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I Definition of "equivalent" probability problems?

  1. Jun 25, 2017 #1

    Stephen Tashi

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    Is there a precise definition for the statement that two differently worded probability problems are "equivalent"?

    One technique of (purportedly) solving a controversial probability problem is to propose an "equivalent" problem whose solution is not controversial. (e.g. The Sleeping Beauty thread: page 22 post #422https://www.physicsforums.com/threads/the-sleeping-beauty-problem-any-halfers-here.916459/page-22 ). However, my impression of this technique is that people just assert they have created an equivalent problem and hope the reader will say Ah-ha!.

    If the specifics of two probability problems are given in the form of probability spaces, can we define their "equivalence" by the existence of some sort of mapping between the two spaces ? Can we define the concept of "homomorphic" and "isomorphic" for probability problems?
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  3. Jun 25, 2017 #2


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    I couldn't see an equivalence in the probability space in that linked post. Further, I could not even see any justification for the claim
    Code (Text):
    But if Beauty is actually wakened, one of the first two possibilities [our of four] is eliminated.
    On my reading, one out of possibilities 3 and 4 is also eliminated, leaving only two possibilities, so I can't see how they get a 1/3 from that.

    The Sleeping Beauty problem, with the agreed interpretation of 'credence', is about expected values, whereas that post only talks about probabilities, so I very much doubt they are equivalent.

    Directly to your question, I propose the following definition of equivalence for probability problems:

    I will couch it in terms of a question about expected values because Qns about probabilities can always be coded as questions about expected values by using indicator functions, but not vice versa.

    For ##j\in\{1,2\}##, let ##Q_j## be the question of whether the statement ##\psi(E^{P_j}[X_j|A_j])## is true where ##\psi## is a unary predicate and ##X_j:\Omega_j\to \mathbb R## is a ##\Sigma_j##-measurable random variable on probability space ##\langle \Omega_j,\Sigma_j,P_j\rangle##, and ##A_j\in\Sigma_j##.

    Then we say that questions ##Q_1## and ##Q_2## are 'equivalent' if
    • there is a map ##f:\Omega_1\to\Omega_2## that generates an inclusion-preserving bijection between ##\Sigma_1## and ##\Sigma_2## (##f## itself need not be a bijection), and
    • ##f## preserves probability measure (ie for all ##S\in\Sigma_1:\ P_2(f(S))=P_1(S)##), and
    • ##A_2=f(A_1)##, and
    • for all ##B\in\Sigma_1##, ##E^{P_2}[X_2|f(B)]=E^{P_1}[X_1|B]##
    I am not sure that we need the second point, because I think it may be implied by the fourth - not in its entirety, but insofar as it relates to the problem at hand. But I've kept it just in case, and because I think the first two points are enough for a workable definition of the two probability spaces being 'isomorphic'.

    The last two points relate to the specific problem, and ensure that ##A_2## and ##X_2## are effectively the 'images under ##f##' of ##A_1## and ##X_1##.

    Note that the predicate ##\psi## that makes the True or False statement about the expected value of the random variable conditional on the specified event, must be the same for both cases. Everything else is indexed by ##j##.
  4. Jun 26, 2017 #3

    Stephen Tashi

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    That definition is plausible - although I always fear strange examples if infinities are present.

    Are there any interesting examples of isomorphisms using probability spaces with finite outcomes? The first examples that come to mind are not interesting because they are merely the same information denoted with different letters - e.g. {H,T} and fair coin vs {even,odd} for fair die.

    Perhaps, we can make an interesting homework problem if the sample spaces are defined by stating a partition into sets that are expressed in terms of other sets.

