I have the equation $(1 − B)y_t = (1 − B)a_t$, can I conclude from this that $y_t=a_t$ for all $t$?

Question

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.

Answer 1

Look at the comment by @Richard Hardy:

What the equation $(1-B)y_t=(1-B)a_t$ says is that the increments in $y_t$ equal the increments in $a_t$. Does that sound sufficient for $y_t\equiv a_t$?

To make it more clear, make a numerical example where two series have the same increment:

1 2 3 4 5 6 7 8  9  10
3 4 5 6 7 8 9 10 11 12  

and the answer should be clear. An analogy with calculus is that two functions might have the same derivative without being equal.