# a constant $k$ of a Weibull Distribution

I am supposed to show the constant of a Weibull Distribution, $$k$$, is a product of $$\alpha$$ and $$\beta$$.

I know $$\alpha = \frac{1}{\theta}$$ ($$\theta$$ being the probability of a success) for the Distribution becomes an exponential distribution when $$\beta = 1$$. But I'm still not sure how $$k = \alpha\beta$$. How do I show this? also, is there a specific name for this "process?" • Since $f$ is a probability distribution you know that $\int_\mathbb{R} f dx = 1$. Use this to solve for $k$. – Digitalis Apr 15 at 12:23

## 1 Answer

\begin{align} \int_{0}^{+ \infty} kx^{\beta - 1}e^{-\alpha x^{\beta}} dx &= 1 \\ \Rightarrow \int_{0}^{+ \infty} x^{\beta - 1}e^{-\alpha x^{\beta}} dx &= \frac{1}{k} \\ \end{align}

and using the transformation:

$$t = x^{\beta} \Rightarrow dt = \beta x^{\beta - 1}dx$$

we have:

\begin{align} \frac{1}{k} &= \int_{0}^{\infty} \frac{1}{\beta} e^{-\alpha t} dt \\ \Rightarrow \frac{\beta}{k} &= \int_{0}^{\infty} e^{-\alpha t} dt \\ \Rightarrow \frac{\beta}{k} &= \left [ -\frac{1}{\alpha} e^{-\alpha t} \right ]_{0}^{\infty} \\ \Rightarrow \frac{\beta}{k} &= \frac{1}{\alpha} \\ \Rightarrow k &= \alpha \beta \\ \end{align}



• Don't think this kind of processes have a specific name, just note that often to answer this kind of questions, exploiting the fact that probability distribution integrated over their support sum up to $1$, instead of directly going on with math it can be useful to manipulate the quantity within the integral in order to obtain a new distribution. An example of this is the computation of the $k^{th}$ moment of a Gamma distribution. – Nicg Apr 15 at 13:04