In this section, we present the methods used in the application to estimate the different types of confidence bounds for exponentially distributed data. The complete derivations were presented in detail (for a general function) in the Confidence Bounds chapter.

At this time we should point out that exact confidence bounds for the
exponential distribution
have been derived, and exist in a closed form, utilizing the *χ*2
distribution. These are described in detail in Kececioglu [20],
and are covered in the section on test design
in the Additional
Reliability Analysis chapter. For most exponential data analyses,
Weibull++ will
use the approximate confidence bounds, provided from the Fisher information
matrix or the likelihood ratio, in order to stay consistent with all of
the other available distributions in the application. The *χ*2
confidence bounds for the exponential distribution are discussed in more
detail in the Recurrent Event
Data Analysis chapter.

This section includes the following subsections:

Fisher Matrix Confidence Bounds for the Exponential Distribution

Likelihood Ratio Confidence Bounds for the Exponential Distribution

For the failure rate
the upper (*λ**U*) and lower (*λ**L*) bounds are estimated
by [30]:

(15)

(16)

where *K**α*
is defined by:

(17)

If *δ* is the
confidence level, then for the two-sided bounds, and
*α* = 1 - *δ*
for the one-sided bounds.

The variance of ,
*Var*(), is
estimated from the Fisher matrix, as follows:

(18)

where Λ is the log-likelihood function of the exponential distribution, described in Appendix C.

Note that no true MLE solution exists for the case of the two-parameter
exponential distribution. The mathematics simply break down while trying
to simultaneously solve the partial derivative equations for both the
*γ* and *λ*
parameters, resulting in unrealistic conditions. The way around this conundrum
involves setting *γ* = *T*1,
or the first time-to-failure, and calculating *λ*
in the regular fashion for this methodology. Weibull++ treats *γ*
as a constant when computing bounds, *i.e.
**Var*() = 0.
(See the discussion in Appendix
C for more information.)

The reliability of the two-parameter exponential distribution is:

(19)

The corresponding confidence bounds are estimated from:

(20)

(21)

These equations hold true for the one-parameter exponential distribution,
with *γ* = 0.

The bounds around time for a given exponential percentile, or reliability value, are estimated by first solving the reliability equation with respect to time, or reliable life:

(22)

The corresponding confidence bounds are estimated from:

(23)

(24)

The same equations apply for the one-parameter exponential with *γ*
= 0.

For one-parameter distributions such as the exponential,
the likelihood confidence bounds are calculated by finding values for
*θ* that satisfy:

(25)

This equation can be rewritten as:

(26)

For complete data, the likelihood function for the exponential distribution is given by:

(27)

where the *xi*
values represent the original time-to-failure data. (Note: Note that the
procedures outlined here are for complete data only. For data with suspensions
or intervals, one merely needs to expand the likelihood function to include
these other data types. See the Data
& Data Types chapter for more information.) For a given value
of *α*, values
for *λ* can be
found which represent the maximum and minimum values that satisfy Eqn.
(16).
These represent the confidence bounds for the parameters at a confidence
level *δ*, where
*α* = *δ*
for two-sided bounds and *α*
= 2*δ*
- 1 for one-sided.

Five units are put on a reliability test and experience failures at 20, 40, 60, 100, and 150 hours. Assuming an exponential distribution, the MLE parameter estimate is calculated to be = 0.013514. Calculate the 85% two-sided confidence bounds on these parameters using the likelihood ratio method.

The first step is to calculate the likelihood function for the parameter estimates:

(28)

where *xi*
are the original time-to-failure data points. We can now rearrange Eqn.
(16)
to the form:

Since our specified confidence level, *δ*,
is 85%, we can calculate the value of the chi-squared statistic, We can now substitute this
information into the equation:

It now remains to find the values of *λ*
which satisfy this equation. Since there is only one parameter, there
are only two values of *λ* that will
satisfy the equation. (Note: This is why one cannot generate a contour
plot for exponential distributions in Weibull++, as there is only one variable to vary, hence
a two-dimensional plot cannot be generated.) These values represent the
*δ* = 85% two-sided
confidence limits of the parameter estimate . For our problem, the confidence limits are:

In order to calculate the bounds on a time estimate for a given reliability, or on a reliability estimate for a given time, the likelihood function needs to be rewritten in terms of one parameter and time/reliability, so that the maximum and minimum values of the time can be observed as the parameter is varied. This can be accomplished by substituting a form of the exponential reliability equation into the likelihood function. The exponential reliability equation can be written as:

This can be rearranged to the form:

This equation can now be substituted into Eqn. (27) to produce a likelihood
equation in terms of *t* and *R*:

(29)

The "unknown" parameter *t*/*R*
depends on what type of bounds are being determined. If one is trying
to determine the bounds on time for a given reliability, then *R*
is a known constant and* t* is the
unknown parameter. Conversely, if one is trying to determine the bounds
on reliability for a given time, then *t*
is a known constant and *R*
is the unknown parameter. Either way, Eqn. (29) can be used to solve Eqn.
(16)
for the values of interest.

For the data given in Example 5, determine the 85% two-sided confidence
bounds on the time estimate for a reliability of 90%. The ML estimate
for the time at *R*(*t*)
= 90% is = 7.797.

In this example, we are trying to determine the 85% two-sided confidence
bounds on the time estimate of 7.797. This is accomplished by substituting
*R *=
0.90 and *α* = 0.85
into Eqn. (29). It now remains to find the values of *t*
which satisfy this equation. Since there is only one parameter, there
are only two values of *t* that will
satisfy the equation. These values represent the *δ*
= 85% two-sided confidence limits of the time estimate . For our problem, the confidence limits are:

For the data given in Example 5, determine the 85% two-sided confidence
bounds on the reliability estimate for a *t*
= 50. The ML estimate for the time at *t*
= 50 is = 50.881%.

In this example, we are trying to determine the 85% two-sided confidence
bounds on the reliability estimate of 50.881%. This is accomplished by
substituting *t* = 50 and
*α* = 0.85into
Eqn. (29). It now remains to find the values of *R*
which satisfy this equation. Since there is only one parameter, there
are only two values of *t* that will
satisfy the equation. These values represent the *δ*
= 85% wo-sided confidence limits of the reliability estimate For our problem, the
confidence limits are:

From the Confidence Bounds chapter,
we know that the posterior distribution of *λ*
can be written as:

where , is the non-informative prior
of *λ*.

With the above prior distribution, *f*(*λ*|*Data*)
can be rewritten as:

(30)

The one-sided upper bound of *λ*
is:

The one-sided lower bound of *λ*
is:

The two-sided bounds of *λ*
are:

The reliable life equation is:

For the one-sided upper bound on time we have:

(31)

Eqn. (21)
can be rewritten in terms of *λ*
as:

From Eqn (30), we have:

(32)

Eqn. (32) is solved w.r.t. *TU*. The same method is
applied for one-sided lower and two-sided bounds on time.

The one-sided upper bound on reliability is given by:

(33)

Eqn. (33) can be rewritten in terms of *λ*
as:

From Eqn (30), we have:

(34)

Eqn. (34) is solved w.r.t. *RU*. The same method can
be used to calculate one-sided lower and two sided bounds on reliability.

See Also:

The Exponential Distribution

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