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 $f(x):$
\begin{equation}
 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}
\end{equation}
If $X$ is multiplied by a constant $c$ what is the distribution of $cX$? Is it still Weibull distributed?
 A: Look at the cumulative distribution of the Weibull:
\begin{equation}
 F(x;\lambda,k) =
 \begin{cases}
  1 - e^{-(x/\lambda)^{k}} & x\geq0 ,\\
  0 & x<0,
 \end{cases}
\end{equation}
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.
A: 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.
\begin{align*}
P(Y \leq y) & = P(cX \leq y)\\
& = P(X \leq y/c)\\
& = F_X\left( \dfrac{y}{c} \right)\\
f_Y(y) & = \dfrac{1}{c} f_X \left(\dfrac{y}{c} \right)\\
& = \dfrac{1}{c} \dfrac{k}{\lambda} \left(\dfrac{y}{c \lambda} \right)^{k-1} \exp \left(- \dfrac{y}{c \lambda} \right)^{k}.
\end{align*}
As, long as $c > 0$, $c \lambda > 0$, and so that is the pdf of a Weibull$(c\lambda, k)$.
