Problem
Assume the we are given a stationary process $x_t$. That is $$E(X) = const$$ $$ Cov(x_t, x_{t+k}) = f(k)$$ It is known that $E(x_t) = 4$, $Var(x_t) = 16$ and $Cov(x_t, x_{t-1}) = 4 $.
We go through each $t$ and with probability $0.3$ delete $x_t$. Let's call new sequence $y_t$. (In other words, $(y_t)$ will contain some of elements of $(x_t)$)
Is $y_t$ a stationary process?
Find $ \Bbb Var(y_t), \Bbb Cov(y_t,y_{t-1})$.
My thoughts
Of course, the expected value will stay the same because it doesn't depent on other $y_t$. However, I think, the covariance won't be the same. My counterexample, assume there are $$(x_t) = x_1, x_2, x_3, x_4, x_5,\dots$$ $$(y_t) = x_1, x_2, x_3, x_4, x_6, \dots $$
that is we droped only $x_5$ among first five variables.
We know that as $(x_t)$ is stationary process
$$ Cov(x_1,x_3) = Cov(x_3,x_5) = f(2)$$
however as if $(y_t)$ is stationary process:
$$ Cov(x_1,x_3) = Cov(x_3,x_6) = f(3) \implies f(2) = f(3)$$
It may happend the equality $f(2) = f(3)$ is not held. Hence, $y_t$ is not stationary process.
Is my logic right? I don't proceed to the variance utill I find out whether $y_t$ is stationary or not.
Hint
We don't know which $x_i$ will be in $(y_t)$ and which won't. Let us use indicator $a_i = a_i(i = j)$:
Then some $y_t$ is $y_t = x_j = \sum_i a_i x_i = 0 \cdot x_1 + \dots + 1\cdot x_j + 0\cdot \dots + \dots$
Then $$\Bbb E(y_t) = \Bbb E(\sum_i a_i x_i) = \sum_i \Bbb E( a_i x_i) = \sum_i \Bbb E( a_i) \Bbb E(x_i)$$ Remember that $\Bbb E(a_j) = (1 - 0.3)\cdot1 + 0.3\cdot0 = 0.7$
This should help to find out what the process is stationary.