Show that a time series process is stationary Consider the time series 
$$
x_t = B_1 + B_{2}t + w_t,
$$
where $B_1$ and $B_2$ are known constants and $w_t$ is a white noise process with variance $\sigma^2$. 
Show that the process 
$y_t = x_t - x_{t-1}$ is stationary.
For finding Covariance I cant seem to understand what to do.
 A: Your discrete stochastic process is defined as:
\begin{equation}
x_t = B_1 + B_2t + w_t~~~~~~~, ~~ w_t \sim WN(0,\sigma^2)
\end{equation}
Clearly it is not stationary since:
$$E[x_t] = B_1+B_2t$$
Now we consider the differentiated process of $x_t$, using the lag operator ($LY_t=Y_{t-1}$):
$$ \Delta Y_t = (1-L)Y_t = Y_t - Y_{t-1} $$
$$ = B_1 + B_2t + w_t - (B_1 + B_2(t-1) + w_{t-1})$$
$$ = B_1 + B_2t + w_t - B_1 - B_2t + B_2 - w_{t-1} $$
$$ \Delta Y_t = B_2 + w_t - w_{t-1} $$
Now it is clearly stationary since we have:
$$ E[\Delta Y_t] = B_2~~,~~VAR[\Delta Y_t]=2\sigma^2 $$
and the covariance depends on time lag only.
A: Since you only need help with the covariance of the process $y_t=x_t-x_{t-1}=\Delta x_t$.
Note that $y_{t+k} = \Delta x_{t+k}=B_2+w_{t+k}-w_{t+k-1} \quad \forall k \in \mathbb R$
Now to find the covariance start by its definition and use its properties i.e.
\begin{align*} cov(y_t,y_{t+k})&=cov(B_2+w_{t}-w_{t-1},B_2+w_{t+k}-w_{t+k-1})\\
&= cov(w_t-w_{t-1},w_{t+k}-w_{t+k-1})
 \end{align*}
From here its straightforward to find what you are looking for.
