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This may seem like a pretty simple question, but I want to make sure I am getting this right because it seems pretty foundational.

I'm reading this note on conformal prediction. In the very first section the scenario is this:

We collect i.i.d pairs $(X_i, Y_i)$, $i=1,\dots,n$ from distribution $P$ on $\boldsymbol{\mathcal{X}} \times \boldsymbol{\mathcal{Y}}$. Given a specified "error tolerance" $\alpha \in (0, 1)$, the goal is to find a prediction set

$$ C_\alpha(X) : \boldsymbol{\mathcal{X}} \to \{\text{subsets of }\boldsymbol{\mathcal{Y}}\} $$

with the property that for a new i.i.d pair ($X_{n+1}, Y_{n+1}$):

$$ \mathbb{P}(Y_{n+1} \in C_{\alpha}(X_{n+1})) \ge 1 - \alpha \tag{1} $$

where the probability is taken over all of our data $(X_1, Y_1) \dots (X_{n+1}, Y_{n+1})$.

This last line is where I want to focus on. This is my current interpretation, and please let me know if it is wildly incorrect or correct!

First let $x_i$ and $y_i$ denote a particular fixed realization of $X_i$ and $Y_i$; for example by going out and collecting data or measuring or observing. How I think about this line is the following: let's say I go out and collect the data $(x_1, y_1), \dots (x_n, y_n), (x_{n+1}, y_{n+1})$. I plug $x_{n+1}$ into my prediction set function $C_{\alpha}$ and get a prediction set. I then check if $y_{n+1}$ is in that prediction set. If it is I write down a check mark; if it is not I write down an x.

The property $(1)$ is saying that if I run this same exact experiment an infinite number of times (i.e. go collect the data, etc.), the proportion between check marks and x's will be $1-\alpha$.

Thanks in advance!

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Your understanding is correct insofar as the probability refers to repeated realisations of the whole sequence $((X_1,Y_1),\ldots,(X_{n+1},Y_{n+1})$. You are probably aware though that property (1) will be fulfilled if data are generated from an idealised model ($P$); what actually happens if you repeat experiments in reality is anyone's guess.

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