# Why transform uniform random variables to standard normal random variables?

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

• 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 ? Mar 17, 2020 at 23:58
• @mlofton, sorry I should have omitted the term residual, have updated. Mar 18, 2020 at 0:03
• Where have you seen someone doing this? Mar 18, 2020 at 0:49
• Updated with relevant reference, but I have seen this same idea in other papers Mar 18, 2020 at 1:12
• 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.
– whuber
Mar 18, 2020 at 12:27