The probability density of y is given by

p(y \mid \mu, \sigma) = \text{normal}(f^{-1}(y) \mid 0, 1) \cdot \left| \frac{ \mathrm{d} f }{ \mathrm{d} x} \right|^{-1}(f^{-1}(y)).

Now

\begin{align*}
\text{normal}(f^{-1}(y) \mid 0, 1)
&=
\frac{1}{\sqrt{2\pi}} \exp \left( -\frac{1}{2} \left( f^{-1}(y) - 0 \right)^{2} \right)
\\
&=
\frac{1}{\sqrt{2\pi}} \exp \left( -\frac{1}{2} \left( \frac{y - \mu}{\sigma} - 0 \right)^{2} \right)
\\
&=
\frac{1}{\sqrt{2\pi}} \exp \left( -\frac{1}{2} \left( \frac{y - \mu}{\sigma} \right)^{2} \right)
\end{align*}

This *almost* looks like a normal density function; the only problem is the normalization. Fortunately we still have that Jacobian determinant term,

\begin{align*}
\left| \frac{ \mathrm{d} f }{ \mathrm{d} x} \right|^{-1}(f^{-1}(y))
&=
\left| \frac{ \mathrm{d} (\mu + \sigma \cdot x) }{ \mathrm{d} x } \right|^{-1}(f^{-1}(y))
\\
&=
\left| \sigma \right|^{-1}(f^{-1}(y))
\\
&=
\frac{1}{\sigma},
\end{align*}

which is exactly the missing normalization!

In other words we’ve just proved that

\begin{align*}
p(y \mid \mu, \sigma)
&= \text{normal}(f^{-1}(y) \mid 0, 1) \cdot \left| \frac{ \mathrm{d} f }{ \mathrm{d} x} \right|^{-1}(f^{-1}(y))
\\
&= \frac{1}{\sqrt{2\pi}} \exp \left( -\frac{1}{2} \left( \frac{y - \mu}{\sigma} \right)^{2} \right) \cdot \frac{1}{\sigma}
\\
&= \frac{1}{\sqrt{2\pi \sigma^{2}}} \exp \left( -\frac{1}{2} \left( \frac{y - \mu}{\sigma} \right)^{2} \right)
\\
&= \text{normal}(y \mid \mu, \sigma).
\end{align*}

In other words

\text{normal}(x \mid 0, 1)
\\
y = \mu + \sigma \cdot x

is equivalent to

\text{normal}(y \mid \mu, \sigma)

exactly *because* of the Jacobian!

We don’t need a Jacobian in the model here because we’re not trying to force any behavior on y = \mu + \sigma \cdot x. Instead we’re forcing behavior on x and letting Stan figure out the induced behavior on y. If we want to work out that induced behavior analytically we have to include the Jacobin as above.

Jacobians are needed when we have a density that depends on a derived quantity, \pi(y). In general this *will not* induce behavior on x such that the behavior of the derived quantity y = f(x) is quantified by \pi(y). Instead the induced behavior is quantified by

\pi_{\text{induced}}(x) = \pi(f(x))

which then implies a density

\begin{align*}
\pi_{\text{induced}}(y)
&= \pi_{\text{induced}}(f^{-1}(y)) \left| \frac{ \mathrm{d} f}{\mathrm{d} x} \right|^{-1} (f^{-1}(y))
\\
&=
\pi(f \circ f^{-1}(y)) \left| \frac{ \mathrm{d} f}{\mathrm{d} x} \right|^{-1} (f^{-1}(y))
\\
&=
\pi(y)) \left| \frac{ \mathrm{d} f}{\mathrm{d} x} \right|^{-1} (f^{-1}(y))
\end{align*}

for the derived quantity. Note that here the Jacobian term skews the behavior of y from the expected \pi(y)! If we had included the inverse Jacobian determinant correct as we should have then it would have canceled with this extra term to give \pi_{\text{induced}}(y) = \pi(y) as desired.