Let $G_1,\dotsc,G_r$ combinatorial impartial games. The selective compound $G=G_1 \vee \dotsc \vee G_r$ is the following game: The positions are the tuples of positions of the $G_i$, and a move is a move in any positive number of $G_i$s. My former question asked for the $\mathcal{P}$-positions of $G$ in terms of the ones for $G_i$. However, there we used implicitly the normal play rule (the one with the last move wins).
Question. What are the $\mathcal{P}$- and $\mathcal{N}$-positions for $G$ under the misère play rule?
Here is a very basic example which already shows that something strange may happen: Let $G_1$ be the subtraction game with one pile and substraction set $\{1\}$. Thus, positions of $G_1$ are natural numbers and the only moves are $n \mapsto n-1$ if $n>0$. The $\mathcal{P}$-positions of $G_1$ under the misère play rule are the odd natural numbers. One verifies that the $\mathcal{P}$-positions in $G_1 \vee \dotsc \vee G_1$ are a) the ones of the form $(0,\dotsc,0,n,0,\dotsc,0)$ for some odd natural number $n$, b) $(n_1,\dotsc,n_s)$, where at least two $n_i$ are positive, and all $n_i$ are even.