Introduction to Weibull Distribution

The Weibull distribution is one of the most widely used continuous probability distributions in reliability engineering, quality control, and materials science. Named after Swedish mathematician Waloddi Weibull, this distribution provides a flexible framework for modeling failure times, lifetimes, and various positive continuous phenomena. Its ability to model increasing, decreasing, or constant failure rates makes it invaluable in engineering applications.

Definition of Weibull Distribution

A continuous random variable $X$ is said to have a Weibull distribution with three parameters if $Y = \left(\frac{X-\mu}{\beta}\right)^\alpha$ has the exponential distribution with PDF $f(y) = e^{-y}, y > 0$.

The probability density function of the three-parameter Weibull distribution is:

$$ \begin{equation*} f(x;\alpha, \beta, \mu)=\left{ \begin{array}{ll} \frac{\alpha}{\beta} \left(\frac{x-\mu}{\beta}\right)^{\alpha-1}e^{-\left(\frac{x-\mu}{\beta}\right)^\alpha}, & \hbox{$x>\mu$, $\alpha, \beta>0$;} \ 0, & \hbox{Otherwise.} \end{array} \right. \end{equation*} $$

Parameters

  • $\alpha$ (Shape Parameter): Controls the shape of the distribution; determines the failure rate behavior ($\alpha > 0$)
  • $\beta$ (Scale Parameter): Controls the scale/spread; determines where the bulk of the distribution lies ($\beta > 0$)
  • $\mu$ (Location Parameter): Shifts the distribution along the x-axis; represents the minimum value ($\mu \geq 0$)

Notation: $X \sim W(\mu, \alpha, \beta)$ or $X \sim \text{Weibull}(\alpha, \beta, \mu)$

Two-Parameter Weibull Distribution

The most commonly used form sets $\mu = 0$. The PDF becomes:

$$ \begin{equation*} f(x;\alpha, \beta)=\left{ \begin{array}{ll} \frac{\alpha}{\beta} \left(\frac{x}{\beta}\right)^{\alpha-1}e^{-\left(\frac{x}{\beta}\right)^\alpha}, & \hbox{$x>0$, $\alpha, \beta>0$;} \ 0, & \hbox{Otherwise.} \end{array} \right. \end{equation*} $$

Notation: $X \sim W(\alpha, \beta)$

Special Case: Exponential Distribution

When $\alpha = 1$, the two-parameter Weibull distribution reduces to the exponential distribution with parameter $1/\beta$:

$$f(x) = \frac{1}{\beta}e^{-x/\beta}, \quad x > 0, \beta > 0$$

Standard Weibull Distribution

If we let $\mu = 0$ and $\beta = 1$, we get the standard Weibull distribution:

$$ \begin{equation*} f(x;\alpha)=\left{ \begin{array}{ll} \alpha x^{\alpha-1}e^{-x^\alpha}, & \hbox{$x>0$;} \ 0, & \hbox{Otherwise.} \end{array} \right. \end{equation*} $$

Graph of Weibull Distribution

The shape of the Weibull distribution varies dramatically with the shape parameter $\alpha$:

  • $\alpha < 1$: Decreasing hazard rate (infant mortality)
  • $\alpha = 1$: Constant hazard rate (exponential, no aging)
  • $\alpha > 1$: Increasing hazard rate (wear-out failures)
  • $\alpha \approx 3.6$: Approximates the normal distribution

Probability Density Function (PDF)

For the two-parameter Weibull distribution:

$$f(x) = \frac{\alpha}{\beta}\left(\frac{x}{\beta}\right)^{\alpha-1}e^{-\left(\frac{x}{\beta}\right)^\alpha}$$

Properties of the PDF

  • Support: $[0, \infty)$ for two-parameter form
  • Shape Control: Fully determined by shape parameter $\alpha$
  • Scale Effect: Larger $\beta$ stretches the distribution horizontally
  • Flexibility: Can represent decreasing (infant mortality), constant, or increasing (wear-out) failure rates

Cumulative Distribution Function (CDF)

The cumulative distribution function of the Weibull distribution is:

$$ \begin{equation*} F(x) = 1 - e^{-\left(\frac{x-\mu}{\beta}\right)^\alpha} \end{equation*} $$

