সম্ভাবনা তত্ত্বটি কি অ-নেতিবাচক ফাংশনগুলির সমীক্ষা যা একত্রে / যোগফলকে সংহত করে?

Question

এটি সম্ভবত একটি নির্বোধ প্রশ্ন, তবে সম্ভাব্যতা তত্ত্ব কি ফাংশনগুলির সমীক্ষা যা একটির সাথে সংহত / যোগফল করে?

সম্পাদনা করুন। আমি নেতিবাচকতা ভুলে গেছি। সুতরাং সম্ভাবনা তত্ত্বটি কি অ-নেতিবাচক ফাংশনগুলির সমীক্ষা যা একত্রে / যোগফলকে সংহত করে?

Comments

হ্যাঁ, সম্ভাবনা সর্বদা এক হতে পারে। Likelihoods অন্যদিকে এই বাধ্যতা হবে না।

— মাইক হান্টার

প্রশ্ন যেমন বিবৃত করার একমাত্র যুক্তিসংগত উত্তর নেই, না অন্তত কারণ অনেক ফাংশন আছে $f$ যে 1 থেকে সংহত কিন্তু যার জন্য $\int_{a}^b f(u) du$ কিছু জন্য সম্ভাব্যতা উপস্থাপন করতে পারবেন না $a$ এবং $b$ । উদাহরণস্বরূপ, 0 এবং 1 এর মধ্যে 1.5 এবং 1 এবং 2 এর মধ্যে -0.5 এবং অন্য কোথাও 0 এর একটি ফাংশন বিবেচনা করুন 0 (তবে এটি অন্যান্য কারণেও তর্কসাপেক্ষে "না")

— গ্লেন_বি -রিনস্টেট মনিকা

সম্পর্কিত: stats.stackexchange.com/questions/214485/…

— ইলমারি করোনেন

নেতিবাচক সম্ভাবনার উপর গুরুতর কাগজপত্র রয়েছে, যেমন মরিস এস বার্টলেট t doi.org/10.1017/S0305004100022398

— নিক কক্স

@ দন্টলু সেখানে আমি কী লক্ষ্য রেখেছিলাম তা এখন চমকনের উত্তরে টাও উক্তিটি দ্বারা বেশ ভালভাবে কভার করা হয়েছে।

— গ্লেন_বি -রেইনস্টেট মনিকা

Answer 1

নিখুঁতভাবে আনুষ্ঠানিক স্তরে, কেউ সম্ভাব্যতা তত্ত্বকে মোট পরিমাপ একের সাথে পরিমাপের জায়গাগুলির অধ্যয়নের ডাক দিতে পারে, তবে এটি সংখ্যার তত্ত্বকে কলকের সংখ্যার স্ট্রিংগুলির অধ্যয়নের মতো বলা হত termin

- এলোমেলো ম্যাট্রিক্স তত্ত্বের টেরি টাওয়ের বিষয়গুলি থেকে ।

আমি মনে করি এটি সত্যই মৌলিক জিনিস। আমরা একটি সম্ভাব্যতা স্থান পেয়েছেন $(\Omega, \mathscr F, P)$ এবং দৈব চলক $X : \Omega \to \mathbb R$ pushforward পরিমাপ সঙ্গে $P_X := P \circ X^{-1}$ , তারপর কারণ একটি ঘনত্ব সংহত এক কারণপি(Ω)=1। এবং এটি পিডিএফএস বনাম পিএমএফএসের চেয়ে বেশি মৌলিক।$f = \frac{\text d P_X}{\text d\mu}$$P(\Omega) = 1$

এখানে প্রমাণ আছে:

$$\int_{\mathbb R} f \,\text d\mu = \int_{\mathbb R} \,\text dP_X = P_X(\mathbb R) = P\left(\{\omega \in \Omega : X(\omega) \in \mathbb R\}\right) = P(\Omega) = 1.$$

