# Use of expression "statistically significantly predicts" based on in-sample analysis

Suppose one estimates a linear time series model $$y_t=\beta_0+\beta_1 x_{t-1}+\varepsilon_t$$ and finds that $$\hat\beta_1>0$$ and the $$p$$-value associated with $$\hat\beta_1$$ is lower than the chosen significance level. Can one say, without any caveats, that

$$x$$ statistically significantly predicts $$y$$?

Similarly, can one say

$$x$$ positively predicts $$y$$?

My main concern is that the claim about prediction is based on in-sample analysis without stating the implicit assumptions that are required to make a conclusion about out-of-sample results from in-sample results. Another, minor concern is that the claim is based on a significance test of $$\beta_1$$ rather than a measure of change in prediction errors of $$y$$ when $$x_1$$ is added to the model. However, the latter is probably not a problem as the significance of $$\beta_1$$ probably implies the prediction errors of $$y$$ will be decreased by use of $$x_1$$ in the model.

• “Significantly predicts” means predictions better than random, which is a truly boring hypothesis. Importantly, any threshold for statistical significance is BS. Commented Oct 1, 2023 at 12:48

In your minor concern, I guess you are essentially comparing "$$\beta_{1}$$" and "$$\beta_{1} * x_{1}$$". In Statistics it often the relation of concern, not prediction. So as long as $$x_{1}$$ significantly affects $$y$$, how much it changes $$y$$ is of less concern.
• What does predict coefficients mean? you need a distribution for coefficients: coefficients are constants in frequentist statistics, and I am not interested in a Bayesian interpretation of this question. You can make a OOS prediction, but technically speaking you cannot say anything about out-of-sample statistical significance. Yes, you can using tests like Diebold-Mariano. how much it changes $y$ is of less concern: my question is specifically about prediction. Commented Oct 19, 2019 at 16:24