For the two-parameter case ($\mu = 0$):

$$F(x) = 1 - e^{-\left(\frac{x}{\beta}\right)^\alpha}$$

Survival Function

The survival function (probability of surviving beyond time $x$) is:

$$S(x) = 1 - F(x) = e^{-\left(\frac{x}{\beta}\right)^\alpha}$$

Proof of CDF

$$ \begin{eqnarray*} F(x) &=& P(X \leq x) \ &=& \int_\mu^x f(t); dt\ &=& \int_\mu^x \frac{\alpha}{\beta}\left(\frac{t-\mu}{\beta}\right)^{\alpha-1}e^{-\left(\frac{t-\mu}{\beta}\right)^\alpha}; dt \end{eqnarray*} $$

Let $u = \left(\frac{x-\mu}{\beta}\right)^\alpha$. Then $du = \frac{\alpha}{\beta}\left(\frac{t-\mu}{\beta}\right)^{\alpha-1}dt$.

$$F(x) = \int_0^{\left(\frac{x-\mu}{\beta}\right)^\alpha} e^{-u}; du = 1 - e^{-\left(\frac{x-\mu}{\beta}\right)^\alpha}$$

Key Properties of Weibull Distribution

Mean of Two-Parameter Weibull Distribution

The mean is:

$$E(X) = \beta \Gamma\left(\frac{1}{\alpha} + 1\right)$$

where $\Gamma$ is the gamma function.

Variance of Two-Parameter Weibull Distribution

The variance is:

$$V(X) = \beta^2\left[\Gamma\left(\frac{2}{\alpha} + 1\right) - \left(\Gamma\left(\frac{1}{\alpha} + 1\right)\right)^2\right]$$

Standard Deviation

$$\sigma = \sqrt{V(X)} = \beta\sqrt{\Gamma\left(\frac{2}{\alpha} + 1\right) - \left(\Gamma\left(\frac{1}{\alpha} + 1\right)\right)^2}$$

Additional Properties

Property Formula
Median $M = \mu + \beta(\ln 2)^{1/\alpha}$
Mode $\mu + \beta\left(\frac{\alpha-1}{\alpha}\right)^{1/\alpha}$ (for $\alpha > 1$)
Coefficient of Variation Depends on $\alpha$
Skewness Decreases as $\alpha$ increases
Kurtosis Approaches 3 as $\alpha$ increases

Raw Moments

The $r^{th}$ raw moment of the two-parameter Weibull distribution is:

$$\mu_r^\prime = E(X^r) = \beta^r \Gamma\left(\frac{r}{\alpha} + 1\right)$$

Proof

Let $Y = \left(\frac{X}{\beta}\right)^\alpha$. Then $X = \beta Y^{1/\alpha}$ and $dx = \frac{\beta}{\alpha}y^{(1/\alpha)-1}dy$.

$$ \begin{eqnarray*} \mu_r^\prime &=& \int_0^\infty x^r f(x); dx\ &=& \int_0^\infty x^r \frac{\alpha}{\beta}\left(\frac{x}{\beta}\right)^{\alpha-1}e^{-\left(\frac{x}{\beta}\right)^\alpha}; dx \end{eqnarray*} $$

Substituting $y = \left(\frac{x}{\beta}\right)^\alpha$:

$$\mu_r^\prime = \beta^r \int_0^\infty y^{\frac{r}{\alpha}}e^{-y}; dy = \beta^r \Gamma\left(\frac{r}{\alpha} + 1\right)$$

Failure Rate (Hazard Function)

The failure rate (or hazard function) is:

$$h(x) = \frac{f(x)}{S(x)} = \frac{\alpha}{\beta}\left(\frac{x-\mu}{\beta}\right)^{\alpha-1}$$

Interpretation

  • $\alpha < 1$: Decreasing failure rate (infant mortality phase)
  • $\alpha = 1$: Constant failure rate (random failures, no aging)
  • $\alpha > 1$: Increasing failure rate (wear-out phase)
  • $\alpha \approx 3.4-3.6$: Approximately constant failure rate (resembles normal distribution)

