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I have the equation $(1 − B)y_t = (1 − B)a_t$, am I able to say that this equation implies that $y_t = a_t$ for all t?

$B$ is the backward shift operator where $Bz_t = z_{t-1}$ and $y_t = y_{t-1} + \mu + \sigma a_t$ where $a_t$ is a sequence of independent random variable $N(0,1)$, $\sigma$ is the volatility, and $\mu$ the mean growth rate.

I know: $y_t = y_{t-1} + \mu + \sigma *a_t$

$y_t -\mu * t= y_{t-1} - \mu * (t-1) + \sigma *a_t$

where $z_t = z_{t-1} + \sigma * a_t$

so $a_t = (z_t - z_{t-1})/ \sigma$

I Guess I just don't know how to show whether or not $y_t$ and $a_t$ are defined for the same t. I know according to time series that t is today where as t-1 was yesterday (or the past) and t+1 is the future.

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  • $\begingroup$ Is yt a subscripted y, like $y_t$? $\endgroup$ – Matthew Drury Jan 21 '16 at 22:33
  • $\begingroup$ yes sorry, I just tried to fix it. I forgot about the notation $\endgroup$ – lee yu Jan 21 '16 at 22:34
  • $\begingroup$ I tried to fix it up for you, but am not toally sure I captured what you intended. $\endgroup$ – Matthew Drury Jan 21 '16 at 22:36
  • $\begingroup$ Is this a question from a course or textbook? If so, please add the [self-study] tag & read its wiki. $\endgroup$ – gung Jan 21 '16 at 23:20
  • $\begingroup$ This is most definitely a self-study, I am preparing for an up coming test, essentially I don't know how to use time series in order to conclude if yt=at for all t. what do you mean read its wiki? @gung $\endgroup$ – lee yu Jan 21 '16 at 23:24

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