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I have a non-stationary series of bond yields $x_{t}$ that are logged and differenced $$y_{t}\equiv ln\left(x_{t}\right)-ln\left(x_{t-4}\right) $$ From that, I get a series of forecasted values $\widetilde{y}_{t} $. Since ultimately what I want is actual yields, I need to back out the unlogged levels.To do this, I need first to un-difference, then take exponentials. The second step is obvious. What about the first?

So I want $$ln\left(\widetilde{x}_{t}\right)=\widetilde{y}_{t}+ln\left(\widetilde{x}_{t-4}\right)$$ but all I have is $\widetilde{y}_{t}$, because I forecasted the $y_{t}$ rather than $ln\left(x_{t}\right)$.

Any thoughts? Is it possible? If so how?

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3 Answers 3

You can recover the levels of $X$ at time $t$ if you have $X(0)$ as well as all first differences until $X(t)$. Then $X(t) = X(0) + \sum_{i=1}^t (X(i)-X(i-1))$.

In your case $X:=ln(Y)$, apply the above algorithm to find $ln(Y(t))$ from which $Y(t)=e^{X(t)}$.

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You have $$ln\left(x_{t-4}\right)$$ so you don't need to get an estimate for $$ln\left(\widetilde{x}_{t-4}\right)$$ just plug that in, add your forecast for $y_{t}$, then take the exponential.

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Simply calculate the cumulative sum. But you still need the intercept/a constant. From differences you cannot get the level of the original series. Tiny example: series: 5,6,7,8 / differences: 1,1,1 / cumulative sum: 1,2,3

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