# The Exponential Distribution

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(→The 2-Parameter Exponential Distribution) |
(→The 1-Parameter Exponential Distribution) |
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===The 1-Parameter Exponential Distribution=== | ===The 1-Parameter Exponential Distribution=== | ||

- | The 1-parameter exponential ''pdf'' is obtained by setting <math>\gamma =0</math>, and is given by: | + | The 1-parameter exponential ''pdf'' is obtained by setting <math>\gamma =0\,\!</math>, and is given by: |

::<math> \begin{align}f(t)= & \lambda {{e}^{-\lambda t}}=\frac{1}{m}{{e}^{-\tfrac{1}{m}t}}, | ::<math> \begin{align}f(t)= & \lambda {{e}^{-\lambda t}}=\frac{1}{m}{{e}^{-\tfrac{1}{m}t}}, | ||

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::<math>\lambda \,\!</math> = constant rate, in failures per unit of measurement, (e.g., failures per hour, per cycle, etc.) | ::<math>\lambda \,\!</math> = constant rate, in failures per unit of measurement, (e.g., failures per hour, per cycle, etc.) | ||

- | ::<math>\lambda =\frac{1}{m}</math> | + | ::<math>\lambda =\frac{1}{m}\,\!</math> |

::<math>m\,\!</math> = mean time between failures, or to failure | ::<math>m\,\!</math> = mean time between failures, or to failure | ||

::<math>t\,\!</math> = operating time, life, or age, in hours, cycles, miles, actuations, etc. | ::<math>t\,\!</math> = operating time, life, or age, in hours, cycles, miles, actuations, etc. | ||

- | This distribution requires the knowledge of only one parameter, <math>\lambda </math>, for its application. Some of the characteristics of the 1-parameter exponential distribution are [[Appendix: | + | This distribution requires the knowledge of only one parameter, <math>\lambda \,\!</math>, for its application. Some of the characteristics of the 1-parameter exponential distribution are discussed in Kececioglu [[Appendix:_Life_Data_Analysis_References| [19]]]: |

- | [19]]]: | + | |

*The location parameter, <math>\gamma \,\!</math>, is zero. | *The location parameter, <math>\gamma \,\!</math>, is zero. | ||

- | *The scale parameter is <math>\tfrac{1}{\lambda }=m</math>. | + | *The scale parameter is <math>\tfrac{1}{\lambda }=m\,\!</math>. |

- | *As <math>\lambda </math> is decreased in value, the distribution is stretched out to the right, and as <math>\lambda </math> is increased, the distribution is pushed toward the origin. | + | *As <math>\lambda \,\!</math> is decreased in value, the distribution is stretched out to the right, and as <math>\lambda \,\!</math> is increased, the distribution is pushed toward the origin. |

*This distribution has no shape parameter as it has only one shape, (i.e., the exponential, and the only parameter it has is the failure rate, <math>\lambda \,\!</math>). | *This distribution has no shape parameter as it has only one shape, (i.e., the exponential, and the only parameter it has is the failure rate, <math>\lambda \,\!</math>). | ||

- | *The distribution starts at <math>t=0\,\!</math> at the level of <math>f(t=0)=\lambda \,\!</math> and decreases thereafter exponentially and monotonically as <math>t</math> increases, and is convex. | + | *The distribution starts at <math>t=0\,\!</math> at the level of <math>f(t=0)=\lambda \,\!</math> and decreases thereafter exponentially and monotonically as <math>t\,\!</math> increases, and is convex. |

- | *As <math>t\to \infty </math> , <math>f(t)\to 0</math>. | + | *As <math>t\to \infty </math> , <math>f(t)\to 0\,\!</math>. |

*The ''pdf'' can be thought of as a special case of the Weibull ''pdf'' with <math>\gamma =0\,\!</math> and <math>\beta =1\,\!</math>. | *The ''pdf'' can be thought of as a special case of the Weibull ''pdf'' with <math>\gamma =0\,\!</math> and <math>\beta =1\,\!</math>. | ||

## Revision as of 17:09, 5 September 2012

The exponential distribution is a commonly used distribution in reliability engineering. Mathematically, it is a fairly simple distribution, which many times leads to its use in inappropriate situations. It is, in fact, a special case of the Weibull distribution where β = 1. The exponential distribution is used to model the behavior of units that have a constant failure rate (or units that do not degrade with time or wear out).

