Weibull distribution parameterization

Question

I have the following Weibull distribution:

$f(x;\lambda,\beta) = (\lambda\beta)x^{(\beta-1)}e^{(-\lambda x^b)} $ where $\lambda$ is scale parameter and $\beta$ is shape parameter.

I have an alternative parameterization for Weibull in r link:

$f(x;a,b) = \frac{a}{b} (\frac{x}{b})^{(a-1)} e^{(- (x/b)^a)} $ where $a$ is shape parameter and $b$ is scale parameter.

How do you equate the shape and scale parameters from above distributions. Below is my solution. Clearly, I'm wrong.

> l=0.5
> b=2.5
> 
> dwei <- function(x,l,b){
+   l*b*x^(b-1)*exp(-l*(x^b))
+ }
> 
> dwei(x=2,l,b)
[1] 0.2089704
> dweibull(x=2, shape=b, scale = 1/l)
[1] 0.4598493

Answer 1

I think your order of operations is wrong. As a simplified example, $$\left(\frac{x}{2}\right)^2 = \frac{x^2}{4},$$ so in your pdf, you need $$(\lambda x)^{\beta - 1}=\lambda ^{\beta -1}x^{\beta - 1}$$ in the first product, and likewise inside the exponential function.

Putting it all together: $$ f(x;\lambda,\beta)= \lambda \beta (\lambda x)^{\beta - 1}\exp\left(-(\lambda x)^\beta\right) $$ and implemented in R to double-check:

dwei <- function(x,l,b){
   l * b * (l * x) ^ (b - 1) * exp( - (l * x) ^ b)
 }
dwei(x=2, l=0.5, b=2.5)
[1] 0.4598493

Which matches the R function output.