Recovering Yield Levels from Forecasts of Log Differences
Summary
The document considers how to turn forecasts of a logged, differenced bond-yield series back into yield levels. Its example uses a four-period log difference: recovering the next logged level requires adding the forecasted change to the logged yield from four periods earlier, then exponentiating to return to the original scale.
The answers distinguish this case from reconstructing an entire series from consecutive first differences. In that setting, cumulative differences recover changes in level, but an initial level is still needed; differences alone cannot identify the absolute level. For the stated lagged difference, the known historical value at the matching lag supplies the anchor. These are algebraic reconstruction steps, not a forecasting model or evidence that forecasts are accurate. The discussion does not address uncertainty, bias from exponentiating forecasts, or how to combine forecasts across horizons.
Key ideas
- A forecast of a logged difference must be combined with an appropriate logged level to reconstruct the next level.
- For a four-period difference, the relevant anchor is the logged observation four periods earlier.
- Exponentiating the reconstructed logged value returns the forecast to the original yield scale.
- Cumulative sums of consecutive differences recover changes, but an initial level is required to identify the series level.
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Full text
# how to back out levels from a forecast of differenced series
# how to back out levels from a forecast of differenced series
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?
## Answer by user2763361 (score 1)
https://quant.stackexchange.com/a/10321
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 $Y(t)=e^{X(t)}$.
## Answer by John (score 1)
https://quant.stackexchange.com/a/10322
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.
## Answer by Konsta (score -1)
https://quant.stackexchange.com/a/10320
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: 1,2,3Shown in full with attribution under the source's licence. Licence: CC BY-SA 4.0 (Stack Exchange)
This summary was written by Stratmill's research agent from the original; it is not a copy of the source.