## Exponential Probability Density Function

### The 2-Parameter Exponential Distribution

The 2-parameter exponential *pdf* is given by:

- The scale parameter is .
- The exponential
*pdf*has no shape parameter, as it has only one shape. - As , .

### The 1-Parameter Exponential Distribution

The 1-parameter exponential *pdf* is obtained by setting , and is given by:

where:

- = constant rate, in failures per unit of measurement, (e.g., failures per hour, per cycle, etc.)

- = mean time between failures, or to failure
- = operating time, life, or age, in hours, cycles, miles, actuations, etc.

- The location parameter, , is zero.
- The scale parameter is .
- As , .
- The
*pdf*can be thought of as a special case of the Weibull*pdf*with and .

## Exponential Distribution Functions

### The Mean or MTTF

The mean, or mean time to failure (MTTF) is given by:

### The Median

The median, is:

### The Mode

The mode, is:

### The Standard Deviation

The standard deviation, , is:

### The Exponential Reliability Function

The equation for the 2-parameter exponential cumulative density function, or *cdf*, is given by:

*cdf*, the reliability function of the 2-parameter exponential distribution is given by:

The 1-parameter exponential reliability function is given by:

### The Exponential Conditional Reliability Function

*memoryless property*.

### The Exponential Reliable Life Function

or:

### The Exponential Failure Rate Function

The exponential failure rate function is:

## Characteristics of the Exponential Distribution

### The Effect of lambda and gamma on the Exponential *pdf*

- The exponential
*pdf*has no shape parameter, as it has only one shape. - The exponential
*pdf*is always convex and is stretched to the right as decreases in value. - The value of the
*pdf*function is always equal to the value of at (or ). - The scale parameter is .
- As , .

- The exponential

### The Effect of lambda and gamma on the Exponential Reliability Function

- As , .

### The Effect of lambda and gamma on the Failure Rate Function

- The 1-parameter exponential failure rate function is constant and starts at .

## Estimation of the Exponential Parameters

### Probability Plotting

*cdf*, the exponential probability plot is the only one with a negative slope. This is because the y-axis of the exponential probability plotting paper represents the reliability, whereas the y-axis for most of the other life distributions represents the unreliability.

*cdf*, which is necessary to construct the exponential probability plotting paper. For the two-parameter exponential distribution the cumulative density function is given by:

Taking the natural logarithm of both sides of the above equation yields:

or:

Now, let:

and:

which results in the linear equation of:

Note that with the exponential probability plotting paper, the y-axis scale is logarithmic and the x-axis scale is linear. This means that the zero value is present only on the x-axis. For *t* = 0, *R* = 1 and *F*(*t*) = 0. So if we were to use *F*(*t*) for the y-axis, we would have to plot the point (0,0). However, since the y-axis is logarithmic, there is no place to plot this on the exponential paper. Also, the failure rate, λ, is the negative of the slope of the line, but there is an easier way to determine the value of λ from the probability plot, as will be illustrated in the following example.

#### Plotting Example

**1-Parameter Exponential Probability Plot Example**

### Rank Regression on Y

and:

In our case, the equations for *y*_{i} and *x*_{i} are:

and:

and the *F*(*t*_{i}) is estimated from the median ranks. Once and are obtained, then and can easily be obtained from above equations.
For the one-parameter exponential, equations for estimating *a* and *b* become:

**The Correlation Coefficient**

The estimator of is the sample correlation coefficient, , given by:

#### RRY Example

**2-Parameter Exponential RRY Example**

14 units were being reliability tested and the following life test data were obtained. Assuming that the data follow a 2-parameter exponential distribution, estimate the parameters and determine the correlation coefficient, , using rank regression on Y (RRY).

Life Test Data | |
---|---|

Data point index | Time-to-failure |

1 | 5 |

2 | 10 |

3 | 15 |

4 | 20 |

5 | 25 |

6 | 30 |

7 | 35 |

8 | 40 |

9 | 50 |

10 | 60 |

11 | 70 |

12 | 80 |

13 | 90 |

14 | 100 |

**Solution**

Construct the following table, as shown next.

The median rank values ( ) can be found in rank tables or they can be estimated using the **Quick Statistical Reference** in Weibull++.
Given the values in the table above, calculate and :

or:

and:

or:

Therefore:

and:

or:

Then:

The correlation coefficient can be estimated using equation for calculating the correlation coefficient:

This example can be repeated using Weibull++, choosing 2-parameter exponential and rank regression on Y (RRY), as shown next.

