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If $X \sim Laplace(\mu, b)$ and $\mu = 0$ such that $X \sim Laplace(0,b)$, it becomes a form of the exponential family. Unfortunately, one source tells me that it becomes $|X| \sim exponential(b)$, another source says that $X \sim exponential(\frac{1}{b})$, and my notes say that $X \sim exponential(b)$.

Can any of y'all help me out on this?

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1 Answer 1

The problem is that there are two common ways to parameterize exponential distributions (... and the same issue occurs for gamma distributions more generally).

One is the rate form of the exponential,

$f_X(x) = \theta e^{-\theta x}; \theta>0, x>0$

the other is the scale form

$f_X(x) = \frac{1}{\mu} e^{-x/\mu}; \mu>0, x>0$

with $\mu = 1/\theta$, these two forms are equivalent.

Usually you can guess which form is being used from context.

For example, for the exponential, Wikipedia first gives the rate form (with rate $\lambda$):

$$f(x;\lambda) = \begin{cases}\lambda e^{-\lambda x}, & x \ge 0, \\0, & x < 0.\end{cases}$$

and then gives the scale form (with scale $\beta$):

$$f(x;\beta) = \begin{cases}\frac{1}{\beta} e^{-x/\beta}, & x \ge 0, \\0, & x < 0.\end{cases}$$

Correspondingly, you can write the Laplace distribution either way.

For the Laplace distribution, Wikipedia gives the scale form - which is the more common parameterization in the case of the Laplace - but the rate form is equally valid.

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