# Explanation of Two Propositions on Random Variables from Rice 3rd Edition

There are two propositions I'm struggling with interpreting what they really mean. Perhaps I'm expecting more even though they appear simplistic:

$$X$$ is defined as a random variable, with density $$f$$ and cdf $$F$$. With the appropriate conditions to make them "nice".

First is my understanding of Proposition C. Z is usually reserved for a standard normal random variable and I took it as such. So in my mind what proposition C is saying is "Let our standard normal random variable be a function of an arbitrary CDF. Then our standard normal random variable has a uniform distribution on $$[0,1]$$." But here is where it get's cloudy for me.....we have this standard normal random variable, this standard normal random variable is going to have a distribution, since it has been declared as a "standard normal random variable" I would assume that its distribution would be "normally distributed". The values for this standard normal random variable, $$Z$$, happen to come from the CDF of some other random variable, $$X$$. I see the proof and the steps of the proof are very simple and sound, but the concept is not clicking with me.

With regards to Proposition D I understand what that is saying. In practice we would start with a CDF, $$F(x)$$ and rearrange things in a way to apply a uniform random generator.

Help with interpreting Proposition C?

• You've correctly identified the dissonance. While you are used to seeing $Z$ as a standard normal variable, it doesn't necessarily always mean such. Here it does not: $Z$ is defined as the random variable $F(X)$. If it helps, just replace every occurrence of $Z$ with $Y$ in the proposition. – Greg Martin Jun 19 '20 at 18:54
• In Proposition C, $Z$ is not assumed to be a standarnd normal random variable. It is just defined by $Z := F(X)$. – user6247850 Jun 19 '20 at 18:54
• @dc3rd No, it is crucial that $F$ is CDF of $X$ to be able to say anything about $F(X)$. Moreover, it does not hold for any $X$, but it is important that $F$ is continuous (it not necessarily needs to have inverse, tho (but the proof goes a little bit different, with generalised inverse function)) – Dominik Kutek Jun 19 '20 at 19:39
• It should be specified that $F$ is CDF of $X$. Writing $F(X)$ to mean that $F$ is CDF of $X$ is incorrect, because $F(X)$ is meant to be a random variable, such that $F(X)(\omega) = F(X(\omega))$. You can always write $F_X$ to denote the CDF of $X$ which is as far as I know, one of possible ways to specify so. – Dominik Kutek Jun 19 '20 at 20:06
• And answering to your second question, yes, if $F_X$ is continuous, then you can conclude that $Z:= F_X(X)$ is a uniform random variable on $[0,1]$ – Dominik Kutek Jun 19 '20 at 20:11

Proposition C should be stated: If $$X$$ is a random variable with continuous CDF $$F_X$$ then random variable $$Z:= F_X(X)$$ is uniformly distributed on $$(0,1)$$ (it's the same as uniformy distributed on $$[0,1]$$)
The proof that is in the book, assumes that $$F$$ has inverse $$F^{-1}$$ (just to have simplier proof, but it isn't necessary for proposition C to hold). Moreover, in the proof, it should be said that those equations are true for $$z \in [0,1]$$, otherwise we have for $$z<0$$: $$\mathbb P(Z \le z) = \mathbb P(F_X(X) \le z) = 0$$ (since $$F_X$$ is a function with values in $$[0,1]$$) and for $$z>1$$ we have $$\mathbb P(Z \le z) = \mathbb P(F_X(X) \le z)=1$$, since as we said before, function $$F_X$$ always takes values less or equal to $$1$$.
Proposition $$D$$ is okay, but we need to know that $$F$$ is right continuous function, non-decreasing with values in $$[0,1]$$. Proposition assumes that $$F^{-1}$$ exists, which again isn't neccesary, but then you need the notion of generalised inverse to write the proposition $$D$$.
• Hi I was just going over the contents of this solution again and wanted to make sure I have the proper comprehension with regards to Proposition C. So in my words Proposition C is saying: "$Z$ is the random variable defined as the set of values that the cdf of $X$ takes on at all points defined for random variable $X$. This random variable $Z$ is uniformly distributed over $(0,1)$" – dc3rd Jul 1 '20 at 22:45