Finding when mean equals variance for Weibull distribution

Question

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!

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 $k$:

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