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I have come across a few studies on the topic of forecasting. let $X_t = Y_t|D_{t - 1}$, where $\{Y_t: t = 1,2,3, \dots\}$ is the series to be forecast one step ahead and where $D_t$ represents all data available to the forecaster up to and including time $t$. in a system closed to external information $D_t = \{D_0, y_i,\dots, y_t\}$ then $X_t$ will have a forecast distribution functions $F_t, t = 1,2,3,\dots$ given by our model. Suppose $F_t$ is continuous it follows immediately that when our model is correct.

$$ U_t = F_t(X_t) $$ with $U_t$ independent uniform [0,1] random variables and

$$ V_t = \Phi^{-1}(U_t) $$ are independent normal random variables with $\mathcal{N}(0,1)$ distribution, where $\Phi$ is the standard normal cdf. My question is, why not just do test on the uniform random variables $U_t$? rather than transforming and doing normality checks on $V_t$

The above is taken from Smith (1985)

Smith, J Q (1985) Diagnostic Checks of Non-standard Time Series Models. Journal of Forecasting. 4, pg 283-291

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  • $\begingroup$ Hi: I don't follow because if you're forecasting residuals, where did the residuals come from ? You need to forecast in the first place in order to obtain residuals ? $\endgroup$ – mlofton Mar 17 '20 at 23:58
  • $\begingroup$ @mlofton, sorry I should have omitted the term residual, have updated. $\endgroup$ – Cyrillm_44 Mar 18 '20 at 0:03
  • $\begingroup$ Where have you seen someone doing this? $\endgroup$ – jbowman Mar 18 '20 at 0:49
  • $\begingroup$ Updated with relevant reference, but I have seen this same idea in other papers $\endgroup$ – Cyrillm_44 Mar 18 '20 at 1:12
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    $\begingroup$ Pure speculation: the hope is the resulting transformed series can be modeled as a Gaussian process. These are characterized by their first two moments only, whereas processes with Uniform marginals do not have such a neat, simple characterization. $\endgroup$ – whuber Mar 18 '20 at 12:27
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On the one hand, it's hard to read the mind of an author of a paper that is 35 years old.

On the other hand, the reason may simply be that there are far more tests for a normal than for a uniform distribution. This in turn may be due either to more interest in the normal than in the uniform distribution (the various probabilistic laws imply that we will more often find a normal than a uniform "in the wild"), or to the fact that the normal distribution may be more easily tractable, though this second point is a wild guess from my side.

And of course, back in 1985, there was relatively more interest in asymptotics than later, which would have reinforced the first point.

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  • $\begingroup$ Thanks, I was thinking along these lines as well, but just wanted to run by the community. If there is no better answer by the end of the week I will give you the prize. $\endgroup$ – Cyrillm_44 Mar 19 '20 at 1:00

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