Tool box for experimenting with TV types.

It mostly contains a Haskell module TVToolBox with a collection of functions to experiment with TVs, in particular TV conversion (Simple <-> Indefinite <-> Distributional), and PLN formula application.

It also contains a bit of Maxima code pasted directly in the document (see Section Maxima further below).

The reason for developing this TV toolbox, and writing it in Haskell, is twofold (letting aside the fact that Haskell is so awesome to work with!):

1. I'm not entirely sure where I'm going. In spite of the PLN book dwelling pretty deep, there are still knowledge holes to fill, so I need a convenient platform to experiment on, which Haskell provides, due to its multi-sided scripting, native, interpreting nature, as well as its gnuplot binding.

2. Haskell has pretty much unlimited integer and floating precision, which is very convenient to tell apart numerical errors from phenomenal discrepencies.

• ghc version 7.8.3 or above
• Libraries
  1. multimap
  2. numbers
  3. memoize
  4. gamma
  5. exact-combinatorics
  6. text-format-simple
  7. gnuplot

To use the module see directly TVToolBox.hs for the functions it implements. Besides that there are several files using it which demonstrate the main functions. Each file can be directly executed as a script (no need to compile them). Here's a short description of each of them. For a indepth description see Section Experiment Report.

• Plot Distributional TVs given STVs

dtv-exp.hs

• Plot Indefinite TVs given STVs

itv-exp.sh

• Plot relationship between the mean of a distribution and it mode

mean-mode-exp.hs

• Plot deduction experiment using Distributional TVs

deduction-exp.hs

This section reports of a series of experiments on TV conversion and deduction formula calculation.

I cannot make any sense of the start of Chapter 6 of the PLN book when 2 levels of sampling using Beta distributions are used to generate a distributional TV estimate (EDIT: I think I'm partially understanding it, see Proof p.60). Chapter 4, introducing the foundation of distributional TV, makes perfect sense, and so this work is based on this chapter.

Regardless of whether I eventually get it, I do intend to compare side by side the methodology introduced in Chapter 6 with the methodology used here. Maybe this will help me to see the light, and if not I'll still gather knowledge to make informed implementational decisions regarding TV conversion and formula calculation in the C++ code.

Conversion from STV to DTV is done using the formula introduced in chapter 4, Section 4.5.1 of the PLN book.

P(x+X successes in n+k trials | x successes in n trials)
=
(n+1)*(choose k X)*(choose n x)
--------------------------------
 (k+n+1)*(choose (k+n) (X+x))

where k is the lookahead. X ranges from 0 to k (included), so the distribution is limited to k+1 data points. For small k, it may be convenient to generate more data points, like in order to get smoother distributions on premises and conclusions, or calculate more precise indefinite intervals, or even just compare with the methodology of Chapter 6. For that we use a continuous version of the binomial coefficient using the beta function (maybe it is a bad idea, although according to my experiments it seems coherent).

choose_beta n k = 1.0 / ((n+1.0) * beta (n-k+1) (k+1))

which gives as probability

P((p*100)% success in n+k trials | (s*100)% success in n trials)
=
               beta(-(n+k)*p+n+k+1, (n+k)*p+1)
------------------------------------------------------------------
(k+1)*beta(-n*s+n+1, n*s+1)*beta(- n*s+(n+k)*p+1, n*s-(n+k)*p+k+1)

It might be equivalent to the formula in Chapter 6. Unfortunately I cannot make any sense of that one (s is multipled by k, while p is multipled by n, contrary to here and previous notations in the PLN book, and it's not just a swap, p should be multipled by (n+k). I don't know how to fix/use it). TODO: attempt to make sense of it anyway, or reimplement here the double sampling in the obsolete C++ PLN code (see the obsolete-C++-PLN tag on the opencog repo).

When n+k it large, this continuous version gets very imprecise (this is an implementation limit, not a limit on Haskell's floating number representation), so we use Haskell's binomial function choose that works well on arbitrary large integers. In other words, when n+k is small we use choose_beta that provides some interpolation, when n+k is large we use choose that remains very accurate. That way we can experiment with both small and large n+k.

See figures (obtained with dtv-exp.hs)

for 3d plots of distributions obtained from simple TVs, varying the strength, the count and the lookahead. As you may see the corresponding distributions are a narrower when k is low. But I don't think this alone is a good reason to set k as a low value while doing inferences. The real question is whether the width of the infered TVs grows at a higher rate as measure as inferences progress when k is high versus low. The PLN book seems to say that is the case, I personally would like to measure that first hand.