The estimated parameters and the correlation coefficient using Weibull++ were found to be:

**Reset if location parameter > T1 on Exponential** option on the Calculations page of the Application Setup window.

The probability plot can be obtained simply by clicking the **Plot** icon.

### Rank Regression on X

Again the first task is to bring our exponential *cdf* function into a linear form. This step is exactly the same as in regression on Y analysis. The deviation from the previous analysis begins on the least squares fit step, since in this case we treat *x* as the dependent variable and *y* as the independent variable. The best-fitting straight line to the data, for regression on X (see Parameter Estimation), is the straight line:

The corresponding equations for and are:

and:

where:

and:

The values of *F*(*t*_{i}) are estimated from the median ranks. Once and are obtained, solve for the unknown *y* value, which corresponds to:

Solving for the parameters from above equations we get:

and:

For the one-parameter exponential case, equations for estimating a and b become:

The correlation coefficient is evaluated as before.

#### RRX Example

**2-Parameter Exponential RRX Example**

** Solution**

or:

and:

or:

Therefore:

and:

The correlation coefficient is found to be:

Note that the equation for regression on Y is not necessarily the same as that for the regression on X. The only time when the two regression methods yield identical results is when the data lie perfectly on a line. If this were the case, the correlation coefficient would be − 1. The negative value of the correlation coefficient is due to the fact that the slope of the exponential probability plot is negative.

The probability plot can be obtained simply by clicking the **Plot** icon.

### Maximum Likelihood Estimation

#### MLE Example

**MLE for the Exponential Distribution**

**Solution**

In this example, we have complete data only. The partial derivative of the log-likelihood function, Λ, is given by:

Complete descriptions of the partial derivatives can be found in Appendix D. Recall that when using the MLE method for the exponential distribution, the value of γ is equal to that of the first failure time. The first failure occurred at 5 hours, thus γ = 5 hours. Substituting the values for *T* and γ we get:

or:

Using Weibull++:

The probability plot is:

## Confidence Bounds

### Fisher Matrix Bounds

#### Bounds on the Parameters

For the failure rate the upper () and lower () bounds are estimated by Nelson [30]:

where is defined by:

If is the confidence level, then for the two-sided bounds, and for the one-sided bounds.

The variance of is estimated from the Fisher matrix, as follows:

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

#### Bounds on Reliability

The reliability of the two-parameter exponential distribution is:

The corresponding confidence bounds are estimated from:

These equations hold true for the 1-parameter exponential distribution, with γ = 0.

#### Bounds on Time

The corresponding confidence bounds are estimated from:

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

### Likelihood Ratio Confidence Bounds

#### Bounds on Parameters

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

This equation can be rewritten as:

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

where the *t*_{i} values represent the original time-to-failure data. For a given value of α, values for λ can be found which represent the maximum and minimum values that satisfy the above likelihood ratio equation. These represent the confidence bounds for the parameters at a confidence level δ, where α = δ for two-sided bounds and α = 2δ − 1 for one-sided.

##### Example: LR Bounds for Lambda

**Solution**

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

where *x*_{i} are the original time-to-failure data points. We can now rearrange the likelihood ratio equation 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. These values represent the two-sided confidence limits of the parameter estimate . For our problem, the confidence limits are:

#### Bounds on Time and Reliability

This can be rearranged to the form:

This equation can now be substituted into the likelihood ratio equation to produce a likelihood equation in terms of *t* and *R*:

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 the equation for the mean and the Bayes's rule equation for single parametera 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, the likelihood ratio function can be solved for the values of interest.

##### Example: LR Bounds on Time

For the data given above for the LR Bounds on Lambda example (five failures at 20, 40, 60, 100 and 150 hours), 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 .

**Solution**

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 the likelihood ratio bound equation. 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:

##### Example: LR Bounds on Reliability

Again using the data given above for the LR Bounds on Lambda example (five failures at 20, 40, 60, 100 and 150 hours), 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 .

**Solution**

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 = 0.85 into the likelihood ratio bound equation. 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% two-sided confidence limits of the reliability estimate . For our problem, the confidence limits are:

### Bayesian Confidence Bounds

#### Bounds on Parameters

From Confidence Bounds, 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, can be rewritten as:

The one-sided upper bound of λ is:

The one-sided lower bound of λ is:

The two-sided bounds of λ are:

#### Bounds on Time (Type 1)

The reliable life equation is:

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

The above equation can be rewritten in terms of λ as:

From the above posterior distribuiton equation, we have:

The above equation is solved w.r.t. *t*_{U}. The same method is applied for one-sided lower and two-sided bounds on time.

