Let $P_{x,w}(q)$ be the Kazhdan Lusztig polynomial. It is well-known that $P_{x,w}(q)\neq 0\iff x\le w$. 

By the interpretation of the Kazhdan Lusztig polynomial in terms of extension group, it holds that $x\le w \iff \mathrm{Ext}_{\mathcal{O}}^i(M(w\cdot(-2\rho)),L(x\cdot(-2\rho)))\neq 0$ for some $i\ge 0$, where $M(\eta)$ is the Verma module, $L(\eta)$ is its unique simple quotient and $\rho$ is the half sum of positive roots. 

> Let $\mu$ be an integral, antidominant weight, $\Delta$ be the set of simple roots, $\Sigma=\{\alpha\in\Delta:\langle\mu+\rho,\alpha^\lor\rangle=0\}$, $W^{\Sigma}=\{w\in W:w<ws_\alpha,\forall \alpha\in\Sigma_\mu\}$ and $x,w\in W^{\Sigma}$.

It is not hard to show $\mathrm{Ext}_{\mathcal{O}}^i(M(w\cdot\mu),L(x\cdot\mu))\neq 0$ for some $i\ge 0\implies x\le w$.

> I would like to know whether the converse: 

>$x\le w\implies \mathrm{Ext}_{\mathcal{O}}^i(M(w\cdot\mu),L(x\cdot\mu))\neq 0$ for some $i\ge 0$ 

>is true or not. If it is true, I would like to know why. If not, any counter-example?