1

I have a process, $Y_t = X_t - X_{t-1}$.

$\{X_t\}$ is a stationary process with autoccovariance function $\gamma_{X}(k)$.

I wish to find the autocovariance function of the process $\{Y_t\}$ in terms of $\gamma_{X}(k)$.

I think I am correct in saying $\mathbb{E} (Y_t) = 0$.

So, $$\gamma_{Y}(k) = \mathbb{E}[\{Y_t - \mathbb{E}(Y_t)\}\{Y_{t+k} - \mathbb{E}(Y_{t+k})\}]$$ $$= \mathbb{E}(Y_tY_{t+k})$$ $$= \mathbb{E}[\{X_t -X_{t-1}\}\{X_{t+k} -X_{t+k-1}\}]$$

From here I have expanded, and can see we get different results for different $k$, however I'm not entirely sure that what I have done here is heading in the right direction.

Any comments would be appreciated,

Much thanks.

Edit: Continuing.

Let's say $\mathbb{E}(X_t) = \mu_X$

Then, $\gamma_X(k) = \mathbb{E}(X_tX_{t+k}) - \mu_X^2$

So, $$\mathbb{E}[\{X_t -X_{t-1}\}\{X_{t+k} -X_{t+k-1}\}]$$ $$= \mathbb{E}(X_tX_{t + k}) - \mathbb{E}(X_{t-1}X_{t + k}) - \mathbb{E}(X_tX_{t + k - 1}) + \mathbb{E}(X_{t-1}X_{t + k - 1})$$

$$= (\gamma_X(k) + \mu_X^2) - (\gamma_X(k+1) + \mu_X^2) - (\gamma_X(k-1) + \mu_X^2) + (\gamma_X(k) + \mu_X^2) $$

$$ = 2\gamma_X(k) -\gamma_X(k+1) - \gamma_X(k-1)$$

  • 1
    Yes, you are heading in the right direction. So far, it looks fine. – saz Oct 19 '14 at 12:58
  • @saz Okay so, another question if you wouldn't mind. Are we able to assume $X_t$ is independent of $X_s$ for $t \neq s$? I am thinking so, since ${X_t}$ is a stationary process. –  Oct 19 '14 at 13:53
  • 1
    No. This would imply that the autocovariance function of any stationary process with $\mathbb{E}X_1=0$ equals $0$. – saz Oct 19 '14 at 14:07
  • @saz I've edited my post to what I think may be the complete solution, if you wouldn't mind looking. –  Oct 19 '14 at 14:26
  • 1
    Yes, your calculation is correct. Note that if you write $$\mathbb{E}((X_t-X_{t-1})(X_{t+k}-X_{t+k-1})) = \mathbb{E}\bigg[((X_t-\mathbb{E}X_t)-(X_{t-1}-\mathbb{E}X_{t-1}))((X_{t+k}-EX_{t+k})-(X_{t+k-1}-\mathbb{E}X_{t+k-1}))\bigg],$$ you don't have to mess around with the $\gamma_X^2$. The result is of course the same. – saz Oct 19 '14 at 20:21

0 Answers0