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I agree a "Probability 101" article shouldn't use sigma algebras. It definitely cuts off 99.9% of the people that might benefit from the article. Not that sigm

I agree a "Probability 101" article shouldn't use sigma algebras. It definitely cuts off 99.9% of the people that might benefit from the article. Not that sigma algebras don't have a place. If you are actually doing advanced things with martingales, sigma algebras are nice clean framework to express what you are doing.

throwaway81523 4y ago
It looks like a good article for those who are already studying theoretical probability. For laypeople, not so much. I haven't yet figured out what the author is doing with the sigma algebra machinery. I had thought that the event set of a probability space was intuitively the power set of the sample set, and that the machinery of sigma algebras was only added to avoid pathologies like non-measurable sets (think Banach-Tarski paradox). Maybe there is more to it? I would like to understand this better and I do plan to read the article, but I had found the Wikipedia article on martingales to be reasonably understandable a while back. The good thing about this article is it shows applications.
- ajkjk 4y ago
  I'm not a layperson (not really, at least) and I strongly feel like the sigma algebra framework is basically a waste of time. Mathematicians love abstracting and formalizing things but it's not useful, paradoxically, if you're trying to actually just do math.
  - jhanschoo 4y ago
    Sigma algebras are primarily useful for modeling less well-behaved probability spaces when you aren't dealing with (a product space of) probability spaces of that can be described by a pdf or pmf
- slopbop 4y ago
  Sigma algebras are useful even in a finite setting where we don't have to worry about pathologies. Think of them as modelling the lack of complete information in a probabilistic setting. If I know exactly which sample point represents my state, then I know the exact value of any random variable. If instead I only know that my state belongs to a given member of my sigma algebra, then I have some information, but not enough to necessarily pinpoint the value of a random variable. In fact, the familiar tools of measure theory can take this intuition further. If a random variable is measurable with respect to a sigma algebra, then knowing which element of that sigma algebra my state is in actually is sufficient to pinpoint the value of a random variable. Maybe to make this more concrete: Let's say I'm going to do two coinflips. My probability space is {HH, HT, TH, TT}. You can check for yourself that the sigma algebra generated by {{HH, HT}}, {TT, TH}} is not the trivial one- this is the sigma algebra that represents "Knowing the value of the first flip, but not the second". If we let X_first and X_second be 1 or 0 if the first or second flip is H or T respectively, then X_first is measurable with respect to this sigma algebra, but X_second is not. With Martingales and other stochastic processes, we don't generally have just one sigma algebra, but a sequence of sigma algebras called a "filtration", where each sigma algebra is finer than the last (ie, contains more sets, therefore gives you more measurable random variables). This filtration sort of defines the stochastic process- it's encoding the slow drip of extra information as the stochastic process evolves over time.
throwaway81523 4y ago
I think this article is not "martingales for laypeople". It is "martingales for math nerds" where the application is solving certain types of math problems to mathematical standards of rigor. For that rigor, you do need the sigma algebras. I would say also, the target reader of the article already knows what a sigma algebra is, so the formal definition is given only for review purposes, plus to motivate something the author does about finding a disjoing minimal cover of the sample space. I haven't yet read the article carefully enough to understand why he does that.
- synergy7 4y ago
  I think the author uses minimal set because it allows constructing any sigma-algebra from it. This (minimal set-based construction) is only possible for finite spaces. It allows, for example, an intuitive explanation of how sigma algebra F2 is coarser than F1. It serves a similar purpose as Borel sigma algebras in the general case, although a minimal set does not need to be a sigma algebra. Using different sigma algebras allows describing easily all events on which the probability is defined, and how those events may change with time (filtration). I am not sure how use of sigma algebras (or algebras for finite case) can be avoided in general.

throwaway81523 4y ago

It looks like a good article for those who are already studying theoretical probability. For laypeople, not so much. I haven't yet figured out what the author is doing with the sigma algebra machinery. I had thought that the event set of a probability space was intuitively the power set of the sample set, and that the machinery of sigma algebras was only added to avoid pathologies like non-measurable sets (think Banach-Tarski paradox). Maybe there is more to it? I would like to understand this better and I do plan to read the article, but I had found the Wikipedia article on martingales to be reasonably understandable a while back. The good thing about this article is it shows applications.

ajkjk 4y ago
I'm not a layperson (not really, at least) and I strongly feel like the sigma algebra framework is basically a waste of time. Mathematicians love abstracting and formalizing things but it's not useful, paradoxically, if you're trying to actually just do math.
- jhanschoo 4y ago
  Sigma algebras are primarily useful for modeling less well-behaved probability spaces when you aren't dealing with (a product space of) probability spaces of that can be described by a pdf or pmf
slopbop 4y ago
Sigma algebras are useful even in a finite setting where we don't have to worry about pathologies. Think of them as modelling the lack of complete information in a probabilistic setting. If I know exactly which sample point represents my state, then I know the exact value of any random variable. If instead I only know that my state belongs to a given member of my sigma algebra, then I have some information, but not enough to necessarily pinpoint the value of a random variable. In fact, the familiar tools of measure theory can take this intuition further. If a random variable is measurable with respect to a sigma algebra, then knowing which element of that sigma algebra my state is in actually is sufficient to pinpoint the value of a random variable. Maybe to make this more concrete: Let's say I'm going to do two coinflips. My probability space is {HH, HT, TH, TT}. You can check for yourself that the sigma algebra generated by {{HH, HT}}, {TT, TH}} is not the trivial one- this is the sigma algebra that represents "Knowing the value of the first flip, but not the second". If we let X_first and X_second be 1 or 0 if the first or second flip is H or T respectively, then X_first is measurable with respect to this sigma algebra, but X_second is not. With Martingales and other stochastic processes, we don't generally have just one sigma algebra, but a sequence of sigma algebras called a "filtration", where each sigma algebra is finer than the last (ie, contains more sets, therefore gives you more measurable random variables). This filtration sort of defines the stochastic process- it's encoding the slow drip of extra information as the stochastic process evolves over time.

throwaway81523 4y ago

I think this article is not "martingales for laypeople". It is "martingales for math nerds" where the application is solving certain types of math problems to mathematical standards of rigor. For that rigor, you do need the sigma algebras. I would say also, the target reader of the article already knows what a sigma algebra is, so the formal definition is given only for review purposes, plus to motivate something the author does about finding a disjoing minimal cover of the sample space. I haven't yet read the article carefully enough to understand why he does that.

synergy7 4y ago
I think the author uses minimal set because it allows constructing any sigma-algebra from it. This (minimal set-based construction) is only possible for finite spaces. It allows, for example, an intuitive explanation of how sigma algebra F2 is coarser than F1. It serves a similar purpose as Borel sigma algebras in the general case, although a minimal set does not need to be a sigma algebra. Using different sigma algebras allows describing easily all events on which the probability is defined, and how those events may change with time (filtration). I am not sure how use of sigma algebras (or algebras for finite case) can be avoided in general.