# Why do we say a probability integral transformed variable is uniform on the interval [0, 1]?

I am reading and trying to learn about the probability integral transform and some of its uses. From the CV question PIT on a sample with m bins, and KS test used to estimate a good value for m, the probability integral transform of random variable $X$ with size $n$ looks very simple:

$$U_{i} = \frac{R_{i}}{n+1}$$

where $R_{i}$ is the rank of the $i^{\text{th}}$ observation in $X$.

I can see where $U$ is uniformly distributed, but in Quantlbex's answer, they write "By construction the distribution of the transformed variables is uniform between 0 and 1."

My doubtlessly naïve question is: shouldn't that be "uniform between $\frac{1}{n+1}$ and $\frac{n}{n+1}$?" Or do we simply say "uniform between 0 and 1" because those are the limiting bounds for arbitrarily high $n$?

The probability integral transformation does not use empirical CDFs; it uses the actual CDF. The transformation is not a function of the sample size. The prob. integral transformation of $X$ is $F(X)$ where $F$ is the CDF of $X$.
• Thanks Frank, that makes sense. Can you help me understand what this formula for $U_{i}$ is then? (Hope I am not being dense, this is new to me). Sep 21 '14 at 20:14
• You appear to be wanting to do a rank transformation, and converting it to an (ultimately) $[0,1]$ range. You didn't tell us why. And marginal ranks are only usually of interest when there is only one other variable of interest, e.g. when computing Wilcoxon or Spearman statistics. The only other example I can think of is that your type of calculation is used in some definitions of sample quantiles (but not exactly as you've stated it). State your ultimate goal. Sep 21 '14 at 20:33