For the 2-parameter Weibull distribution,

The mean $E(X)$ is given as $E(X) = \lambda \Gamma(1+\frac{1}{k})$

The variance is given as $var(X) = \lambda^2 [\Gamma(1+\frac{2}{k})-(\Gamma(1+\frac{2}{k}))^2]$

where $k,\lambda>0$ are the shape and scale parameters respectively.

Solving for $k$, I would like to find analytically when the mean equals the variance for this distribution. But I don't have much experience working with Gamma functions. The only solution I have found so far is $k = \lambda = 1$.

Any help would be appreciated!


1 Answer 1


You want the combinations of $k$ and $\lambda$ for which the Weibull mean and variance are equal:

$$\lambda \Gamma \left(1+\frac{1}{k}\right)=\lambda ^2 \left(\Gamma \left(1+\frac{2}{k}\right)-\Gamma \left(1+\frac{1}{k}\right)^2\right)$$

Just solve for $\lambda$ in terms of $k$:

$$\lambda=\frac{\Gamma \left(1+\frac{1}{k}\right)}{\Gamma \left(1+\frac{2}{k}\right)-\Gamma \left(1+\frac{1}{k}\right)^2}$$

  • $\begingroup$ Thanks for your comment! Apologies. What I want is your equation, but now solving for $k$! $\endgroup$ Oct 22, 2022 at 12:54
  • $\begingroup$ You can just set $k$ to whatever you want and the get the corresponding value of $\lambda$. $\endgroup$
    – JimB
    Oct 22, 2022 at 14:26
  • $\begingroup$ Gotcha! Though I am wondering if there a clean formula for the reverse where I am setting $\lambda$ to be whatever and getting out the corresponding $k$ $\endgroup$ Oct 22, 2022 at 17:57
  • 2
    $\begingroup$ I would bet that there's no clean or even horrendously complicated formula. However, you might be able to find a good approximation . $\endgroup$
    – JimB
    Oct 22, 2022 at 20:54

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