I need to find the variance and autocovariance or process $W_t = Y_t-Y_{t-}$ with $Y_t = c + \phi_1Y_{t-1} + E_t$ an AR(1) process with $-1 < \phi_1 < 1$
I end up with the following calculation : \begin{align*} \gamma_k &= Cov(W_t, W_{t-k}) \\ &= Cov(Y_t-Y_{t-1}, Y_{t-k}-Y_{t-1-k}) \\ &= Cov(Y_t, Y_{t-k}) - Cov(Y_t, Y_{t-1-k}) - Cov(Y_{t-1}, Y_{t-k}) + Cov(Y_{t-1}, Y_{t-1-k}) \\ &= Cov(Y_t, Y_{t-k}) - Cov(Y_t, Y_{t-(1+k)}) - Cov(Y_{t-1}, Y_{t-1-(1+k)}) + Cov(Y_{t-1}, Y_{t-1-k}) \\ &= \phi^k\frac{\sigma^2}{1-\phi_1^2} - \phi^{1+k}\frac{\sigma^2}{1-\phi_1^2} - \phi^{1+k}\frac{\sigma^2}{1-\phi_1^2} - \phi^k\frac{\sigma^2}{1-\phi_1^2} \\ &= (\phi_1^k - \phi_1^{1+k})\frac{2\sigma^2}{1-\phi_1^2} \\ &= (1 - \phi_1)\frac{2\phi_1^k\sigma^2}{1-\phi_1^2} \end{align*}
And then I need to find the variance of $W_t$ which equal in the question of the exercise to $\frac{2}{1+\phi_1}$. However, I end up by replacing $k=0$ to :
\begin{align*} Var(W_t) &= \gamma_0 \\ &= \frac{(1-\phi_1)2\sigma^2}{(1-\phi_1)(1+\phi_1)} \\ &= \frac{2\sigma^2}{1+\phi_1} \end{align*}
I don't know how to remove the $\sigma^2$