In studying the exponential distribution, I have encountered a number of resources that have stated that the skewness of the exponential distribution does not depend not its parameter $\lambda$.
Now, I realize that $\lambda$ for $X \sim \text{exp}(\lambda)$ is the number of events per unit time. However, in my studies of the exponential distribution, it has not been clear to me why the skewness is independent of this?
I'm wondering if someone could please take the time to explain why this is the case; and, if possible, mathematically show that this is the case.
The skewness or skew of a random variable $X$ with mean $\mu$ and variance $\sigma^2$ is $$\operatorname{Skew}[X] = \frac{\operatorname{E}[(X-\mu)^3]}{\sigma^3}.$$ Now suppose $Y = aX$ for some $a > 0$. Then $$\operatorname{E}[Y] = \operatorname{E}[a X] = a \mu,$$ and $$\operatorname{Var}[Y] = \operatorname{Var}[a X] = a^2 \sigma^2.$$ Hence $$\operatorname{Skew}[Y] = \frac{\operatorname{E}[(aX - a\mu)^3]}{(a^2 \sigma^2)^{3/2}} = \frac{\operatorname{E}[a^3(X - \mu)^3]}{a^3 \sigma^3} = \frac{a^3 \operatorname{E}[(X - \mu)^3]}{a^3 \sigma^3} = \operatorname{Skew}[X].$$ So what we have shown is that skew is invariant with respect to nondegenerate order-preserving scaling transformations of a random variable. This is not a coincidence: it is built into the definition of skew because, unlike the third central moment (the expectation in the numerator of the skew), the skew divides by $\sigma^3$ to ensure that as a measure of distribution asymmetry, it is not influenced by the scale on which the random variable is observed.
Now, this also means that for an exponential distribution, its skew is necessarily independent of its parameter, because that parameter is a scale/rate parameter (rate being the reciprocal of scale). In other words, in $$X \sim \operatorname{Exponential}(\lambda), \quad f_X(x) = \lambda e^{-\lambda x}, \quad x > 0,$$ $\lambda$ is a parameter that results from scale transformations of the form $X = \frac{1}{\lambda} W$, where $W \sim \operatorname{Exponential}(1)$. To see this, we note $$f_W(w) = e^{-w}, \quad w > 0,$$ hence $$f_X(x) = f_W(\lambda x) \left|\frac{d}{dx}[\lambda x]\right| = \lambda e^{-\lambda x}.$$ So every exponential distribution is a scaling of the exponential distribution with mean $1$, with scale parameter $1/\lambda$ (or equivalently, rate parameter $\lambda$). Therefore, all exponential distributions have the same skew.
This also generalizes immediately to distributions in which one or more, but not all, parameters are scale. The skew of such a distribution is independent of the choice of scale. For example, take the gamma distribution, parametrized by shape $\alpha$ and scale $\theta$: $$X \sim \operatorname{Gamma}(\alpha, \theta), \quad f_X(x) = \frac{x^{\alpha-1} e^{-x/\theta}}{\theta^\alpha \Gamma(\alpha)}, \quad x > 0.$$ Because $\theta$ is a scale parameter arising from transformations of the form $$X = \theta W, \quad W \sim \operatorname{Gamma}(\alpha, 1), \quad f_W(w) = \frac{w^{\alpha-1} e^{-x}}{\Gamma(\alpha)},$$ it follows that $$\operatorname{Skew}[X] = \operatorname{Skew}[W].$$ I leave it as an exercise to the reader to show that there exist constants $c, p$ such that $\operatorname{Skew}[W] = c a^p$ and that the values of these constants do not depend on $\theta$ or $a$.