 60000 × 2 = 10000 hours.
  2. We know that V ( X ) = 1 α Γ ( 1 + 2 β ) − Γ ( 1 + 1 β ) 2 so that V ( X ) = ( 10000 ) 2 { Γ ( 1 + 6 ) − ( Γ ( 1 + 3 ) ) 2 } = ( 10000 ) 2 { Γ ( 7 ) − ( Γ ( 4 ) ) 2 } = ( 10000 ) 2 { 6 ! − ( 3 ! ) 2 } = 684 ( 10000 ) 2 = 6.84 × 1 0 10
  3. The probability that one bearing will last at least 50,000 hours is given by the calculation P ( X > 50000 ) = 1 − F ( 50000 ) = 1 − ( 1 − e − ( 0.0001 × 50000 ) 0.5 ) = e − 5 0.5 = 0.107

    Assuming that the four bearings have independent distributions, the probability that all four bearings last at least 50,000 hours is ( 0.107 ) 4 = 0.0001

Exercises
  1. The lifetimes in hours of certain machines have Weibull distributions with probability density function

    f ( t ) = 0 ( t < 0 ) α β ( α t ) β − 1 exp { − ( α t ) β } ( t ≥ 0 )

    1. Verify that ∫ 0 t α β ( α u ) β − 1 exp { − ( α u ) β } d u = 1 − exp { − ( α t ) β } .
    2. Write down the distribution function of the lifetime.
    3. Find the probability that a particular machine is still working after 500 hours of use if α = 0.001 and β = 2.
    4. In a factory, n of these machines are installed and started together. Assuming that failures in the machines are independent, find the probability that all of the machines are still working after t hours and hence find the probability density function of the time till the first failure among the machines.
  2. If the lifetime distribution of a machine has hazard function h ( t ) , then we can find the reliability function R ( t ) as follows. First we find the “cumulative hazard function” H ( t ) using

    H ( t ) = ∫ 0 t h ( t ) d t then the reliability is R ( t ) = e − H ( t ) .

    1. Starting with R ( t ) = exp { − H ( t ) } , work back to the hazard function and hence confirm the method of finding the reliability from the hazard.
    2. A lifetime distribution has hazard function h ( t ) = θ 0 + θ 1 t + θ 2 t 2 . Find
      1. the reliability function.
      2. the probability density function.
    3. A lifetime distribution has hazard function h ( t ) = 1 t + 1 . Find
      1. the reliability function;
      2. the probability density function;
      3. the median.

      What happens if you try to find the mean?

    4. A lifetime distribution has hazard function h ( t ) = ρ θ ρ t + 1 , where ρ > 0 and θ > 1.

      Find
      1. the reliability function.
      2. the probability density function.
      3. the median.
      4. the mean.
  3. A machine has n components, the lifetimes of which are independent. However the whole machine will fail if any component fails. The hazard functions for the components are h 1 ( t ) , … , h n ( t ) . Show that the hazard function for the machine is ∑ i = 1 n h i ( t ) .
  4. Suppose that the lifetime distributions of the components in Exercise 3 are Weibull distributions with scale parameters ρ 1 , … , ρ n and a common index (i.e. “shape parameter”) γ so that the hazard function for component i is γ ρ i ( ρ i t ) γ − 1 . Find the lifetime distribution for the machine.
  5. (Difficult). Show that, if T is a Weibull random variable with hazard function γ ρ ( ρ t ) γ − 1 ,
    1. the median is M ( T ) = ρ − 1 ( ln 2 ) 1 ∕ γ ,
    2. the mean is E ( T ) = ρ − 1 Γ ( 1 + γ − 1 )
    3. the variance is V ( T ) = ρ − 2 { Γ ( 1 + 2 γ − 1 ) − [ Γ ( 1 + γ − 1 ) ] 2 } .

    Note that Γ ( r ) = ∫ 0 ∞ x r − 1 e − x d x . In the integrations it is helpful to use the substitution u = ( ρ t ) γ .

