Mean and Variance of the Weibull Distribution

For the standard two-parameter Weibull distribution, both the mean and variance can be written using the gamma function. The key is to distinguish the shape parameter from the scale parameter, since Weibull notation varies between sources.

The formulas below use shape $k>0$ and scale $\lambda>0$ . If your source uses a different parameterization, check its density before substituting values.

Standard two-parameter form

A random variable $X$ has a Weibull distribution with shape $k$ and scale $\lambda$ if its density is

f(x)=\frac{k}{\lambda}\left(\frac{x}{\lambda}\right)^{k-1}\exp\left[-\left(\frac{x}{\lambda}\right)^k\right], \qquad x\ge 0.

For $x\ge 0$ , the cumulative distribution function is $F(x)=1-\exp[-(x/\lambda)^k]$ ; for $x<0$ , $F(x)=0$ . The scale $\lambda$ sets the units and general size of the observations; the shape $k$ controls the distribution’s form.

With this convention, the mean and variance are

\mathbb{E}[X]=\lambda\,\Gamma\left(1+\frac{1}{k}\right)

and

\operatorname{Var}(X)=\lambda^2\left[\Gamma\left(1+\frac{2}{k}\right)-\Gamma\left(1+\frac{1}{k}\right)^2\right].

Here $\Gamma$ is the gamma function, defined for $z>0$ by $\Gamma(z)=\int_0^\infty u^{z-1}e^{-u}\,du$ . It is not the same thing as saying that $X$ itself has a gamma distribution.

Why these formulas hold

The moment calculation follows from a change of variable in the expectation integral. Set $u=(x/\lambda)^k$ . Then $du=\frac{k}{\lambda}(x/\lambda)^{k-1}dx$ , so the density times $dx$ becomes $e^{-u}du$ . For any real $r>-k$ ,

\begin{aligned} \mathbb{E}[X^r] &=\int_0^\infty x^r f(x)\,dx\\ &=\lambda^r\int_0^\infty u^{r/k}e^{-u}\,du\\ &=\lambda^r\Gamma\left(1+\frac{r}{k}\right). \end{aligned}

The restriction $r>-k$ ensures that the integral is finite near zero. Setting $r=1$ gives the mean, and setting $r=2$ gives the second raw moment. Then use $\operatorname{Var}(X)=\mathbb{E}[X^2]-\mathbb{E}[X]^2$ .

Worked example: shape 2, scale 3

Suppose $X$ has shape $k=2$ and scale $\lambda=3$ . Since $\Gamma(3/2)=\sqrt{\pi}/2$ and $\Gamma(2)=1$ ,

\mathbb{E}[X]=3\Gamma\left(\frac{3}{2}\right)=\frac{3\sqrt{\pi}}{2}\approx 2.659.

The second moment is

\mathbb{E}[X^2]=3^2\Gamma(2)=9.

Therefore,

\begin{aligned} \operatorname{Var}(X) &=9-\left(\frac{3\sqrt{\pi}}{2}\right)^2\\ &=9\left(1-\frac{\pi}{4}\right)\\ &\approx 1.931. \end{aligned}

The standard deviation is the square root of the variance, approximately $1.390$ . Notice that the mean has the same units as the scale, while the variance has squared units.

Parameterization and location variations

Some references write the Weibull density with an inverse scale parameter, such as $\exp[-(\alpha x)^k]$ . In that convention, $\alpha=1/\lambda$ , so the mean is $\alpha^{-1}\Gamma(1+1/k)$ and the variance is $\alpha^{-2}[\Gamma(1+2/k)-\Gamma(1+1/k)^2]$ . Identify the exponent in the density before using a formula; treating an inverse scale as a scale reverses the parameter’s role.

A three-parameter Weibull adds a location or threshold $\mu$ , so that $X=\mu+Y$ where $Y$ is a two-parameter Weibull variable. Its mean is $\mu+\lambda\Gamma(1+1/k)$ , and its variance is the same as that of $Y$ , because adding a constant shifts every value without changing its spread.

As a quick check on the formulas, setting $k=1$ gives the exponential distribution with scale $\lambda$ : the mean becomes $\lambda$ and the variance becomes $\lambda^2$ .

Common mistakes

Swapping shape and scale: The shape is the exponent $k$ ; the scale is $\lambda$ in $(x/\lambda)^k$ .
Using the second moment as the variance: $\mathbb{E}[X^2]$ is not the variance. Subtract the square of the mean.
Reporting standard deviation as variance: Standard deviation is $\sqrt{\operatorname{Var}(X)}$ and has the original measurement units.
Ignoring a location parameter: A threshold shifts the mean but does not alter the variance.