    For example, suppose ##\Sigma_1## is defined as the sigma algebra on subsets of ##\Omega_1## given by ##\{ X\cup Y, X \cap Y, Y \cup Z\}## with respective probabilities { 1/2, 1/3, 1/4}. I think the mechanical way to compare this probability space to another one that was defined in a similar manner would be to "atomize" ##\Omega_1## into the sets:
    ##a_1 = X \cap Y \cap Z##
    ##a_2 = X \cap Y \cap Z^c##
    ##a_3 = X \cap Y^c \cap Z##
    ## a_4 = X \cap Y^c \cap Z^c##
    ##a_5 = X ^c \cap Y \cap Z##
    ##a_6 = X^c \cap Y \cap Z^c##
    ##a_7 = X^c \cap Y^c \cap Z##
    ##a_8 = X^c \cap Y^c \cap Z^c##

    Then state the probabilities as information about the atoms:
    ##P_1( X \cup Y ) = P_1(a_1 \cup a_2 \cup a_3 \cup a_4 \cup a_5 \cup a_6) = 1/2##
    ##P_1( X \cap Y) = P(a_1 \cup a_2) = 1/3##
    ##P_1( Y \cup Z) = P(a_1 \cup a_2 \cup a_5 \cup a_6 \cup a_3 \cup a_7) = 1/4##

    Suppose the problem is to compare this to a probability space defined in terms of a different number of events ##A,B,C,D## so we get a different number of atoms ##b_1, b_2,...b_{16}##. (I won't give the specifics.)

    I think that comparing the two spaces (in search of an isomorphism) would amount to trying to match up the groupings of atom ##a##-atoms with groupings of ##b##-atoms that have the same probability.
    Last edited: Jun 26, 2017
  5. Jun 26, 2017 #4


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    Yes, searching for matches is the search for the isomorphism ##f## above. Only measurable groups of atoms need be matched. For instance ##Z## is not measurable in the example, as a consequence of which none of the individual atoms ##a_k## are measurable (ie they have no probability). But they can be put together to create measurable sets as shown.

    A simple, but not terribly useful, example of an isomorphism on a finite probability space is to take a space ##\langle \Omega_2,\Sigma_2,P_2\rangle## that has ##\Omega_2=\{a_1,...,a_n\}## and then create a space ##\langle \Omega_1,\Sigma_1,P_1\rangle## that duplicates the atoms of ##\Omega_2## but for which ##\Sigma_1## cannot split pairs. That is
    $$\Omega_1=\left\{b_{kj} :\ 1\leq k\leq n\wedge j\in\{1,2\}\right\}$$
    $$\textrm{for }S\in\Sigma_2:\ f(S)=\left\{b_{kj}\ |\ a_k\in S\wedge j\in\{1,2\}\right\}$$

    ie, as long as the differences between the two spaces are not measurable, the spaces are isomorphic. We can split atoms in either space to our heart's content as long as the splits are not reflected in the Sigma Algebras.

    Perhaps testing whether two problems are equivalent amounts to finding the coarsest probability space that can be used to formalise each one, then testing whether an isomorphism can be found between the two that preserves the problem premises and conclusion.

    By the way, looking at things in this very formal way has helped me understand why I found the thirder arguments sometimes intuitively appealing, yet unable to be formally proven - and hence ultimately rejected. The problem is that 'Beauty wakes' is treated as an event, with the formal probability-theoretic meaning of a measurable subset of the sample space. But it is either not an event, or it is the trivial event that is the entire sample space (because Beauty surely wakes in both of the two events Coin-Heads and Coin-Tails). In a verbal argument, it sounds so reasonable to say that 'Beauty wakes' is an event, because it is an event in the natural language sense of the word. But not in the probability-theoretic sense.
  6. Jun 26, 2017 #5


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    I agree that there could be an isomorphism between two probability spaces. But if one problem is controversial and another is not, I have my doubts that there could be such an isomorphism. Maybe I just do not have enough imagination to see how one could be controversial and the other not be. If someone presented a claimed example to me, I would really have to take a hard look at that isomorphism.
  7. Jun 27, 2017 #6

    Stephen Tashi

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    I don't think we should side-step the (perhaps temporary) closure of the Sleeping Beauty thread by discussing all issues that problem presents. If we do, this thread will get chaotic. I suggest we only discuss technical matters here. It would be appropriate to have a detailed look at some "equivalent" problems that have been proposed, to write ot their probability spaces in detail and to compare them.
  8. Jun 27, 2017 #7