In order to convert a Distributional TV to an Indefinite TV we need to find an indefinite interval [L, U] so that U-L, the width, is as small as possible, yet takes up about the portion set by the indefinite confidence level, b.

To do that we perform a optimization on L so that U-L is minimized and the cumulative probability between L and U is equal to b or greater. The optimization algo is a mere bisective search (with some parameters to overcome the presence of noise).

See figures (obtained with itv-exp.hs) showing the indefinite intervals for different strengths, counts and lookaheads.

In order to convert a DTV back to an STV we need to find the strength and the count that best fits a given distributional TV.

It may be quite important to get an accurate conversion from DTV to STV when applying formula using DTV internally but STV externally. The problem of course is that the DTV of a conclusion may not fit well the an STV. However, according to preliminary experiments it seems it does fit well, indicating that STV inference on steroid (see Section STV Inference on Steroid) might work well enough.

For the strength we take the mode of the distributional TV, rather than its mean. For an explanation as to why we take the mode see the following section.

For the count we find it by optimization. The optimization here again is a mere bisective search, and no effort is put on run-time efficiency at that point. All fitness functions rely on generating an STV with strength equal to the mode of the DTV, and the count being evaluated for fitness. Three fitnesses have been tried:

1. Jensen-Shannon divergence between the DTV and the generated STV.
2. Standard deviation distance between the DTV and the generated STV.
3. Width distance between the corresponding ITV and the generated STV.

The Jensen-Shannon divergence based fitness function is supposed to be optimal. However it is noisy (EDIT: I think this is due to some shit I need to fix) and I suppose would only be truly optimal if the STV strength was adjusted as well (perhaps micro-adjusted, but still).

The indefinite TV width based fitness function is the one suggested in the PLN book, it works but it is somewhat noisy.

The standard deviation based fitness function is not noisy at all. It may also be computationally cheaper than the other two. For that reason it is the fitness of choice for count search.

Although our optimization algo can cope with noise, it requires a bit of tweaking on the user's part, while optimizing a non noisy function just works.

See figures (obtained with deduction-exp.hs)

for profile comparisons of the 3 fitness functions for a deduction conclusion AC. See figures

for distributional TV and then estimated nearest STVs according to these 3 fitness functions. As you may see the stdDev based count optimization yield the best result in this case, clearly just due to the fact that it doesn't have to deal with noise.

The mean of a distribution associated to an STV differs from it's strength. This difference is very significant when n is low, as shown in the figure below (obtained with mean-mode-exp.hs).

It's been numerically observed that strength of an STV is equal to the mode of its DTV (whenever k*x is divisible by n, or k tends to infinity).

Here's a half-finished proof of it.

Given

    x
s = -
    n

    x+X
p = ---
    n+k

we want to proove that P (from Section 4.5.1 of the PLN book) is maximized when p = s, that is

    k*x
X = ---
     n

Let's look at P:

P(x+X successes in n+k trials | x successes in n trials)
=
(n+1)*(choose k X)*(choose n x)
-------------------------------
 (k+n+1)*(choose (k+n) (X+x))

Removing constant factors we obtain

    choose k X
------------------
choose (k+n) (X+x)

We want to show that following

    k*x
X = ---
     n

maximizes it. That is for all X in [0, k]

    choose k X            choose k (k*x/n)
------------------ <= ------------------------
choose (k+n) (X+x)    choose (k+n) ((k*x/n)+x)

which is equivalent to

     k!/(X!(k-X)!)                   k!/((k*x/n)!(k-k*x/n)!)
------------------------- <= -------------------------------------
(k+n)!/((X+x)!(k+n-X-x)!)    (k+n)!/(((k*x/n)+x)!(k+n-(k*x/n)-x)!)

equivalent to

(X+x)!(k+n-X-x)!    ((k*x/n)+x)!(k+n-(k*x/n)-x)!
---------------- <= ----------------------------
    X!(k-X)!             (k*x/n)!(k-k*x/n)!

TODO: complete the proof (not very important though).

We focus solely on the deduction rule for now. It is probably the most important one in PLN, but also one of the hardest to "put on steroid" because it has many premises, 5 in total, AB, BC, A, B, C.

