# Is the multiple of a Weibull distributed variable also Weibull distributed?

A Weibull distributed variable $X \sim \textrm{Weibull}(\lambda, k)$ has probability density function $f(x):$

$$f(x;\lambda,k) = \begin{cases} \frac{k}{\lambda}\left(\frac{x}{\lambda}\right)^{k-1}e^{-(x/\lambda)^{k}} & x\geq0 ,\\ 0 & x<0, \end{cases}$$

If $X$ is multiplied by a constant $c$ what is the distribution of $cX$? Is it still Weibull distributed?

• Whenever a parameter, such as $\lambda$, appears consistently in a formula for the PDF either dividing or multiplying the variable, it is a scale (or, respectively, a rate) parameter. That immediately implies any change of units of measurement of the variable will have the same distributional shape. The only subtlety is that you really ought to think in terms of the probability element $f(x)dx$: that missing $dx$ also needs its factor of $1/\lambda$ or $\lambda$, respectively, which you can observe in the normalizing constant for $f$.
– whuber
Commented Jul 21, 2016 at 13:59

The easiest way to find out is to find the distribution of $cX$. Let $Y = cX$. You can find the distribution of $Y$ using the change of formula formula, however I personally don't like to use the formula. So this is what I do. I first find the CDF of $Y$, and then differentiate the CDF.
As, long as $c > 0$, $c \lambda > 0$, and so that is the pdf of a Weibull$(c\lambda, k)$.
Look at the cumulative distribution of the Weibull: $$F(x;\lambda,k) = \begin{cases} 1 - e^{-(x/\lambda)^{k}} & x\geq0 ,\\ 0 & x<0, \end{cases}$$ Obviously, a positive constant multiplier $c$ on a Weibull distributed random variable can be absorbed into (by changing) $\lambda$. So the answer is yes, presuming $c$ is positive.