Distributions with undefined parameters I am studying the Bayesian Lasso and noticed something interesting on Page 682 at the bottom of the second column here
Some background: a hierarchical setup for data $X, y$, regression coefficients, $\beta_i$, variance $\sigma^2$ and additional parameters $\tau^2_i$ is setup leading to an intractable posterior. The goal is to implement a Gibbs sampler to sample from the posterior, and thus this requires full conditionals.
The full conditional for each $\tau^2_i$ is
$$
\dfrac{1}{\tau^2_i} \mid (\beta, \sigma^2, y) \sim \text{Inverse Gaussian} \left(\sqrt{\dfrac{\lambda^2 \sigma^2}{\beta_i^2}}, \lambda^2  \right)
$$
where $\lambda^2$ is a fixed quantity $ > 0$. The support of $\sigma^2$ and $\tau^2_i$ is $\mathbb{R}^+$, but the support of $\beta_i$ is $\mathbb{R}$. Thus, $\beta_i$ could be 0, leading to an undefined first argument in the Inverse Gaussian.
My question is: How is this not a problem? 
 A: It turns out that the answer lies in looking at the pdf of the Inverse Gaussian.
The full conditional for each $\tau^2_i$ is
$$\dfrac{1}{\tau^2_i}| \beta, \sigma^2, y \sim \text{Inverse Gaussian}\left(\sqrt{\dfrac{\lambda^2 \sigma^2}{\beta_i^2}}, \lambda^2  \right). $$
\begin{align*}
f(\tau^2_i | \beta, \sigma^2, y) & = \sqrt{\dfrac{\lambda^2}{2 \pi}} \left( \tau^2_i\right)^{-1/2} \exp \left\{\dfrac{-\beta_i^2 \left(1 - \tau^2_i \sqrt{\frac{\lambda^2 \sigma^2}{\beta^2_i}} \right)}{2 \sigma^2\tau_i^2}  \right\}\\
& = \sqrt{\dfrac{\lambda^2}{2 \pi}} \left( \tau^2_i\right)^{-1/2} \exp \left\{\dfrac{-\beta_i^2 - (\tau^2_i)^2\lambda^2 \sigma^2 + 2\tau^2_i \lambda \sigma\beta_i}{2 \sigma^2 \tau^2_i}  \right\}
\end{align*}
If $\beta_i = 0$, then 
\begin{align*}
f(\tau^2_i | \beta, \sigma^2, y) & =\sqrt{\dfrac{\lambda^2}{2 \pi}} \exp \left\{ -\dfrac{\lambda^2}{2}\tau^2_i \right\}.
\end{align*}
Thus, $$\tau^2_i | \beta, \sigma^2, y \sim \text{Gamma}(1/2, \lambda^2/2) \text{ and } \dfrac{1}{\tau^2_i} | \beta, \sigma^2, y \sim \text{Inverse Gamma}(1/2, \lambda^2/2).  $$
So when the mean parameter of the Inverse Gaussian distribution is undefined, or in particular of the form $1/0$, the distribution turns out to be an Inverse Gamma distribution!I am sure it has something to do with the generalized inverse Gaussian distribution, but exactly what, I am not sure. This is an extremely surprising result.