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    Okay, but I don't actually think that there is any other issue involved.
  9. Jun 27, 2017 #8

    Stephen Tashi

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    Yes, but (invoking whatever limited authority an OP has to define the topic of a thread !) I'm genuinely interested in how to define and evaluate whether two probability problems are "equivalent" in a technical manner - how to do this in general, not just considering probability problems related to the SB problem. So I'm interested defining what technical technique can be used to compare two well-posed problems. I agree the SB problem is ill-posed. However, two problems proposed as "equivalent" to it might each be well-posed. We could investigate in a technical manner whether two such well-posed problems are equivalent to each other.
  10. Jun 27, 2017 #9


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    The reason that people invoke equivalence is as a step in figuring out the probability space. We might have an informally stated problem involving probability, and in order to solve it, we have to translate the informal problem into a more mathematical form that is suitable for calculation. Often, we don't know how to do that translation. But if we argue that the problem is essentially the same as another problem that we do know how to analyze mathematically, then we substitute the equivalent problem.

    But this notion of equivalence is prior to formalization. So I think that trying to say when two formal probability problems are equivalent is missing the whole point of invoking the equivalence in the first place.

    In the thread that won't be named, the issue was about how to formulate the thought experiment as a problem in probability. Once you've done that, the problem is trivially solved. So I don't see that the thought experiment had anything to do with equivalences among formalized probability problems.
  11. Jun 29, 2017 #10

    Stephen Tashi

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    I'll use Boolean algebra-like notation.
    "~A" denotes the set complement ##A^c##
    "A + B" denotes the set union ##A \cup B##
    "(A)(B)" denotes the set intersection ##A \cap B##.

    I consider two balls-in-urns problems.

    Problem 1:
    Urn H contains one amber-colored ball and one sienna-colored ball. Urn T contains two sienna-colored balls. A fair coin is flipped. If it lands heads, Urn H is selected. If the coin lands tails, Urn T is selected. Exactly one ball is drawn from the chosen urn.

    Question 1-1: If the ball is amber-colored, what is the probability that Urn H was selected?

    Definition of events:
    H = the coin lands heads
    ~H = the coin lands tails
    A = the ball draw is amber-colored
    ~A = the ball drawn is sienna-colored

    The given information implies the "atomic" events:
    a_1= (H)(A), P(a_1) = 1/4
    a_2 = (H)(~A), P(a_2) = 1/4
    a_3 = (~H)(A), P(a_3) = 1/2
    a_4 = (~H)(~A), P(a_4) = 0

    Find P(H|A) = P( (H)(A)) / P(A) = (1/4)/ (1/4 + 1/2) = 1/3

    Problem 2:
    Urn H contains exactly two balls, h1 and h2. Ball h1 is amber-colored and has a "1" marked on it. Ball h2 is sienna-colored and has a "2" marked on it.
    Urn T: contains exactly two balls, t1 and t2. Both balls are amber colored. Ball t1 has a "1" marked on it and ball t2 has a "2" marked on it.
    A fair coin is flipped. If the coin lands heads urn H is selected. If the coin lands tails, urn T is selected. All the balls are drawn out of the selected urn.

    Definition of events:

    H: urn H is slected
    ~H: urn T is selected
    t1: ball t1 is drawn
    t2: ball t2 is drawn
    h1: ball h1 is drawn
    h2: ball h2 is drawn.