#### Bounds on Reliability (Type 2)

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

The above equaation can be rewritten in terms of λ as:

From the equation for posterior distribution we have:

The above equation is solved w.r.t. *R*_{U}. The same method can be used to calculate one-sided lower and two sided bounds on reliability.

## Exponential Distribution Examples

### Grouped Data

20 units were reliability tested with the following results:

Table - Life Test Data
| |

Number of Units in Group | Time-to-Failure |
---|---|

7 | 100 |

5 | 200 |

3 | 300 |

2 | 400 |

1 | 500 |

2 | 600 |

1. Assuming a 2-parameter exponential distribution, estimate the parameters by hand using the MLE analysis method.

2. Repeat the above using Weibull++. (Enter the data as grouped data to duplicate the results.)

3. Show the Probability plot for the analysis results.

4. Show the Reliability vs. Time plot for the results.

5. Show the *pdf* plot for the results.

6. Show the Failure Rate vs. Time plot for the results.

**Solution**

2. Enter the data in a Weibull++ standard folio and calculate it as shown next.

3. On the Plot page of the folio, the exponential Probability plot will appear as shown next.

4. View the Reliability vs. Time plot.

5. View the *pdf* plot.

6. View the Failure Rate vs. Time plot.

Note that, as described at the beginning of this chapter, the failure rate for the exponential distribution is constant. Also note that the Failure Rate vs. Time plot does show values for times before the location parameter, , at 100 hours.

7. In the case of grouped data, one must be cautious when estimating the parameters using a rank regression method. This is because the median rank values are determined from the total number of failures observed by time where indicates the group number. In this example, the total number of groups is and the total number of units is . Thus, the median rank values will be estimated for 20 units and for the total failed units () up to the group, for the rank value. The median ranks values can be found from rank tables or they can be estimated using ReliaSoft's Quick Statistical Reference tool.

For example, the median rank value of the fourth group will be the rank out of a sample size of twenty units (or 81.945%).

The following table is then constructed.

Given the values in the table above, calculate and :

or:

and:

or:

Therefore:

and:

or:

Then:

Using Weibull++, the estimated parameters are:

### Using Auto Batch Run

Table - Leukemia Treatment Results
| |||

Time (weeks) | Number of Patients | Treament | Comments |
---|---|---|---|

1 | 2 | placebo | |

2 | 2 | placebo | |

3 | 1 | placebo | |

4 | 2 | placebo | |

5 | 2 | placebo | |

6 | 4 | 6MP | 3 patients completed |

7 | 1 | 6MP | |

8 | 4 | placebo | |

9 | 1 | 6MP | Not completed |

10 | 2 | 6MP | 1 patient completed |

11 | 2 | placebo | |

11 | 1 | 6MP | Not completed |

12 | 2 | placebo | |

13 | 1 | 6MP | |

15 | 1 | placebo | |

16 | 1 | 6MP | |

17 | 1 | placebo | |

17 | 1 | 6MP | Not completed |

19 | 1 | 6MP | Not completed |

20 | 1 | 6MP | Not completed |

22 | 1 | placebo | |

22 | 1 | 6MP | |

23 | 1 | placebo | |

23 | 1 | 6MP | |

25 | 1 | 6MP | Not completed |

32 | 2 | 6MP | Not completed |

34 | 1 | 6MP | Not completed |

35 | 1 | 6MP | Not completed |

Create a new Weibull++ standard folio that's configured for grouped times-to-failure data with suspensions. In the first column, enter the number of patients. Whenever there are uncompleted tests, enter the number of patients who completed the test separately from the number of patients who did not (e.g., if 4 patients had symptoms return after 6 weeks and only 3 of them completed the test, then enter 1 in one row and 3 in another). In the second column enter F if the patients completed the test and S if they didn't. In the third column enter the time, and in the fourth column (Subset ID) specify whether the 6MP drug or a placebo was used.

Next, open the Batch Auto Run utility and select to separate the 6MP drug from the placebo, as shown next.

The software will create two data sheets, one for each subset ID, as shown next.

Calculate both data sheets using the 2-parameter exponential distribution and the MLE analysis method, then insert an additional plot and select to show the analysis results for both data sheets on that plot, which will appear as shown next.