Scale versus rate parameterization in exponential and gamma distributions

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In some cases, I've seen the PDFs of the exponential and gamma distribution presented as:

$$X\sim Exp(\lambda) \rightarrow f(x) = \lambda e^{-\lambda x}$$ $$X\sim Gamma(\alpha, \lambda) \rightarrow f(x) = \frac{\lambda^\alpha x^{\alpha - 1} e^{-\lambda x}}{\Gamma(\alpha)}$$

Other times, I've seen them presented as:

$$X\sim Exp(\lambda) \rightarrow f(x) = \frac{1}{\lambda} e^{\frac{-x}{\lambda}}$$ $$X\sim Gamma(\alpha, \lambda) \rightarrow f(x) = \frac{x^{\alpha - 1} e^{\frac{-x}{\lambda}}}{\lambda^\alpha\Gamma(\alpha)}$$

I understand very loosely that the first method is parmeterization by rate and the second method is parameterization by scale, though I do not understand what this means beyond taking the inverse of $\lambda$ which therefore results in different CDFs, moments, etc for each.

My questions are:

  1. What are the benefits of using one parameterization over the other?
  2. What exactly does it mean to say rate versus scale?
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A brief answer (focusing on the exponential first):

One benefit of the first form is that the $\lambda$ there is exactly the arrival rate of the corresponding Poisson arrival process. That is if there is an arrivals process that happens at the rate of say 4 every unit time period, then the inter-arrival times have exponential distribution $\text{Exp}(\lambda=4)$ with PDF $4 e^{-4x}$. This is possibly the reason why it is called the "rate" parametrization.

One benefit of the second form is that the $\lambda$ there is exactly the mean of the distribution. A possible reason why it is called the "scale" parametrization could be the following. Suppose $X \sim \text{Exp}(\lambda=1)$. Then consider a scaled random variable $Y = c \, X$ for some $c > 0$. The CDF of $Y$ is $P(Y \leq y) = P(X \leq y/c) = 1 - e^{-y/c}$ and notice how the scaling factor becomes the parameter in this case.

Since Gamma is a generalization of exponential (that is Exp$(\lambda)$ is $\text{Gamma}(\alpha=1, \lambda)$, the points above can be directly extended.