How to find ROC of system when input is two sided and output is one-sided Given the $\mathcal Z$-transform of input $x[n]$ and output $y[n]$, how can I find the ROC of the system function $H(z) = Y(z)/X(z)$? I have
$$X(z) = \frac{2z\left(z-\frac{10}{3}\right)}{\left(z-\frac 13\right)(z-3)},\quad\text{ROC}: \frac 13 &lt; |z| &lt; 3$$
and 
$$Y(z) = \frac{-1}{4}\frac{z}{\left(z-\frac 13\right)\left(z-\frac 14\right)},\quad\text{ROC}: |z| &gt; \frac 13$$

If I substitute $\hat{X}(z) = 1/X(z)$, then $ H(z) = Y(z)\hat{X}(z)$. Then, I know that the ROC of $H(z)$ contains the intersection of the ROCs of $Y(z)$ and $\hat{X}(z)$. But how can I find the ROC of $\hat{X}(z)$? I don't know if the corresponding sequence $\hat{x}[n]$ is causal or not.
The transfer function is

$$H(z)=\frac{Y(z)}{X(z)}=-\frac18\frac{z-3}{\left(z-\frac14\right)(z-\frac{10}{3})}\tag{1}$$

It has three possible ROCs:


$|z|&lt;\frac14$ (left-sided, unstable)
$\frac14&lt;|z|&lt;\frac{10}{3}$ (two-sided, stable)
$|z|&gt;\frac{10}{3}$ (right-sided, unstable)


Only the second ROC, corresponding to a two-sided and stable impulse response, overlaps with the ROC of $X(z)$. The intersection, and, consequently, the ROC of $Y(z)$ would be $\frac13&lt;|z|&lt;3$ (equal to the ROC of $X(z)$). However, the pole at $z=3$ is removed by pole-zero cancellation, which results in the final ROC $|z|&gt;\frac13$ for $Y(z)$.