CDF(X)^2 versus CDF(X^2): What's the difference? To explain clearly what my question is imagine this simple example.
"There are 101 boxes each numbered 1,2,3... and so on. An android can randomly choose one of these boxes, but each one with a probability proportional to its number."
import numpy as np
from empiricaldist import Pmf, Cdf


NB: The function Pmf() from empiricaldist simply creates a pandas series with a convenient .normalize() method.
Cdf() works similarly. You don't need to know these functions to follow the question.

The PMF of the event of choosing a box is Px
Px = np.linspace(0,1, 101)
Px = Pmf(Px)
Px.normalize()

While the PMF of the event of choosing 2 times the same box is Px2
Px2 = Pmf(px**2)  


Now I can compute 2 CDF one for each event
Cx = Px.make_cdf()
Cx2 = Px2.make_cdf()

But I can also calculate the following distribution
Cx_2 = Cx**2


Now my question is: how to interpret Cx2 and Cx_2?
Each probability of a quantity 'Q' of Cx can be interpreted as the probability of choosing a box with value equal or lower than Q.
For Cx2 each value is the probability of choosing twice a box with the same value lower than Q.
How can be Cx_2 interpreted since it is different from Cx2?
Can it be the probability of choosing twice a box with a value lower then Q but not necessarily the same value?

Your question is not quite clear, but seems to ask about the difference between:

*

*The probability of seeing $x$ or less in both of the two draws, your yellow Cx_2 CDF curve, i.e. $\mathbb P(\max(X_1,X_2) \le x)$

*The conditional probability of seeing $x$ or less in both of the two draws given that both draws are the same, your green Cx2 curve, i.e. $\mathbb P(\max(X_1,X_2) \le x\mid X_1=X_2)$
They are different distributions, and you are more likely to see a larger number in the first (i.e. the yellow curve is to the right of the green curve).  In the first $\mathbb P(\max(X_1,X_2)=1)\approx 0.0000000377$ and $\mathbb P(\max(X_1,X_2)=101)\approx 0.0388$, while in the second $\mathbb P(\max(X_1,X_2)=1 \mid X_1=X_2)\approx 0.00000287$ and $\mathbb P(\max(X_1,X_2)=101\mid X_1=X_2)\approx 0.0293$
Note that neither of these are the CDF of $X^2$, which would have support on the values $1,4,9,\ldots, 10201$, and would have $\mathbb P(X^2=1)=\mathbb P(X=1)\approx 0.000194$ and $\mathbb P(X^2=10201)=\mathbb P(X=101)\approx 0.0196$
