# Finding when mean equals variance for Weibull distribution

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!

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 $$k$$:

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

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