    1. d d u ( − exp { − ( α u ) β } )

      so

      ∫ 0 t α β ( α u ) β − 1 exp { − ( α u ) β } d u = − exp { − ( α u ) β } 0 t = 1 − exp { − ( α t ) β } .

    2. The distribution function is F ( t ) = 1 − exp { − ( α t ) β } .
    3. R ( 500 ) = exp { − ( 0.001 × 500 ) 2 } = 0.7788
    4. The probability that all of the machines are still working after t hours is

      R n ( t ) = { R ( t ) } n = exp { − ( α t ) n β } .

      Hence the time to the first failure has a Weibull distribution with β replaced by n β . The pdf is

      f n ( t ) = 0 ( t < 0 ) α n β ( α t ) n β − 1 exp { − ( α t ) n β } ( t ≥ 0 )

    1. The distribution function is

      F ( t ) = 1 − R ( t ) = 1 − e − H ( t ) so the pdf is f ( t ) = d d t F ( t ) = h ( t ) e − H ( t ) .

      Hence the hazard function is

      h ( t ) = f ( t ) R ( t ) = h ( t ) e − H ( t ) e − H ( t ) = h ( t ) .

    2. A lifetime distribution has hazard function h ( t ) = θ 0 + θ 1 t + θ 2 t 2 .
      1. Reliability R ( t ) . H ( t ) = θ 0 t + θ 1 t 2 ∕ 2 + θ 2 t 3 ∕ 3 0 t = θ 0 t + θ 1 t 2 ∕ 2 + θ 2 t 3 ∕ 3 R ( t ) = exp { − ( θ 0 t + θ 1 t 2 ∕ 2 + θ 2 t 3 ∕ 3 ) } .
      2. Probability density function f ( t ) . F ( t ) = 1 − R ( t ) f ( t ) = ( θ 0 + θ 1 t + θ 2 t 2 ) exp { − ( θ 0 t + θ 1 t 2 ∕ 2 + θ 2 t 3 ∕ 3 ) } .
    3. A lifetime distribution has hazard function h ( t ) = 1 t + 1 .
      1. Reliability function R ( t ) . H ( t ) = ∫ 0 t 1 u + 1 d u = ln ( u + 1 ) 0 t = ln ( t + 1 ) R ( t ) = exp { − H ( t ) } = 1 t + 1
      2. Probability density function f ( t ) . F ( t ) = 1 − R ( t ) , f ( t ) = 1 ( t + 1 ) 2
      3. Median M . 1 M + 1 = 1 2 so M = 1.

    To find the mean:   E ( T + 1 ) = ∫ 0 ∞ 1 t + 1 d t  which does not converge, so neither does E ( T ) .

    1. A lifetime distribution has hazard function

      h ( t ) = ρ θ ρ t + 1 , where ρ > 0 and θ > 1.

      1. Reliability function R ( t ) . H ( t ) = ∫ 0 t ρ θ ρ t + 1 d t = θ ln ( ρ t + 1 ) 0 t = θ ln ( ρ t + 1 ) R ( t ) = exp { − θ ln ( ρ t + 1 ) } = ( ρ t + 1 ) − θ
      2. Probability density function f ( t ) . F ( t ) = 1 − R ( t ) f ( t ) = θ ρ ( ρ t + 1 ) − ( 1 + θ )
      3. Median M . ( ρ M + 1 ) − θ = 2 − 1 ρ M + 1 = 2 1 ∕ θ ∴ M = ρ − 1 ( 2 1 ∕ θ − 1 )
      4. Mean. E ( ρ T + 1 ) = θ ρ ∫ 0 ∞ ( ρ t + 1 ) − θ d t = θ θ − 1 − ( ρ t + 1 ) 1 − θ 0 ∞ = θ θ − 1 E ( T ) = ρ − 1 θ θ − 1 − 1 = ρ − 1 ( θ − 1 ) − 1
  3. For each component H i ( t ) = ∫ 0 t h i ( t ) d t R i ( t ) = exp { − H i ( t ) }

    For machine

    R ( t ) = ∏ i = 1 n R i ( t ) = exp { − ∑ i = 1 n H i ( t ) }

    the probability that all components are still working.

