On the optimal distribution for importance sampling Let's say $X$ is a rv, $p(x)$ is its pmf.
I want to importance-sample $\mu := \mathbb E[f(X)]$,
for some bounded function $0<f<1$,
using another distribution $q(x)$.
Then what I should do is summing
$$\hat\mu := \frac{1}{n} \sum_{k=1}^n \frac{f(Y_k)p(Y_k)}{q(Y_k)}$$
where $Y_k$ are generated using $q$.
According to
this lecture note,
there is an optimal $q$, denoted by $q'$,
that minimizes the variance of $\hat\mu$:
$$q'(x)\propto f(x)p(x)$$
If I sample using $q'$,
then all summands in the definition of $\hat\mu$ becomes constant.
So yeah, it does give me a variance-zero experience.
Really appreciate.
What I don't seem to understand is that,
if I now want to importance-sample $g := 1 - f$ instead,
then suddenly $q'$ is not the best distribution,
as $q'$ is no longer $\sim gp = (1 - f)p$.
Yet another example is that, if I want to sample $h := 100 + f$,
then $hp$ is just $100(p \pm \epsilon)$,
so the optimal $q$ is very close to $p$.
Or maybe it's telling me that I shouldn't bother.
How is it possible that the optimal choice of $q$
depends on an affine transformation I applied to $f$?
Does that mean that I should optimize over
the affine transformations that can be applied to $f$?

There's more than one optimal choice of $q$.
Suppose $p(x)=2x$ on $[0,1]$ and 0 otherwise, and $f(x)=x$. You can optimally choose $q(x)=p(x)$ and get zero variance or you can choose $q(x)=2-2x$ and estimate
$$E[f(X)]=1-\frac{1}{2}E_Q[2-2x]$$
and get zero variance that way.  There's a whole affine family of zero-variance estimators. The optimality proofs don't claim the optimum is unique.
