Deriving the Weibull Distribution using the Exponential

Question

I need to derive the Weibull Distribution using the Exponential, I can see that the CDF's of the two are very similar.

The CDF of the exponential is: $$ F(x) = 1-e^{(-x/b)^a} $$

And the CDF of the Weibull is: $$ F(x) = 1-e^{(-x.b)} $$

Answer 1

If we define the cumulative distribution function of the Weibull as:

$$F_{W}(x)=1-\exp\bigg(-\bigg(\frac{x}{\lambda}\bigg)^{k}\bigg)$$

and the cumulative distribution function of the standard exponential as:

$$F_{E}(x)=1-\exp(-x)$$

If we assume $X$ is a standard exponential random variable

$$X\sim \text{Exp}(1)$$

Then, by applying the transform

$$W=\lambda\, X^{1/k}$$

We can derive the relationship as:

$$\begin{align} \text{Pr}(W\le x)&=\text{Pr}(\lambda \,X^{1/k}\le x)\\ &=\text{Pr}(X\le (x/\lambda)^k)\\ &=1-\exp\bigg(-\bigg(\frac{x}{\lambda}\bigg)^{k}\bigg) \end{align}$$