To obtain the DTV of a conclusion we

1. Generate the DTV corresponding to each premises (see Subsection STV to DTV).
2. Sample the DTVs of all premises.
3. Check the consistency of the sampled probabilities.
4. If they are consistent, calculate the resulting strength using the deduction formula. Use this value as sample of the DTV's conclusion.
5. Convert the resulting DTV back to an STV (see Subsection DTV to STV).

Note that for now sampling the DTV merely is done in brute force, just enumerating all discretized strengths weighted by their second order probabilities, thus the resulting strength is also weighted by the product of their second order probabilities.

TODO: possibly implement Monte Carlos sampling.

The idea of STV inference on steroid is to generate simple, efficient and hopefully accurate inference formulae over STVs based on DTVs internal computations. Think of a digital mixing consoles performing internal signal processing in 48-bit while inputting and outputting signal in 24-bit coupled with super-compilation.

Given a certain inference rule accompanied with its formula f it would

1. Randomly pick input STVs, varying strengths and confidences and compute the resulting STV
   1. Turn them into DTVs
   2. Compute the resulting DTV using f
   3. Converte the resulting DTV into an STV
2. Record in a dataset all input STVs (strength and confidence) and their corresponding output STV. Try to gather a large number of individuals, ideally around r*exp(2*m), with r=10 and m equal to the number of premises.
3. Given the dataset use multivariate polynomial regression to find a model that accurately compute the resulting STV given the STVs of the premises. The model would be composed of 2 multivariate polynomials, one for the strength, one for the confidence.

Here's a description of the work that remains to be done, in more or less the right order to go about them.

Naively computing the mode of a sampled distribution (picking up the most frequent one) is lot less robust than the mean (averaging). I noticed that this leads to more jitters and bader fits than using the mean when the distribution is narrow. Yet the mode clearly is the correct statistics for the strength of the corresponding STV.

One way to improve that would probably to try to fit a beta distribution and use it's estimated mode, in fact there might just be a formula relating the mode, the mean and standard deviation of a beta distribution that could be used, see Subsection Relating parameters of a Beta distribution with its mean and mode. Note that one wouldn't need to compute the beta function at all, bypassing the beta function limited precision issue.

The other simpler but less accurate option could be to use the mean as an estimate of the mode when the standard deviation is sufficiently low. But that's a bit of a hack, better go with the suggestion above.

So far the discretization has a constant step-size 1/resolution. We could make it more efficient, with larger step-sizes for low probability regions, and shorter step-size for high probability regions. This would increase the resolution where it matters and speed up formula computation.

When the count of an STV is very high, that is corresponding to a confidence that approaches 1.0, calculating the binomial and thus the probability distribution takes a very long time (about 10 minutes for n=4M and k=10K). It would be good to optimize that. Clearly when a distribution is so narrow you don't need as many datapoints as k. Maybe this could be sampled. Alternatively the binomial itself could be approximated and optimized. In the same vein it would be useful to have a very efficient and accurate gamma or beta functions, possibly replacing the need for an exact binomial.

So far the DTV of a conclusion is calculated by combining virtually all consistent discretized strengths, which is very costly.

Instead on may perform Monte-Carlos simulations, i.e. using second order probability of a strength as the probability to pick it up.

In Section 7.2. of the PLN book it is explained how to use matrix product and matrix inversion to perform DTV deduction and inversion. I can't honnestly make sense of it. I supposed it is not contradictory with the approach implemented here so far, but it feels it could simplify and and possibly optimize computation a lot. So understanding that part and implementing it here would be a must.

The obsolete C++ PLN uses a double sampling of beta distributions to generate the DTVs. The code has been removed a long while ago but can be found in the opencog git repository under the tag obsolete-C++-PLN, under the directory opencog/reasoning/pln. It would interesting to reimplement the idea here and compare it with current single sampling approach.

The PLN book seems to imply that the indefinite intervals tend to widen at a more rapid rate if k, the lookahead, is large. It would be interesting to verify that firt hand. For that we would just chain n-1 inference steps using the conclusions of the previous steps as premises of the following steps and calculate w1/wn, where w1 corresponds to the standard deviation of the premise (or it's average if there are more than one premise) of step 1, and wn would correspond to the standard deviation of the conclusion of the last step.

We would obtain an average of multiple inferences of w1/wn varying the STV of the initial premises, then look at how mean(w1/wn) varies w.r.t. k.

See Section STV Inference on Steriod for the details.

Here's a record of some work done with Maxima, symbolic manipulations and plotting. Mostly trying to calculate and approach

and see how far ahead we can go.

