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I need to sample values from a Weibull distribution whose mean and variance are provided (respectively 62 and 4275). I am running a Matlab code, therefore if I want to use wblrnd(shape,scale) I need to estimate these two parameters. The problem is that,according to wikipedia, mean and variance are related to shape and scale parameters via a gamma function, and this makes the calculation non-trivial. Is there a simple way to sample values in Matlab via mean and variance, or to easily move from these two parameters to the shape and scale parameters?

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1 Answer 1

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This can be accomplished with monovariate root finding. First solve for the shape parameter, $\lambda$, in terms of the scale parameter, $k$, using the expression for the mean, $\mu$, of the standard Weibull distribution:

$$\mu = \lambda \Gamma\left(1+\frac{1}{k}\right) \rightarrow \lambda = \frac{\mu}{\Gamma\left(1+\frac{1}{k}\right)}$$

Then plug this in to the expression for the variance, $\sigma^2$, to eliminate $\lambda$ and simplify:

$$\sigma^2 = \lambda^2 \left(\Gamma\left(1+\frac{2}{k}\right) - \left(\Gamma\left(1+\frac{1}{k}\right)\right)^2\right) \rightarrow \frac{\sigma^2}{\mu^2} - \frac{\Gamma\left(1+\frac{2}{k}\right)}{\left(\Gamma\left(1+\frac{1}{k}\right)\right)^2} + 1 = 0$$

This appears to be a monotonically increasing function for $k \in (0,+\infty)$ and the specified values of $\mu$ and $\sigma^2$. The root of this function can be found in Matlab by using fzero:

mu = 62;
sig2 = 4275;
f = @(k)sig2/mu^2-gamma(1+2./k)./gamma(1+1./k).^2+1;
k0 = 1;               % Initial guess
k = fzero(f,k0)       % Solve for k
lam = mu/gamma(1+1/k) % Substitue to find lambda

which yields

k =

   0.948647322786540


lam =

  60.540561349132474

You can then verify that this works by sampling with wblrnd:

rng(1); % Set seed to make repeatable
n = 1e7;
r = wblrnd(lam,k,[n 1]);
mu_r = mean(r)
sig2_r = var(r)
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