Suppose $ X_t = (\alpha + \beta{t})S_{t} + e_{t}$.

Define $C_{12} = 1 - B^{12}$ as the backshift operator.

Assume $ S_{t}=S_{t-12}$ for all $t$.

Assume that $e_{t}$ is white noise.

Is $C*X_{t}$ stationary? If not, give a different backshift operator that makes the series stationary.

When I solved this I got: $E[CX_{t}] = 12*\beta*S_{t}$ which is a constant.

Also I got that

$$var(CX_{t})= var((\alpha + \beta{t})S_{t} + e_{t} - (\alpha*S_{t-12} + \beta*t*S_{t} - S_{t}*12*\beta + e_{t-12})) $$ $$ = var(e_t +12 \beta S_t - e_{t-12}) = cov(e_{t},e_{t+12})$$

I use $S_{t}$ since $S_{t} = S_{t-12}$ for all $t$.

Apparently this is wrong as the series is nonstationary using this backshift operator.

What am I doing wrong? How is this series nonstationary and how do I find the correct backshift operator that makes it stationary?

  • $\begingroup$ Your variance looks like it has a typo, or rather two typos. $\endgroup$
    – Glen_b
    Mar 11, 2013 at 4:17
  • $\begingroup$ What do you mean by $var(e_{t},e_{t+12})$? $\endgroup$
    – Glen_b
    Mar 11, 2013 at 4:54
  • $\begingroup$ It's the covariance between the white noise at time t and the white noise at time t+12. And this obviously is not dependent on t only on the lag between t and t+12. I don't how I can make myself more clear $\endgroup$
    – phil12
    Mar 11, 2013 at 4:56
  • $\begingroup$ How do you get $var((\alpha + \beta{t})S_{t} + e_{t} - (\alpha*S_{t+12} + \beta*t*S_{t} + S_{t}*12*\beta + e_{t+12})) = cov(e_{t},e_{t+12})$? $\endgroup$
    – Glen_b
    Mar 11, 2013 at 4:57
  • $\begingroup$ After differencing, you get $var(e_{t} -S_{t}12\beta - e_{t+12})$ and this is constant for all t. I realize my mistake My notation was bad. $\endgroup$
    – phil12
    Mar 11, 2013 at 5:00

1 Answer 1


Edited to reflect the changed question:

$$C_{12}X_{t} = 12 \beta S_{t-12} + e_t - e_{t-12}$$

The distribution of $e_t - e_{t-12}$ is not changing with time, but note that the mean is changing with time, since $S_t$ isn't constant.

To figure out what operator would make the original series stationary, consider what you'd do to $12 \beta S_{t-12} + e_t - e_{t-12}$ to make it stationary

  • $\begingroup$ What convinces you that the solution is correct? $\endgroup$
    – Glen_b
    Mar 11, 2013 at 4:28
  • $\begingroup$ "I lost points" only indicates that the person allocating points thought it was non stationary. Do you have some convincing explanation as to why you should agree? (NB: Are you sure you transcribed everything correctly? Please double check, especially your variance.) $\endgroup$
    – Glen_b
    Mar 11, 2013 at 4:31
  • $\begingroup$ et is white noise. I think my explanation makes sense. The variance simply depends on the lag and the mean is constant since St = St-1 $\endgroup$
    – phil12
    Mar 11, 2013 at 4:34
  • $\begingroup$ That's really weird. Thanks for clarifying my understanding of my problem. All the information I gave you was correct. $\endgroup$
    – phil12
    Mar 11, 2013 at 21:57
  • $\begingroup$ It's always possible I've made an error too (I do that sometimes), but in this case I doubt it. Mistakes creep in to even the best-run courses. Can you get a worked solution? $\endgroup$
    – Glen_b
    Mar 11, 2013 at 21:59

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