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I have a time series of a variable V1 with seasonality and a strong trend. The trend however seems to be closely related to (and caused by) the trend of another time varying variable (V2). As V2 grows i.e shows a positive trend, the trend on V1 declines (see chart). enter image description here

I want to use V2 as an exogenous variable in the time series forecast of V1. I am using prophet for this where I forecast V2 and then use it as an exogenous variable to forecast V1. But, since V2 has a trend that affects the trend of V1 does it make sense to use V2 as an exogenous variable at all? I can simply run a forecast with and without the exogenous variable to see what difference it makes. It might not be much, but I am more interested in knowing if there is a fundamental flaw in using exogenous variables with trends.

The question is similar to this one, but it was not very clear to me.

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Lets say you want to predict $y_t$ based on its past $y_{0},y_{1},...,y_{t-1}$, as well as based on exogenous variable $x_t$. Models such as the additive Prophet model treat $x_t$ as independent over time. Hence, Prophet will not notice a trend in the process $\{x_0,x_1,...,x_t\}$ because it only looks at $x_t$ and then immediately forgets $x_t$ (see here)

However, this does not mean that you cannot use $x_t$ to predict $y_t$. If $x_t$ is a good external regressor for $y_t$, then there must exist some function $f$ such that $f(x_t)\approx y_t$. This does not change if there is a trend in $\{x_0,x_1,...,x_t\}$. For example, if your exogenous variable is actually a straight line with slope $k_1$ and y-intercept $d_1$, and your target variable is a straight line with slope $k_2$ and y-intercept $d_2$, then the solution is $f(x_t)=\frac{k_2}{k_1}(x_t-d_1)+d_2=y_t$. Since Prophet treats all external variables as linear regressors, it can solve this particular problem.

For the data you posted, I do not think that there is a linear function from the exogenous variable to the target variable. So Prophet should not be able to learn this exactly - in fact, it cannot be more exact than the best linear function from $x_0,...,x_t$ to $y_0,...,y_t$.

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