We have a stationary time series {X_t} with zero mean and ACVF $\gamma_x(h)$. At each time $t$ with probability $p$ an error occurs, recording the value of $X_{t-1}$ instead of $t$. This happens independent each $t$ and probability $p$ does not vary.
We want to show that data series {$Y_t$} is stationary and compute it's ACVF $\gamma_Y(h)$.
We suppose that {$X_t$} is a causal AR(1) process with an ACF satifying $\rho_x(1) = 0.5$. Compute the ACF $\rho_Y(h)$for the value of $\rho$ that maximizes $\rho_Y(h)$
Using this information i find that
$\gamma_Y(h) = \gamma_x(h)$ for $h = 0$
$\gamma_Y(h) = \gamma_x(h) + p(1-p)(\gamma_x(h-1) + \gamma_x(h+1)-2\gamma_x(h))$ for $h \ge 1$
$\gamma_Y(h) = \gamma_y(-h)$ for $h \le -1$
For an AR(1) process with $\phi \in (0,1)$ for $h\ge1$. We have
Can anyone explain what we are doing below?
$\gamma_y(h) - \gamma_x(h) = p(1-p)(\phi^{-1} + \phi -2)\phi^{h}(1-\phi^2)^{-1}\sigma^2$
Since $\rho_x(1) = \phi = 0.5$ and $p(1-p)$ is maximized for $p = 0.5$ for $h \ge 1$ we get
$\rho_y(h) = (0.5)^h + 0.25(2+0.5-2)(0.5)^h = 1.125(0.5)^h = 1.125\rho_x(h)$
Why do we set $\gamma_y(h)-\gamma_x(h)$ and how do we find $\rho_y(h)?$
The notation is slighly confusing for me. I assume $\sigma^2$ is the variance of the noise that feeds the AR process.
You have $$\gamma_y(h) = \gamma_x(h) + p(1-p)\left(\gamma_x(h-1) + \gamma_x(h+1)-2\gamma_x(h)\right) \tag{1}$$ for $h \ge 1$ Now, if $X$ is AR(1), its ACVF must has the form of a two-sided decaying exponential - further, because we are given $\gamma_x(1)=0.5$ we can already write $$\gamma_x(h) = \sigma^2_X \, \phi^{|h|} \tag{2}$$
with $\phi = 0.5$ , and $\sigma^2_X = \sigma^2_n/(1-\phi^2)$ (see eg). Replacing, we get
$$\begin{align} \gamma_y(h) &= \gamma_x(h) + p(1-p) \sigma^2_n \frac{ (\phi^{h+1}+\phi^{h-1}-2\phi^h)}{1-\phi^2}\\ &= \gamma_x(h) + p(1-p) \sigma^2_n \frac{\phi^h (\phi -\phi^{-1} -2)}{1-\phi^2} \end{align} \tag{3}$$
Because we want to maximize $\gamma_y(h)$, and because the fraction in $(3)$ is positive, we want $p(1-p)$ to be maximum, which occurs at $p=1/2$. I think you can go on from here.