Questions tagged [kolmogorov-axioms]

Constitute Kolmogorov's mathematical definition of a probability space. It is a triplet $(\Omega, \mathcal F, P)$ where $P$ has to satisfy three axioms: non-negativity, adding to one, and countable addivity.

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Definition of recurrent stochastic process, in general

This interesting question: Recurrence definition for a Markov chain gives the definition of a recurrent state for some discrete process. I was wondering, in the case of a continuous (time) process, ...
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Do uncalibrated "probability" predictions satisfy Kolmogorov's axioms?

Let's say we have some binary variable of interest and fit a model to predict the probability of the two classes, say a logistic regression or a "classification" neural network. This model ...
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Kolmogorov axioms consequences

Earlier today in my stochastic processes lecture, the prof mentioned that there does not exist a probability measure P(A) defined for all subsets of [0,1] which would satisfy all the 3 Kolmogorov ...
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Axiomatized formal foundation of statistics

I'm learning statistics in foreign language. Being aware of possibly different foundations (e.g. frequentists v.s. Bayesian), I hope to acquire one formally axiomatized foundation of statistics. That ...
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Show that if A and B are disjoint, then A ∩ C and B ∩ C are also disjoint

I need to show using Venn diagram. Is my solution correct, I understand from the Venn diagram it would be right, but can someone explain in a more formal/mathematical way why this implication is true? ...
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How to prove $P(\emptyset)=0$ from the axioms of probability?

A text states that you can prove that a probability of a null set is 0 through one of the axioms of probability. I know the three axioms, but I fail to employ these axioms to prove the above. I ...
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Explain to my intuition: New axioms for rigorous Bayesian probability

I stumbled upon this very interesting article : New axioms for rigorous Bayesian probability Whose aim is to provide new axioms for probability theory ...
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Can we get uniform distributions on infinite spaces by giving up infinite additivity

I am wondering whether it is possible to translate the idea of drawing a number randomly from the set of all natural numbers. If we have infinite additivity as an axiom this obviously does not work. ...
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If the sum of the probabilities of events is equal to the probability of their union, does that imply that the events are disjoint?

Axiomatically, probability is a function $P$ that assigns a real number $P(A)$ to each event $A$ if it satisfies the three fundamental assumptions (Kolmogorov's assumptions): $P(A) \geq 0 \ \text{for ...
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How do we characterize probabilities on this infinite sample space

Consider the following set, $S$, of infinite sequences: $$S = (a_n)_{n \in \mathbb{N}} \\\text{where } a_n \in \mathbb{N} \text{ and } a_n \leq nk+1 \text{ with }k\text{ a positive integer}$$ in ...
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Variations in Bayes' theorem in the denominator

I've got a question regarding the different variations of the denominator in Bayes' Theorem. I'm led to believe the denominator of this equation is = P(B) I did some research and read about a rule ...
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Is it necessary for the random variables to be defined on the same sample space to calculate joint distribution function?

Is it necessary for the random variables say X and Y to be defined on the same sample space to calculate their joint distribution function? I do not think so though I read it that way in a book.
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Is there any *mathematical* basis for the Bayesian vs frequentist debate?

It says on Wikipedia that: the mathematics [of probability] is largely independent of any interpretation of probability. Question: Then if we want to be mathematically correct, shouldn't we disallow ...
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Do Bayesians accept Kolmogorov's axioms?

Usually probability theory is taught with Kolgomorov's axioms. Do Bayesians also accept Kolmogorov's axioms?
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Question on the consequences of the Kolmogorov axioms

I am reading the (german) Applied Statistics and on page 140 as a consequence of the Kolmogorov axioms it is stated that if $P(A)=0$ one cannot conclude that $A=\emptyset$ . Similarly if $P(A)=1$ one ...
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