এটি প্রায় অ্যাডমোর জবাব (+1) এর পুনরাবৃত্তি কারণ সমস্ত সিডিএফ càdlàg, এবং সিডিএফগুলির সেট এবং আর ( R , B ) এর সমস্ত সম্ভাব্যতার ব্যবস্থার সেটের মধ্যে একের মধ্যে সম্পর্ক রয়েছে , তবে একটি আরভি এর সিডিএফ তার বিতরণের ক্ষেত্রে সংজ্ঞায়িত করা হয়, আমি সম্ভাবনার স্থানগুলি এই ধরণের প্রচেষ্টা দিয়ে "শুরু" করার জায়গা হিসাবে দেখি।$\mathbb R$$(\mathbb R, \mathbb B)$

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আমি সিডিএফ এবং সম্ভাব্যতা ব্যবস্থার মধ্যে চিঠিপত্রের বিশদভাবে আপডেট করছি এবং উভয়ই কীভাবে এই প্রশ্নের যুক্তিসঙ্গত উত্তর।

আমরা দুটি সম্ভাব্য ব্যবস্থা নিয়ে শুরু করে এবং সম্পর্কিত সিডিএফ বিশ্লেষণ করে শুরু করি। পরিবর্তে একটি সিডিএফ দিয়ে শুরু করে এবং এর দ্বারা অনুপ্রাণিত পরিমাপটি দেখে আমরা শেষ করি।

যাক এবং আর উপর সম্ভাব্যতা পরিমাপ করে হতে ( আর , বি ) দিন এফ কিউ এবং এফ আর তাদের নিজ নিজ CDFs হতে (অর্থাত এফ কিউ ( একটি ) = প্রশ্নঃ ( ( - ∞ , একটি ] ) এবং জন্য একভাবে আর ।) প্রশ্নঃ এবং আর উভয়ই এলোমেলো ভেরিয়েবলের (যেমন বিতরণ) ধাক্কা ধরণের পদক্ষেপের প্রতিনিধিত্ব করবে তবে তারা আসলে কোথা থেকে এলো তা বিবেচ্য নয়।$Q$$R$$(\mathbb R, \mathbb B)$$F_Q$$F_R$$F_Q(a) = Q\left((-\infty, a]\right)$$R$$Q$$R$

মূল ধারণাটি হ'ল: যদি ও আর সেটগুলি সমৃদ্ধ পর্যায়ে সংগ্রহের বিষয়ে একমত হয় তবে তারা সেগুলি দ্বারা উত্পন্ন al- বীজগণিতকে সম্মত করে । স্বজ্ঞাতভাবে, যদি আমরা এমন একটি ইভেন্টের একটি ভাল আচরণের সংগ্রহ পেয়েছি যা গণনাযোগ্য সংখ্যার পরিপূরক, ছেদগুলি এবং ইউনিয়নগুলি সমস্ত বি দ্বারা তৈরি করে , তবে সেগুলির সমস্তগুলির সাথে একমত হয়ে কোনও বোরেল সেটের সাথে দ্বিমত পোষণ করার কোনও সুযোগ নেই।$Q$$R$$\sigma$$\mathbb B$

এটি আনুষ্ঠানিক করা যাক। যাক দিন এল = { একটি ⊆ আর : প্রশ্নঃ ( একটি ) = আর ( একটি ) } , অর্থাত্ এল এর উপসেট পি ( আর ) যার উপর প্রশ্ন এবং আর সম্মত হন (এবং সংজ্ঞায়িত করা হয়) নোট করুন যে আমরা এল -বোরেল সেটগুলিতে তাদের সম্মতি জানাতে দিচ্ছি যেহেতু এল হিসাবে সংজ্ঞায়িত করা অগত্যা একটি উপসেট নয়$\mathscr S = \{(-\infty, a] : a \in \mathbb R\}$$\mathcal L = \{A \subseteq \mathbb R : Q(A) = R(A)\}$$\mathcal L$$\mathcal P(\mathbb R)$$Q$$R$$\mathcal L$ । আমাদের লক্ষ্য দেখানোর জন্য যে বি ⊆ এল ।$\mathbb B$$\mathbb B \subseteq \mathcal L$

