Laplace: Add mean argument to helper distributions

[WardBrian]

Summary

Closes #3202. Note that this is branched off of #3209, since doing the two changes independently would lead to a great many merge conflicts.

Tests

Existing tests updated to pass 0 as the mean. It would be nice to add some tests
with a non-zero mean, if there are suggestions (@avehtari @charlesm93)

Side Effects

Release notes

Checklist

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• the basic tests are passing

  • unit tests pass (to run, use: ./runTests.py test/unit)
  • header checks pass, (make test-headers)
  • dependencies checks pass, (make test-math-dependencies)
  • docs build, (make doxygen)
  • code passes the built in C++ standards checks (make cpplint)

• the code is written in idiomatic C++ and changes are documented in the doxygen

• the new changes are tested

[avehtari]

This example https://discourse.mc-stan.org/t/embedded-laplace-numerical-problem/39700 uses non-zero mean (mu) and Poisson. So with the helper it should look something like (not certain what is the argument order now)

laplace_marginal_poisson_log_lpmf(y | mu, [1], cov_fun, (sigmaz, N), theta_0)

[WardBrian]

Thanks @avehtari.

Writing that test made me realize something else: The y_index argument to laplace_marginal_poisson_log_lpmf is currently 0-indexed rather than 1-indexed. Is that intentional @charlesm93 @SteveBronder?

[WardBrian]

My instinct (and @avehtari's above) was to make this {1}, but that led to an index-out-of-bounds, because this variable is used as a 0-based index. This comment is to remind me that I want to verify this is intentional