Thinking about what continuous density that would approximate is a good way to start thinking about Jacobians :-)

No. Theyâ€™re all given equal density. You need to integrate density to get back to probability. What happens is that every interval of the same width has the same probability in a simple upper or lower-bounded or unconstrained distribution. But if you do that with a non-compact set of \mathbb{R}^N, things blow up, so probability and density arenâ€™t properly defined. But we just act as if they were and continue (hoping the posterior will be proper after seeing some data).

So if you do something like

```
real<lower = 0, upper = 1> sigma;
```

thatâ€™s fine, because \int_0^1 c \, \mathrm{d}\sigma = c. But if we do this,

```
real<lower = 0> sigma;
```

we run into problems because \int_0^{\infty} c \, \mathrm{d}\sigma= \infty. So thatâ€™s why itâ€™s â€śimproperâ€ť.

Correct, not affected by density of floating point numbers other than in terms of precision. Thereâ€™s much higher precision in a small neighborhood around 0 than the same sized neighborhood around 1, for example.