দেখা যাচ্ছে যে ( এস দ্বারা উত্পাদিত σ- বীজগণিত ) আসলে বি , তাই আমরা আশা করি যে এস ইভেন্টের যথেষ্ট পরিমাণে বড় সংগ্রহ যা যদি এস এর সর্বত্র Q = R হয় তবে তারা সমান হতে বাধ্য হয় বি এর সমস্ত উপর ।$\sigma(\mathscr S)$$\sigma$$\mathscr S$$\mathbb B$$\mathscr S$$Q = R$$\mathscr S$$\mathbb B$

নোট করুন যে সীমাবদ্ধ ছেদগুলির অধীনে বন্ধ রয়েছে এবং এল সম্পূর্ণরূপে এবং গণনাযোগ্য বিচ্ছিন্ন ছেদগুলির অধীনে বন্ধ রয়েছে (এটি σ- সংবেদনশীলতা থেকে অনুসরণ করে )। এর অর্থ এই যে , S একটি হল π -system এবং এল একটি হল λ -system । দ্বারা π - λ উপপাদ্য সুতরাং আমরা যে আছে σ ( এস ) = বি ⊆ এল । এস এর উপাদানগুলি$\mathscr S$$\mathcal L$$\sigma$$\mathscr S$$\pi$$\mathcal L$$\lambda$$\pi$$\lambda$$\sigma(S) = \mathbb B \subseteq \mathcal L$$\mathscr S$ are nowhere near being as complex as an arbitrary Borel set, but because any Borel set can be formed from a countable number of complements, unions, and intersections of elements of $\mathscr S$, if there is not a single disagreement between $Q$ and $R$ on elements of $\mathscr S$ then this will be followed through to there being no disagreements on any $B \in \mathbb B$.

We have just shown that if $F_Q = F_R$ then $Q = R$ (on $\mathbb B$), which means that the map $Q \mapsto F_Q$ from $\mathscr P := \{P : P \text { is a probability measure on } (\mathbb R, \mathbb B)\}$ to $\mathcal F := \{F : \mathbb R \to \mathbb R : F \text { is a CDF}\}$ is an injection.

Now if we want to think about going the other direction, we want to start with a CDF $F$ and show that there is a unique probability measure $Q$ such that $F(a) = Q\left((-\infty, a]\right)$. This will establish that our mapping $Q \mapsto F_Q$ is in fact a bijection. For this direction, we define $F$ without any reference to probability or measures.

We first define a Stieltjes measure function as a function $G : \mathbb R \to \mathbb R$ such that

1. $G$ is non-decreasing
2. $G$ is right-continuous

(and note how being càdlàg follows from this definition, but because of the extra non-decreasing constraint "most" càdlàg functions are not Stieltjes measure functions).

It can be shown that each Stieltjes function $G$ induces a unique measure $\mu$ on $(\mathbb R, \mathbb B)$ defined by

$$\mu\left((a, b]\right) = G(b) - G(a)$$ (see e.g. Durrett's Probability and Random Processes for details on this). For example, the Lebesgue measure is induced by $G(x) = x$.

Now noting that a CDF is a Stieltjes function $F$ with the additional properties that $\lim_{x\to-\infty} F(x) := F(-\infty) = 0$ and $\lim_{x\to\infty} F(x) := F(\infty) = 1$, we can apply that result to show that for every CDF $F$ we get a unique measure $Q$ on $(\mathbb R, \mathbb B)$ defined by

$$Q\left((a, b]\right) = F(b) - F(a).$$

Note how $Q\left((-\infty, a]\right) = F(a) - F(-\infty) = F(a)$ and $Q\left((-\infty, -\infty]\right) = F(\infty) - F(-\infty) = 1$ so $Q$ is a probability measure and is exactly the one we would have used to define $F$ if we were going the other direction.

All together we have now seen that the mapping $Q \mapsto F_Q$ is 1-1 and onto so we really do have a bijection between $\mathscr P$ and $\mathcal F$. Bringing this back to the actual question, this shows that we could equivalently hold up either CDFs or probability measures as our object which we declare probability to be the study of (while also recognizing that this is a somewhat facetious endeavor). I personally still prefer probability spaces because I feel like the theory more naturally flows in that direction but CDFs are not "wrong".