    The given information:
    p(H) = 1/2
    e1 = (H)(h1)(h2), P(e1) = 1/2
    e2 = H( t1 + t2)
    = (H)(t1) + (H)(t2), P(e2) = 0
    e3 = (~H)(t1)(t2), p(e3) = 1/2
    e4 = (~H)(h1 + h2) =
    = (~H)(h1) + (~H)(h2), p(e4) = 0

    The "atomic" events:
    a_1 = (H)(e1)(e2)(e3)(e4), P(a_1) = 0
    a_2 = (H)(e1)(e2)(e3)(~e4), P(a_2) = 0
    a_3 = (H)(e1)(e2)(~e3)(e4), P(a_3) = 0
    a_4 = (H)(e1)(e2)(~e3)(~e4), P(a_4) = 1/2
    a_5 = (H)(e1)(~e2)(e3)(e4), p(a_5) = 0
    a_6 = (H)(e1)(~e2)(e3)(~e4), p(a_6) = 0
    a_7 = (H)(e1)(~e2)(~e3)(e4), p(a_7) = 0
    a_8 = (H)(e1)(~e2)(~e3)(~e4), p(a_8) = 0
    a_9 = (H)(~e1)(e2)(e3)(e4), P(a_9) = 0
    a_10 = (H)(~e1)(e2)(e3)(~e4), P(a_10) = 0
    a_11 = (H)(~e1)(e2)(~e3)(e4), P(a_11) = 0
    a_12 = (H)(~e1)(e2)(~e3)(~e4), P(a_12) = 0
    a_13 = (H)(~e1)(~e2)(e3)(e4), p(a_13) = 0
    a_14 = (H)(~e1)(~e2)(e3)(~e4), p(a_14) = 0
    a_15 = (H)(~e1)(~e2)(~e3)(e4), p(a_15) = 0
    a_16 = (H)(~e1)(~e2)(~e3)(~e4), p(a_16) = 0

    a_17 = (~H)(e1)(e2)(e3)(e4), P(a_17) = 0
    a_18 = (~H)(e1)(e2)(e3)(~e4), P(a_18) = 0
    a_19 = (~H)(e1)(e2)(~e3)(e4), P(a_19) = 0
    a_20 = (~H)(e1)(e2)(~e3)(~e4), P(a_20) = 0
    a_21 = (~H)(e1)(~e2)(e3)(e4), p(a_21) = 0
    a_22 = (~H)(e1)(~e2)(e3)(~e4), p(a_22) =0
    a_23 = (~H)(e1)(~e2)(~e3)(e4), p(a_23) = 0
    a_24 = (~H)(e1)(~e2)(~e3)(~e4), p(a_24) = 0
    a_25 = (~H)(~e1)(e2)(e3)(e4), P(a_25) = 0
    a_26 = (~H)(~e1)(e2)(e3)(~e4), P(a_26) = 0
    a_27 = (~H)(~e1)(e2)(~e3)(e4), P(a_27) = 0
    a_28 = (~H)(~e1)(e2)(~e3)(~e4), P(a_28) = 0
    a_29 = (~H)(~e1)(~e2)(e3)(e4), p(a_29) = 1/2
    a_30 = (~H)(~e1)(~e2)(e3)(~e4), p(a_30) = 0
    a_31 = (~H)(~e1)(~e2)(~e3)(e4), p(a_31) = 0
    a_32 = (~H)(~e1)(~e2)(~e3)(~e4), p(a_32) = 0

    Question 2-1: An observer sees an amber-colored ball is drawn,but does not see the number marked on it and does not see from which urn it was drawn. What is the probability that the urn that was selected is urn H?

    A = an amber-colored ball is drawn on one of the draws = h1 + t1 + t2
    P(H| A) = P( (H)(A))/ P(A) = P(a_4)/ (P(a_4) + P(a_29)) = (1/2)/1 = 1/2

    Question 2-2: Exactly one ball is drawn from the selected urn. That ball is amber-colored. What is the probability that the urn selected was urn H?
    E = exactly one ball is drawn from the chosen urn and that ball is amber-colored
    = (h1)(~h2)(~t1)(~t2) + (~h1)(~h2)(t1)(~t2) + (~h1)(~h2)(t1)(~t2)
    P(E) = 0
    P( (H)(E) ) = 0
    P(H | E) is not defined
    Last edited: Jun 29, 2017
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