Density of a linear transformation of an inverse gaussian variable Suppose that $y$ is a random variable with the inverse Gaussian distribution, i.e. it has density
\begin{align}
    f(y) = \sqrt{ \frac{\lambda}{ 2 \pi y^3} } \exp \left( \frac{-\lambda\left( y - \mu\right)^2}{2 \mu^2 y} \right).
\end{align}
Now, suppose that I create a new random variable as $x = ay + b$, what would be the density of that variable? If it helps, I am very interested in a specific case where
$x = b_0 + b_1 h + a y$, $\mu = h/a^2$, $\lambda = \mu^2$ (so, we'll have $E(y) = Var(y) = \mu$) and we have $a < 0$ and $b = b_0 + b_1 h > 0$. I know I should get a density that is reflected, scaled and shifted, but I can't seem to figure out how to make sense of that. Of course, the more general case for arbitrary $(a,b)$ would be even more of interest to me, if an answer does exist.
 A: I did find a general answer to my problem.
Suppose that you have two random variables, $X, Y: \Omega \rightarrow \mathbb{R}$ and two non random parameters $(a,b) \in \mathbb{R}$ related by $Y = aX + b$. Furthermore, let $f_X$ and $f_Y$ be their respective density functions. Then,
\begin{align}
   f_Y(y) = \frac{1}{|a|} f_X(x).
\end{align}
The proof is actually fairly simple. I'll do it for $a > 0$:
\begin{align}
   \mathbb{P}(u < Y < v) &= \mathbb{P}\left( u < aX + b < v \right) \\
                         &= \mathbb{P}\left( \frac{u-b}{a} < X < \frac{v-b}{a} \right) \\
                         &= \int_{\frac{v-b}{a}}^{\frac{u-b}{a}} f_X(x) dx
\end{align}
Then, we proceed with a change of variable: $y = ax + b$ implies $x = \frac{y-b}{a}$ and $dx = \frac{1}{a}$ such that
\begin{align}
   \int_{\frac{v-b}{a}}^{\frac{u-b}{a}} f_X(x) dx = \int_{v}^{u} \frac{1}{|a|} f_X\left(\frac{y-b}{a}\right) dy := \int_u^v f_Y(y) dy
\end{align}
where we thus have defined the density $f_Y(y) := \frac{1}{|a|} f_X\left(\frac{y-b}{a}\right) = \frac{1}{|a|} f_X(x)$. The case for $a < 0$ follows the same logic, except one negative sign appears when using the survival function and another when using $-|a| = a$, so they cancel out and you recover the same relationship.