Comments

+1 for a broader perspective on the matter; You correctly note that Skorokhod's càdlàg function-space is only a present notion of what probability theory entails, radically different from Borel's, and Skorokhod's discoveries only date back ~40 years or so. Who knows what the next century may uncover?

— AdamO

@AdamO absolutely, and there’s the weirder ones like non-Archimedean probability, where even if they never become the dominant view (and to my knowledge no one is seriously trying to do that) I find they help me to better understand the standard formulation (eg how serious of a thing sigma additivity is)

— jld

I read the question title and thought of that quote from Terence Tao; must have read it years ago (2010) but it's really memorable. As he goes on to say, At a practical level, the opposite is true…

— ShreevatsaR

See my comment on the question: How do alternative theories of probability, such as Bayesian (and Dempster-Shafer and the Transferable Belief Model and Dezert-Smarandache Theory), imprecise probabilities, plausibility theory, etc. relate to this question and discussion ?

— E. Douglas Jensen

@E.DouglasJensen I'm not sure, i'm addressing this in terms of the standard Kolmogorov axioms so in that context I think my answer is "right", but if we're changing the axioms then I suppose all bets are off. Also i'm not being philosophical at all about this so if we're trying to connect this to the real world in any way, e.g. with questions like "what is the probability that the sun rises", then i'm sure it gets more complicated. Nevertheless, it seems a pretty safe bet that the probability that "anything" happens is the maximal value (probably $1$) and that there's no uncertainty in that

— jld

Answer 2

No; the Cantor distribution is just such a counterexample. It's a random variable, but it has no density. It has a distribution function, however. I would say, therefore, that probability theory is the study of càdlàg functions, inclusive of the Cantor DF, that have left limits of 0 and right limits of 1.

Comments

Nice, I never heard of cadlag functions. However, these still assume a real and a metric space. Not all probability theory is done on such spaces.

— HRSE

You may for example go back to Terrence Fine, Theories of Probability. Also note that cadlag functions (at least according to the wikipedia article) have the real numbers as a domain. LJ Savage's "Foundations of Statistics" gives an account of (subjective) probability theory on spaces that are not necessarily real.

— HRSE

@jwg Some other comments in this post address negative probability, which seems to be of some use in quantum physics though my simple mind cannot fathom such a thing.

— AdamO

@HRSE thanks for the references. I couldn't find either of them online but I skimmed some other papers by those authors although I didn't find any examples of this. If we're defining a random variable $X$ as $X : \Omega \to \mathbb R^n$ then the CDF is defined in terms of the pushforward measure $P_X := P \circ X^{-1}$ (not the measure $P$ on $(\Omega, \mathscr F)$) and since $X$ is real valued $P_X$ is necessarily a measure on $(\mathbb R^n, \mathbb B^n)$ which means we can feed it sets like $(-\infty, a]$ so $F$ has $\mathbb R^n$ as its domain. Am i missing something?

— jld

I think well ordered means every subset has a least element while totally ordered means for all $x$ and $y$, exactly one of $x<y$, $x>y$, or $x=y$ holds, so $\mathbb N$ is both, $\mathbb R$ is just totally ordered, and $\mathbb C$ is neither. We absolutely need to multiply and add probabilities so at the very least the codomain of $P$ ought to be a field, but I don’t think it has to be totally ordered or complete. Complex valued measures are an example of the first and hyperreal valued measures are an example of the second. All of these are metric spaces though (or can be)

— jld

Answer 3

I'm sure you'll get good answers, but will give you a slightly different perspective here.

You may have heard mathematicians saying that physics is pretty much mathematics, or just an application of mathematics to the most basic laws of nature. Some mathematicians (many?) actually do believe that this the case. I've heard that over and over in university. In this regard you're asking a similar question, though not as wide sweeping as this one.

Physicist usually don't bother even responding to this statement: it's too obvious to them that it's not true. However, if you try to respond it becomes clear that the answer is not so trivial, if you want to make it convincing.