Median of Weibull Distribution

The median of the three-parameter Weibull distribution is:

$$M = \mu + \beta(\ln 2)^{1/\alpha}$$

Proof

From $F(M) = 0.5$:

$$1 - e^{-\left(\frac{M-\mu}{\beta}\right)^\alpha} = 0.5$$

$$e^{-\left(\frac{M-\mu}{\beta}\right)^\alpha} = 0.5$$

$$\left(\frac{M-\mu}{\beta}\right)^\alpha = \ln 2$$

$$M = \mu + \beta(\ln 2)^{1/\alpha}$$

Mode of Weibull Distribution

The mode of the three-parameter Weibull distribution is:

$$x_0 = \mu + \beta\left(\frac{\alpha-1}{\alpha}\right)^{1/\alpha}, \quad \text{for } \alpha > 1$$

For $\alpha \leq 1$, the mode is at $x = \mu$.

Properties Summary Table

Property Formula
PDF (2-param) $f(x) = \frac{\alpha}{\beta}\left(\frac{x}{\beta}\right)^{\alpha-1}e^{-(x/\beta)^\alpha}$
CDF (2-param) $F(x) = 1 - e^{-(x/\beta)^\alpha}$
Survival $S(x) = e^{-(x/\beta)^\alpha}$
Support $(0, \infty)$
Mean (2-param) $\beta\Gamma(1/\alpha + 1)$
Variance (2-param) $\beta^2[\Gamma(2/\alpha + 1) - (\Gamma(1/\alpha + 1))^2]$
Median (2-param) $\beta(\ln 2)^{1/\alpha}$
Mode (2-param, $\alpha > 1$) $\beta(((\alpha-1)/\alpha)^{1/\alpha}$
Failure Rate $h(x) = \frac{\alpha}{\beta}(x/\beta)^{\alpha-1}$

Examples with Solutions

Example 1: Vacuum Tube Lifetime

Problem: The lifetime $X$ (in hundreds of hours) of a certain type of vacuum tube has a Weibull distribution with parameters $\alpha = 2$ and $\beta = 3$. Compute:

a. $E(X)$ and $V(X)$ b. $P(X \leq 5)$ c. $P(1.8 \leq X \leq 5)$ d. $P(X \geq 3)$

Solution:

Given: $X \sim W(\alpha = 2, \beta = 3)$

Part (a): Mean and Variance

$$E(X) = \beta\Gamma\left(\frac{1}{\alpha} + 1\right) = 3\Gamma\left(\frac{1}{2} + 1\right) = 3\Gamma(3/2)$$

Since $\Gamma(3/2) = \frac{1}{2}\Gamma(1/2) = \frac{\sqrt{\pi}}{2} \approx 0.8862$:

$$E(X) = 3 \times 0.8862 \approx 2.66 \text{ hundred hours}$$

$$V(X) = \beta^2\left[\Gamma\left(\frac{2}{2} + 1\right) - \left(\Gamma\left(\frac{1}{2} + 1\right)\right)^2\right]$$ $$= 9[\Gamma(2) - (\Gamma(3/2))^2]$$ $$= 9[1 - (0.8862)^2] = 9(1 - 0.785) = 1.935$$

Part (b): $P(X \leq 5)$

$$P(X \leq 5) = F(5) = 1 - e^{-(5/3)^2} = 1 - e^{-(1.667)^2}$$ $$= 1 - e^{-2.778} = 1 - 0.0624 = 0.9376$$

Part (c): $P(1.8 \leq X \leq 5)$

$$P(1.8 \leq X \leq 5) = F(5) - F(1.8)$$ $$= [1 - e^{-(5/3)^2}] - [1 - e^{-(1.8/3)^2}]$$ $$= e^{-(0.6)^2} - e^{-(1.667)^2}$$ $$= e^{-0.36} - e^{-2.778} = 0.6977 - 0.0624 = 0.6353$$

Part (d): $P(X \geq 3)$

$$P(X \geq 3) = 1 - F(3) = 1 - [1 - e^{-(3/3)^2}]$$ $$= e^{-1} = 0.3679$$

Example 2: Magnetic Disk Failure

Problem: The life of a packaged magnetic disk exposed to corrosive gases has a Weibull distribution with $\alpha = 300$ hours and $\beta = 0.5$. Calculate:

a. The probability that a disk lasts at least 600 hours b. The probability that a disk fails before 500 hours

Solution:

Given: $X \sim W(\alpha = 300, \beta = 0.5)$

Note: This appears to have parameters reversed; treating as $\alpha = 0.5$ (shape), $\beta = 300$ (scale).

