I'm preparing the exam for "stochastic models" and I encountered this exercise which is giving me a lot of problems:

Let $$X_t=\phi X_{t-1}+\epsilon_t, ~~~~~~~~~~\epsilon_t \sim WN(0, \sigma^2)$$ with $\phi=0.5$, $\sigma^2=1.21$ and define $W_t=X_t-X_{t-1}$

  1. Compute the expectation and the variance of $W_t$
  2. Compute $\operatorname{Corr}(W_t,W_{t-1})$
  3. Suppose that $\phi$ and $\sigma^2$ are unknown, with $-1<\phi<1$. Provide the expression of the autocovariance and the autocorrelation functions of $W_t$.

My attempts are the followings:

Expectation of $W_t$

Knowing that $E(X_t)=E(X_{t-1})$,


Variance of $W_t$

If $\operatorname{Var}(X_t)=\operatorname{Var}(X_{t-1})$


Correlation between $W_t$ and $W_{t-1}$

If $\operatorname{Var}(W_t)=\operatorname{Var}(W_{t-1})$


replacing the values we have $$\rho(1)=\text{a number divided by 0}$$

Is it possible??

Autocovariance and autocorrelation function if $\phi$ and $\sigma^2$ are unknown

I have no problems with finding the autocorrelation function. But first, I have to find the autocovariance one and I tried in this way:

$$\operatorname{Cov}(W_t,W_{t-1})=E[W_t-E(W_t)W_{t-1}-E(W_{t-1})~~~~~~~~~~~~~~~~~$$ $$=E[W_t W_{t-1}]~~~~~~~~~~~~~~~~~~~~~~~~~$$ $$=E[X_t-X_{t-1} X_{t-1}-X_{t-2}]$$

From this point, I have no idea how to continue (if there are no mistakes but I think there are a lot of ones).

I tried to put my effort into this exercise but I'm stuck in several aspects as you can see. If someone would explain to me my mistakes and some steps to follow in order to conclude this exercise, would be fantastic.

  • 1
    $\begingroup$ Your expressions for $\mbox{Var}W_t$ and $\mbox{Cov}(W_t,W_{t-1})$ don't agree with basic formulas for variance of linear combinations of correlated variables en.wikipedia.org/wiki/Variance#Basic_properties and en.wikipedia.org/wiki/… $\endgroup$ – Jarle Tufto Jan 18 at 19:29
  • $\begingroup$ @Xi'an --- $X_{t-1}$ and $X_{(t-1)}$ are the same object. Initially it was a typo but now I edited the function. About your 2nd question " Is $W$ in $\epsilon_t \sim WN(0, \sigma^2)$ the same as $W_t$", the answer is: No, because in this case, $WN$ means "white noise" while $W_t$ is a function ($W_t=X_t - X_{t-1})$. About your 3rd question, the answer is always no. About you last question, in $X_t=\phi X_{t-1}+\epsilon_t$, $\epsilon_t$ is not conditional on the function $W_t$ $\endgroup$ – Martina Marty Jan 18 at 19:34
  • $\begingroup$ Why do you assume second order stationarity? It does not appear in the model. $\endgroup$ – Xi'an Jan 18 at 19:50

You solution collapses after the calculation of the variance of $W_t$: $$var(W_t)=var(X_t-X_{t-1})=2var(X_t)-2cov(X_t,X_{t-1})=2(1-\phi)var(X_t)$$

since $C_x(h)=cov(X_t,X_{t-h})=\phi^{|h|}var(X_t)$ based on your other question. $var(X_t)$ also can be found as $\frac{\sigma^2}{1-\phi^2}$, and let's call this $V_x$.

So, $cov(W_t,W_{t-1})=cov(X_t-X_{t-1},X_{t-1}-X_{t-2})=-(\phi-1)^2V_x$, correlation follows easily.

And, finally, $cov(W_t,W_{t-h})=cov(X_t-X_{t-1},X_{t-h}-X_{t-h-1})=2C_x(h)-C_x(h-1)-C_x(h+1)$.

P.S. $X_t$ is obtained via LTI filtering of a WSS process, so it is also WSS.

  • 1
    $\begingroup$ I made some stupid mistakes but, after this explanation, I understood them (hoping to avoid them in the future). Thank you so much for your help $\endgroup$ – Martina Marty Jan 18 at 20:18

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.