# New Stan data type: zero\_sum\_vector

**URL:** https://discourse.mc-stan.org/t/new-stan-data-type-zero-sum-vector/26215
**Category:** Developers
**Tags:** specification
**Created:** [February 3, 2022, 8:50pm UTC](https://discourse.mc-stan.org/t/new-stan-data-type-zero-sum-vector/26215 "2022-02-03T20:50:15Z")
**Posts on this page:** 1
**Showing post:** 4

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### Author: ![Bob\_Carpenter](https://yyz2.discourse-cdn.com/flex030/user_avatar/discourse.mc-stan.org/bob_carpenter/32/9230_2.png) [@Bob\_Carpenter](https://discourse.mc-stan.org/u/Bob_Carpenter)
#### Post date: [February 15, 2022, 8:11pm UTC](https://discourse.mc-stan.org/t/new-stan-data-type-zero-sum-vector/26215/4 "2022-02-15T20:11:20Z")

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> [@jsocolar](#):
>
> Just to note (and I don’t expect that this is news to @Bob_Carpenter) that the Stan Users Guide has a section on parameterizing vectors of this sort as well as some discussion of how to set priors on the margins of a centered vector:

Thanks, @jsocolar. This is super cool. I didn’t do the math to figure out the correction and I don’t even think I reviewed the PR.

> [@jsocolar](#):
>
> Instead, I would say that there is no useful adjustment to perform because the Jacobian isn’t square.

There is a Jacobian, but it’s constant (i.e., doesn’t depend on parameters).

The Jacobian is always square in a change of variables but it’s not always apparent what those variables are (and sometimes there’s a choice). For example, if I have variables (x, y) and want a transformed variable \exp(x + y), then I can’t just put a prior on \exp(x + y) and get a distribution over (x, y)—the result is the same for (x - c, \ y + c), so I need more constraint. For example, I could put a prior on x or y in addition to \exp(x + y) and get a proper prior on (x, y). The typical fix is to think of the transform as mapping both x and y, which means you need two outputs. A simplex choice is (x, y) \rightarrow (\exp(x + y), y) and you get a nice triangular Jacobian and easy determinant calculation and then you can put a prior on \exp(x + y) and on y and get a proper prior on (x, y).

Let’s look at a simple example, where we map (x, y) \rightarrow (x, y, -(x + y)). This is indeed not square. We have too many degrees of freedom in the output. So instead we need to pick and choose. (x, y) \rightarrow (-(x + y), \ y) works. Then what’s happening in this model is what @betanalpha is complaining about—we’re overconstraining the solution by putting priors on all of x, y, and -(x + y). That’s why we need the adjustment to the scale of the prior indicated in the user’s guide.

> [@jsocolar](#):
>
> … then they’d write down the wrong Stan model without realizing it.

Is there an example you had in mind?

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