The variances are saved in oo_.variance.

For the difference between conditional and unconditional welfare, see Section 3.3 “The Welfare Measure” in Schmitt-Grohe/Uribe (2004) “OPTIMAL SIMPLE AND IMPLEMENTABLE MONETARY AND FISCAL RULES”, NBER Working Paper 10253. It is important to look at this version of the paper and not later ones. In a nutshell, for one version you use the conditional expectations operator (given that you are in the determinstic steady state) and in the other one the unconditional one. This makes a difference because at higher order the system transitions from the deterministic steady state to the ergodic distribution. This transition is absent in the unconditional welfare measure as the expectation is taken over the ergodic distribution.

SGU (2001) “Stabilization Policy and the Costs of Dollarization” uses unconditional welfare.

For unconditional welfare, there is no difference between the two. The reason is linearity of the expectations operator. You want to compare unconditional lifetime utility in the steady state and under some policy. You have

\mathbb{E}\sum_{t=0}^{\infty}\beta^tU((1-\xi)\bar c,\bar n)=\mathbb{E}\sum_{t=0}^{\infty}\beta^tU(c_t,n_t)

This is

\sum_{t=0}^{\infty}\beta^t\mathbb{E}(U((1-\xi)\bar c,\bar n))=\sum_{t=0}^{\infty}\beta^t\mathbb{E}(U(c_t,n_t))

which is

1/(1-\beta)U((1-\xi)\bar c,\bar n)=1/(1-beta)\mathbb{E}(U(c_t,n_t))

and thus

U((1-\xi)\bar c,\bar n)=\mathbb{E}(U(c_t,n_t))

Hence, it is sufficient to compare the felicity functions because everything is stationary. The \xi you compute would be the same. Things are different if you use recursive utility or conditional welfare.

As you are not interested in a log approximation to Welfare itself, only in one for consumption and labor, you can use either representation. For extracting \xi, it does not matter whether your endogenous variable is log(Welfare) or Welfare itself (the same applies to U itself). But one warning: using a variable substitution with exp() implies that e.g. the argument of exp(U) is actually log(U) so that exp(log(U))=U. When doing this, you need to make sure that the original U which is now logged is not negative.