My answer is that physics is not just a bunch of models and equations and theories. It's a field with its own set of approaches and tools and heuristics and the ways of thinking. That's one reason why although Poincare developed relativity theory before Einstein, he didn't realize all the implications and didn't pursue to get everyone on board. Einstein did, because he was a physicist and he got what it meant immediately. I'm not a fan of the guy, but his work on Brownian motion is another example of how a physicist builds a mathematical model. That paper is amazing, and is filled with intuition and traces of thinking that are unmistakenly physics-ey.

So, my answer to you is that even if it were the case that probability deals with the kind of functions you described, it would still not have been the study of those function. Nor it is a measure theory applied to some subclass of measures. Probability theory is the distinct field that studies probabilities, it's linked to a natural world through radioactive decay and quantum mechanics and gases etc. If it happens so that certain functions seem to be suitable to model probabilities, then we'll use them and study their properties too, but while doings so we'll keep an eye on the main prize - the probabilities.

Comments

+1 for bringing reality to a math fight and actually answering the question with the only reasonable answer, i.e. that any such reductionism misses the point

— jld

@Chaconne I learned a useful word today reductionism, will incorporate it in my vocabulary :)

— Aksakal

+1, this is what I was trying to say with my answer, but I said it less effectively than you I think.

— Nathaniel

Answer 4

Well, partially true, it lacks a second condition. Negative probabilities do not make sense. Hence, these functions have to satisfy two conditions:

• Continuous distributions:

  $$\int_{\mathcal{D}}f(x) dx = 1 \quad \text{and} \quad f(x)>0 \; \forall x \in \mathcal{D}$$

• Discrete distributions:

  $$\sum_{x \in \mathcal{D}}P(x) = 1 \quad \text{and} \quad 0<P(x) \leq 1 \; \forall x \in \mathcal{D}$$

Where $\mathcal{D}$ is the domain where probability distribution is defined.

Comments

Thanks a lot Carlos for the answer, actually I want to know what if the non negative condition was added?

— dontloo

I would say that reducing probability field to study of probability density/mass functions (fulfilling the upper properties) is too bare. Moreover, as it has been stated by @AdamO, there are some cases of random variables which do not have probability density function, even though they have a well defined cdf.

— Carlos Campos

@CarlosCampos: Regarding negative probabilities: They actually do make sense in some contexts, e.g. half coins. See en.wikipedia.org/wiki/Negative_probability for a bit more information.

— Inkane

Answer 5

I would say no, that's not what probability theory fundamentally is, but I would say it for different reasons than the other answers.

Fundamentally, I would say, probability theory is the study of two things:

1. Stochastic processes, and

2. Bayesian inference.

Stochastic processes includes things like rolling dice, drawing balls from urns, etc., as well as the more sophisticated models found in physics and mathematics. Bayesian inference is reasoning under uncertainty, using probabilities to represent the value of unknown quantities.

These two things are more closely related than they might at first appear. One reason we can study them under the same umbrella is that important aspects of both of them can be represented as non-negative functions that sum/integrate to one. But probability isn't just the study of those functions - their interpretation in terms of random processes and inference is also an important part of it.

For example, probability theory includes concepts such as conditional probabilities and random variables, and quantities such as the entropy, the mutual information, and the expectation and variance of random variables. While one could define these things purely in terms of normalised non-negative functions, the motivation for this would seem pretty weird without the interpretation in terms of random processes and inference.

Moreover, one sometimes comes across concepts in probability theory, particularly on the inference side, which cannot be expressed in terms of a non-negative function that normalises to one. The so-called "improper priors" come to mind here, and AdamO gave the Cantor distribution as another example.

There certainly are some areas of probability theory in which the main interest is in the mathematical properties of normalised non-negative functions, for which the two application domains I mentioned are not important. When this is the case, we often call it measure theory rather than probability theory. But probability theory is also - indeed, I would say mostly - an applied field, and the applications of probability distributions are in themselves a non-trivial component of the field.

Comments

You made the domain of topics in probability theory pretty narrow...

— Tim

@Tim not on purpose - I divided it into two areas, but intended each of them to be interpreted very broadly. Can you give me some other topics that don't fit under either heading?

— Nathaniel