Part (a): $P(X \geq 600)$

$$P(X \geq 600) = 1 - F(600) = e^{-(600/300)^{0.5}}$$ $$= e^{-(2)^{0.5}} = e^{-1.414} = 0.2431$$

There’s approximately a 24.31% chance the disk lasts at least 600 hours.

Part (b): $P(X < 500)$

$$P(X < 500) = F(500) = 1 - e^{-(500/300)^{0.5}}$$ $$= 1 - e^{-(1.667)^{0.5}} = 1 - e^{-1.291}$$ $$= 1 - 0.275 = 0.725$$

There’s approximately a 72.5% chance the disk fails before 500 hours.

Characterization of Weibull Distribution

If $X_i, i = 1, 2, \ldots, n$ are i.i.d. random variables, then $\min{X_1, X_2, \ldots, X_n}$ has a Weibull distribution if and only if each $X_i$ has a Weibull distribution.

This property is important in reliability theory for systems with components in series.

Moment Generating Function

The moment generating function of the Weibull distribution does not exist in closed form. However, the MGF and all moments can be derived from the raw moments formula:

$$\mu_r^\prime = \beta^r \Gamma\left(\frac{r}{\alpha} + 1\right)$$

When to Use Weibull Distribution

The Weibull distribution is appropriate when:

  1. Failure Time Analysis: Equipment, components, and systems lifetime modeling
  2. Reliability Engineering: Predicting failure rates and system reliability
  3. Materials Testing: Strength and durability of materials
  4. Weather Data: Wind speed, rainfall intensity
  5. Manufacturing: Quality control and product lifetime
  6. Medical: Survival times and disease progression
  7. Engineering: Component wear-out and infant mortality modeling

Applications

Reliability Engineering

  • Equipment failure analysis
  • Predictive maintenance planning
  • System reliability assessment
  • Component design and testing

Quality Control

  • Product lifetime warranties
  • Burn-in testing
  • Failure rate analysis
  • Acceptance sampling

Other Fields

  • Wind speed distribution (renewable energy)
  • Rainfall intensity (hydrology)
  • Fiber strength (materials science)
  • Vehicle failure rates (automotive)

Advantages and Disadvantages

Advantages

  • Flexibility: Can model increasing, constant, or decreasing failure rates
  • Three Parameters: Allows location, scale, and shape adjustment
  • Theoretical Properties: Well-developed mathematical foundation
  • Practical Relevance: Extensive use in reliability analysis
  • Special Cases: Reduces to exponential and Rayleigh distributions

Disadvantages

  • No Closed-Form MGF: Complex for some theoretical derivations
  • Parameter Estimation: Requires specialized methods (Maximum Likelihood, etc.)
  • Interpretation: Shape parameter less intuitive than others

Connection to Other Distributions

  • Exponential ($\alpha = 1$): Special case of Weibull
  • Rayleigh ($\alpha = 2$): Special case of Weibull
  • Normal (Large $\alpha$): Weibull approaches normal as $\alpha \to \infty$
  • Gamma: Related through transformations
  • Log-normal: Both model positive values with flexibility

Conclusion

The Weibull distribution is indispensable in reliability engineering and quality control due to its flexibility in modeling various failure rate patterns. Its three-parameter formulation (shape, scale, and location) provides the versatility needed for practical applications, from infant mortality phase through wear-out phase modeling. The ability to represent different failure behaviors through the shape parameter makes it the preferred choice for reliability analysts and engineers working with time-to-failure data.

Understanding the Weibull distribution;its properties, parameters, and failure rate behavior;is essential for anyone involved in reliability engineering, quality assurance, or system design.