Kendall's tau and independence Kendall's tau is an association measure between two random variables, say $X$ and $Y$. 
If $X$ and $Y$ are independent, then $\tau = 0$. However, $\tau = 0$ does not imply independence because kendall's tau only captures monotonic association. Is there some conditions under which one has equivalence?
Edit to account for @Cardinal's comment
Say that $X > 0$ and $Y > 0$.
 A: I am going to interpret your question as one regarding a hypothesis on the population quantity $\tau$. If this is not what you intended, please comment to that effect and I will revise  the answer accordingly.
Definition and equivalent expressions
Let $(X,Y)$ by a bivariate random vector with a continuous joint distribution function and let $(X',Y')$ be an independent copy. Kendall's tau is defined as
$$
\renewcommand{\Pr}{\mathbb P}
\tau = \Pr( (X-X')(Y-Y') > 0 ) - \Pr( (X-X')(Y-Y') < 0) \>.
$$
Here are some equivalent definitions under the stated hypotheses that help elucidate this quantity.


*

*$\newcommand{\sgn}{\mathrm{sgn}}\tau = \Pr( \sgn(X-X')  = \sgn(Y-Y') ) - \Pr( \sgn(X-X')) \neq \sgn(Y-Y'))$.

*$\tau = \mathbb E \sgn(X-X')\sgn(Y-Y') = \mathrm{Cov}(\sgn(X-X'),\sgn(Y-Y'))$.

*$\tau = 4 \Pr(X<X',Y<Y') - 1$.  


Note that $X-X'$ and $Y-Y'$ are symmetric about zero. From this and  either (1) or (3), it is clear that if $X$ and $Y$ are independent then $\tau = 0$.
The question at hand, as I understand it, is to characterize the reverse implication.
A first characterization
Using (3) above, we have that $\tau = 0$ implies that
$$
\Pr(X-X' < 0, Y-Y' < 0) = \Pr(X-X'<0)\Pr(Y-Y'<0) = 1/4 \>.
$$
By symmetry, this implies that the probability in each of the four quadrants of the distribution of $(X-X',Y-Y')$ is 1/4. 
We can give the following interpretation: $\tau = 0$ if and only if the event that $X-X'$ is positive or negative is independent of the event that $Y-Y'$ is positive or negative.
This is, at least on the surface, much weaker than what (full) independence of the random variables $X-X'$ and $Y-Y'$ would require, which is that, for each $(a,b) \in \mathbb R^2$,
$$
\Pr(X-X'<a,Y-Y'<b) = \Pr(X-X'<a)\Pr(Y-Y'<b) \>.
$$
A useful reduction by sufficiency
Since $X$ is assumed to have continuous distribution $F$, say, and $Y$ has continuous  distribution $G$, say, then we can restate each of the above in terms of $U = F(X)$ and $V = G(Y)$. For example, (3) above becomes
$$
\tau = 4 \Pr(X < X', Y < Y' ) - 1 = 4 \Pr( F(X) < F(X'), G(Y) < G(Y') )  - 1 \>,
$$ 
which by substitution gives
$$
\tau = 4 \Pr( U  < U', V < V' ) - 1 \>.
$$
Note that $U$ and $V$ are both (marginally) $\mathcal U(0,1)$ random variables.
Thus, without loss of generality we can consider bivariate distributions on the unit cube with uniform marginals. Such a distribution, usually denoted $C(u,v)$ in the bivariate case, is called a copula. 
This also shows why Kendall's tau is invariant to monotone transformations.
A (partially) explicit answer in terms of copulas
Let $(U,V)$ have distribution function $C(u,v)$ with uniform marginals. The reformulation of (3) can be written in terms of $C$  as
$$
\tau  = 4 \Pr(U<U',V<V')-1 = 4 \mathbb E C(U,V) - 1 \>,
$$
which provides us with a fairly explicit answer:

Lemma: For $(X,Y)$ with continuous marginals, $\tau = 0$ if and only if $\mathbb E C(U,V) = 1/4$ where $(U,V) = (F(X),G(Y))$ and $C$
  is the distribution function of $(U,V)$.

Relation to independence and a counterexample
If $X$ and $Y$ are independent, so are $U  = F(X)$ and $V = G(Y)$, in which case $C(u,v) = uv$. Clearly $\mathbb E C(U,V) = \iint_{[0,1]^2} u v \,\mathrm du \,\mathrm dv = 1/4$. But do other examples exist?

Counterexample: Consider the family of copulas
  $$
C(u,v) = uv + \alpha u(u-1)(2u-1)v(v-1)(2v-1)
$$
  indexed by the parameter $-1 \leq \alpha \leq 2$. For every value of $\alpha \in [-1,2]$, we have $\tau = 0$. Note that the subcase of independence corresponds to $\alpha = 0$.

