Consider a group of Gromov-hyperbolic $ Gamma $ and let $ mu $ to be a measure of probability finely supported on $ Gamma $. Suppose the support of $ mu $ generates $ Gamma $ in a semi-group, in other words, a random walk $ X_n $ driven by $ mu $ can visit the whole group $ Gamma $.

Fact: Random walking $ X_n $ almost surely converges to a point on the border of Gromov $ partial X $ of $ X $. Let $ nu $ to be the output measure on $ partial X $. Then, $ ( partial X, nu) $ is a model for the so-called Poisson limit. Measurement $ nu $ is called the harmonic measure (as far as $ mu $).

Now consider the opposite measure $ check { mu} $ Defined by $ check { mu} (g) = mu (g ^ {- 1}) $ and let $ check { nu} $ to be the corresponding harmonic measure.

**Question** : Are $ nu $ and $ check { nu} $ equivalent?

It seems that the answer is positive. In the document Harmonic versus Quasi-Formal Measurements for Hyperbolic Groups, Blachère, Haïssinsky and Mathieu prove that $ h = lv $ on a hyperbolic group if and only if the harmonic measure $ nu $ and the Patterson-Sullivan measure $ rho $ are equivalent. Right here, $ h $, is the asymptotic entropy, $ l $ is the asymptotic drift and $ v $ is the growth in volume. Very roughly, the proof goes like this.

First, the two measures are ergodic and are therefore equivalent or singular. The Hausdorff dimension of $ nu $ is $ h / l $ and the Hausdorff dimension of $ rho $ is $ v $. Using the Lebesgue Differentiation Theorem, they show that they are equivalent if and only if they have the same Hausdorff dimension. This part is very technical, the basic ingredients being the shadow lemmas for the Patterson-Sullivan measure and for the harmonic measure.

Since the lemmas for harmonic measurements $ nu $ and $ check { nu} $ hold and since $ nu $ and $ check { nu} $ have the same dimension of Hausdorff, I think we can adapt their evidence to show that $ nu $ and $ check { nu} $ are equivalent.

However, I would like to prove a similar result in a broader context and this strategy seems too complicated and too technical. So I wonder if there is a much simpler proof.

Note that you really need to use hyperbolicity somewhere. For example, consider a non-centered probability measure $ mu $ sure $ mathbb {Z} ^ d $, C is

$$ p = sum_ {x in mathbb {Z} ^ d} x mu (x) neq 0. $$

Then, $ p $ it's called drifting the random walk. According to Ney and Spitzer's theorem, Martin's limit of the random walk is a sphere: a sequence of points $ x_n $ converges to a point on the boundary if and only if $ x_n $ goes to infinity and $ x_n $ converge in direction, that is to say $ x_n / | x_n | $ converges. The harmonic measure is based on the direction given by $ p $. In particular, the harmonic measure associated with $ check { nu} $ is supported on $ -p $ and so $ nu $ and $ check { nu} $ are singular.