The response of a discrete time system $y\left[n\right]$ obeys the following relation:

$y\left[n\right]=\frac{5}{6}y[n-1]-\frac{1}{6}y[n-2]+x\left[n\right]$

The input to the system is $x\left[n\right]=\delta \left[n\right]-\frac{1}{3}\delta [n-1]$

Which of the following options is TRUE for $y\left[n\right]$ ?

The correct option is: Stable and causal response

Let us analyze the system to determine its causality and stability step-by-step.

First, we are given the system's difference equation:
$y\left[n\right]=\frac{5}{6}y[n-1]-\frac{1}{6}y[n-2]+x\left[n\right]$
We can rewrite this in terms of the shift operator or take the Z-transform of both sides under zero initial conditions to find the transfer function $H\left(z\right)=\frac{Y\left(z\right)}{X\left(z\right)}$.

Taking the Z-transform of the difference equation:
$Y\left(z\right)=\frac{5}{6}{z}^{-1}Y\left(z\right)-\frac{1}{6}{z}^{-2}Y\left(z\right)+X\left(z\right)$
Rearranging the terms:
$Y\left(z\right)[1-\frac{5}{6}{z}^{-1}+\frac{1}{6}{z}^{-2}]=X\left(z\right)$
Therefore, the transfer function of the system is:
$H\left(z\right)=\frac{Y\left(z\right)}{X\left(z\right)}=\frac{1}{1-\frac{5}{6}{z}^{-1}+\frac{1}{6}{z}^{-2}}$

To find the poles of the system, we solve for the roots of the denominator polynomial:
$1-\frac{5}{6}{z}^{-1}+\frac{1}{6}{z}^{-2}=0$
Factoring the denominator:
$(1-\frac{1}{2}(1-\frac{1}{3}=\mathrm{0<!--\; -->}$
This yields two poles:
$z_1=\frac{1}{2},z_2=\frac{1}{3}$

Next, let us analyze the input $x\left[n\right]$ to the system:
$x\left[n\right]=\delta \left[n\right]-\frac{1}{3}\delta [n-1]$
Taking the Z-transform of $x\left[n\right]$:
$X\left(z\right)=1-\frac{1}{3}{z}^{-1}$

Now, we can find the Z-transform of the output $y\left[n\right]$:
$Y\left(z\right)=H\left(z\right)X\left(z\right)=\frac{1-\frac{1}{3}{z}^{-1}}{(1-\frac{1}{2}{z}^{-1})(1-\frac{1}{3}{z}^{-1})}$
Simplifying $Y\left(z\right)$ by canceling the common term $(1-\frac{1}{3}{z}^{-1})$ from the numerator and denominator:
$Y\left(z\right)=\frac{1}{1-\frac{1}{2}{z}^{-1}}$

For a causal response, the region of convergence (ROC) of the Z-transform extends outward from the outermost pole:
$\text{ROC: }\left|z\right|>\frac{1}{2}$
Since the ROC is of the form $\left|z\right|>r$, the response is causal. Taking the inverse Z-transform gives:
$y\left[n\right]={\left(\frac{1}{2}\right)}^{n}u\left[n\right]$
where $u\left[n\right]$ is the unit step function, confirming causality because $y\left[n\right]=\mathrm{0<!--\; -->}$ for $n<0$.

To evaluate stability, we check if the unit circle $\left|z\right|=1$ is contained within the ROC. Since the ROC is $\left|z\right|>\frac{1}{2}$, the unit circle is indeed inside the ROC. Equivalently, the single remaining pole of the system response is at $z=\frac{1}{2}$, which lies strictly inside the unit circle $\left(\right|z|<1)$. Thus, the response $y\left[n\right]$ is stable.

Consequently, the system response $y\left[n\right]$ is both stable and causal.