Complement problems are not in the same class in parametrized complexity hierarchy? If not in $P$ By "complement problems", I mean the two problems' objective functions are complement. For example, the vertex cover and its complement independent set in this sense. For a graph $G(V,E)$, their answers add up to $|V|$. Another example is a problem I am working on. The dominating set problem and its complement "packing as many edges as possible so that there is no path or circle whose length is more than 2". Dominating set is $W[2]$-complete in the sense, and the edge packing problem is in $W[1]$-complete( whether it is complete in $W[1]$, I am still looking). 

So, I come to the following audacious conjecture.
If problem A and problem B are complement, then A and B are not in the same class in the parameterized complexity structure, i.e. $FPT \subseteq W[1] \subseteq W[2] \subseteq \cdots W[SAT] $. Of course, there is another condition : A and B are not $P$ in the classical complexity hierarchy.

I am not very familiar with parameterized complexity theory. So, any advice is welcome. Thank you very much.
The complement problems Minimum (Edge) Bipartization and MaxCut are both in FPT, so the conjecture is false.