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

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$X$ 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?

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    $\begingroup$ 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. $\endgroup$ – Greg Martin Jun 19 '20 at 18:54
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    $\begingroup$ In Proposition C, $Z$ is not assumed to be a standarnd normal random variable. It is just defined by $Z := F(X)$. $\endgroup$ – user6247850 Jun 19 '20 at 18:54
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    $\begingroup$ @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)) $\endgroup$ – Dominik Kutek Jun 19 '20 at 19:39
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    $\begingroup$ 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. $\endgroup$ – Dominik Kutek Jun 19 '20 at 20:06
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    $\begingroup$ 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 $[0,1]$ $\endgroup$ – Dominik Kutek Jun 19 '20 at 20:11
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In proposition C and D, what is missing are the assumptions.

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 $D$.

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  • $\begingroup$ 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 $(0,1)$" $\endgroup$ – dc3rd Jul 1 '20 at 22:45

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