    F ( t ) = 1 − R ( t ) f ( t ) = exp { − ∑ i = 1 n H i ( t ) } ∑ i = 1 n H i ( t ) h ( t ) = f ( t ) ∕ R ( t ) = ∑ i = 1 n h i ( t ) .
  4. For each component h i ( t ) = γ ρ i ( ρ i t ) γ − 1 H i ( t ) = ∫ 0 t γ ρ i ( ρ i t ) γ − 1 d t = ( ρ i t ) γ

    Hence, for the machine,

    ∑ i = 1 n H i ( t ) = t γ ∑ i = 1 n ρ i γ R ( t ) = exp − t γ ∑ i = 1 n ρ i γ F ( t ) = 1 − R ( t ) f ( t ) = γ t γ − 1 ∑ i = 1 n ρ i γ exp − t γ ∑ i = 1 n ρ i γ h ( t ) = γ ρ ( ρ t ) γ − 1

    so we have a Weibull distribution with index γ and scale parameter ρ such that

    ρ γ = ∑ i = 1 n ρ i γ .

  5. Suppose the distribution function is

    F ( t ) = 1 − e − ( ρ t ) γ .

    Then the pdf is

    f ( t ) = d d t F ( t ) = γ ρ ( ρ t ) γ − 1 e − ( ρ t ) γ

    and the hazard function is

    H ( t ) = f ( t ) R ( t ) = f ( t ) 1 − F ( t ) = γ ρ ( ρ t ) γ − 1

    as required.

    1. At the median, M , F ( M ) = 0 . 5 . So 1 − e − ( ρ M ) γ = 0.5 e − ( ρ M ) γ = 0.5 ( ρ M ) γ = − ln ( 0.5 ) = ln ( 2 ) ρ M = ( ln 2 ) 1 ∕ γ M = ρ − 1 ( ln 2 ) 1 ∕ γ
    2. The mean is E ( T ) = ∫ 0 ∞ t f ( t ) d t = ∫ 0 ∞ t γ ρ ( ρ t ) γ − 1 e − ( ρ t ) γ d t = ∫ 0 ∞ γ ( ρ t ) γ e − ( ρ t ) γ d t

      Let u = ( ρ t ) γ , so d u ∕ d t = γ ρ ( ρ t ) γ − 1 and d u = γ ρ ( ρ t ) γ − 1 d t . Then

      E ( T ) = ∫ 0 ∞ ρ − 1 u 1 ∕ γ e − u d u = ρ − 1 ∫ 0 ∞ u 1 + 1 ∕ γ − 1 e − u d u = ρ − 1 Γ ( 1 + 1 ∕ γ )
    3. To find the variance we first find E ( T 2 ) = ∫ 0 ∞ t 2 f ( t ) d t = ∫ 0 ∞ t 2 γ ρ ( ρ t ) γ − 1 e − ( ρ t ) γ d t = ∫ 0 ∞ γ ρ − 1 ( ρ t ) γ + 1 e − ( ρ t ) γ d t

      Let u = ( ρ t ) γ , so d u ∕ d t = γ ρ ( ρ t ) γ − 1 and d u = γ ρ ( ρ t ) γ − 1 d t . Then

      E ( T 2 ) = ∫ 0 ∞ ρ − 2 u 2 ∕ γ e − u d u = ρ − 2 ∫ 0 ∞ u 1 + 2 ∕ γ − 1 e − u d u = ρ − 2 Γ ( 1 + 2 ∕ γ )

      So

      V ( T ) = E ( T 2 ) − { E ( T ) } 2 = ρ − 2 Γ ( 1 + 2 ∕ γ ) − ρ − 2 { Γ ( 1 + 1 ∕ γ ) } 2 = ρ − 2 { Γ ( 1 + 2 ∕ γ ) − { Γ ( 1 + 1 ∕ γ ) } 2 }