Here's the maxima code to estimate the pdf of DTV using the binomial coefficients.

P(n, x, k, X) := ((n+1)*binomial(k, X)*binomial(n, x))
              / ((k+n+1)*binomial(k+n, X+x));

// Using s=x/n and p=(x+X)/(n+k)
//
// s = x/n
// x = s*n
// p = (x+X) / (n+k) = (s*n+X) / (n+k)
// p*(n+k) = s*n+X
// X = p*(n+k) - s*n
// X+x = p*(n+k)
// p(X) = (s*n+X) / (n+k)
// p(X+1) = (s*n+X+1) / (n+k)
// p(X) - p(X+1) = 1/(n+k)

P_prob(n, s, k, p) := ((n+1)*binomial(k, p*(n+k) - s*n)*binomial(n, s*n))
                        / ((k+n+1)*binomial(k+n, p*(n+k)));

pdf_k(n, s, k, p) := P_prob(n, s, k, p) * (n+k);
pdf(n, s, p) = lim k->inf pdf_k(n, s, k, p)

limit(pdf(n, s, k, p), k, inf);

Same as above but the Beta function is used instead of the binomial coefficients. Maxima is able to perform some symbolic simplifications along the way.

binomial_beta(n, k) := 1 / ((n+1) * beta(n-k+1, k+1));
P_beta(n, x, k, X) := ((n+1)*binomial_beta(k, X)*binomial_beta(n, x))
                  / ((k+n+1)*binomial_beta(k+n, X+x));

P_beta_prob(n, s, k, p) := ((n+1)*binomial_beta(k, p*(n+k) - s*n)*binomial_beta(n, s*n))
                        / ((k+n+1)*binomial_beta(k+n, p*(n+k)));

// Discard nonsensical p
P_beta_prob_condition(n, s, k, p) := if (0 <= (p*(n + k) - s*n) and (p*(n + k) - s*n <= k)) then (((n + 1)*binomial_beta(k, p*(n + k) - s*n)*binomial_beta(n, s*n))/((k + n + 1)*binomial_beta(k + n, p*(n + k)))) else 0;

// Simplified by Maxima
(%i44) P_beta_prob(n, s, k, p);
(%o44) beta(- (n + k) p + n + k + 1, (n + k) p + 1)
/((k + 1) beta(- n s + n + 1, n s + 1) beta(- n s + (n + k) p + 1, 
n s - (n + k) p + k + 1))

pdf_beta_k(n, s, k, p) := P_beta_prob(n, s, k, p) * (n+k);
pdf_beta_condition_k(n, s, k, p) := P_beta_prob_condition(n, s, k, p) * (n+k);

pdf_beta(n, s, p) = lim k->inf pdf_beta(n, s, k, p)

Given maxima limited large integer support we tried the following approximation https://en.wikipedia.org/wiki/Binomial_coefficient#Approximations

binomial_approx(n, k) := 2^n / sqrt(1/2*n*%pi) * %e^(-(k-(n/2))^2/(n/2));
P_approx(n, x, k, X) := ((n+1)*binomial_approx(k, X)*binomial_approx(n, x))
                  / ((k+n+1)*binomial_approx(k+n, X+x));

turns out it is really bad, especially when n is high and k is low. Not worth redoing.

plot3d(pdf_beta_k(500, 0.3, k, p), [p, 0, 1], [k, 1, 500], [grid, 100, 100], [z, 0, 100]);
plot3d(pdf_beta_k(100, 0.3, k, p), [p, 0, 1], [k, 1, 500], [grid, 100, 100], [z, 0, 100]);
plot3d(pdf_beta_k(50, 0.3, k, p), [p, 0, 1], [k, 1, 500], [grid, 100, 100], [z, 0, 100]);
plot3d(pdf_beta_k(10, 0.3, k, p), [p, 0, 1], [k, 1, 500], [grid, 100, 100], [z, 0, 100]);
plot3d(pdf_beta_k(5, 0.3, k, p), [p, 0, 1], [k, 1, 500], [grid, 100, 100], [z, 0, 100]);
plot3d(pdf_beta_k(n, 0.3, 500, p), [p, 0, 1], [n, 1, 500], [grid, 100, 100], [z, 0, 100]);

This may turn out to be handy.

TODO: Do the same for the standard deviation. The mean and standard deviation are rather robust statistics, so this could be then handy for estimating the mode given the beta function parameters.

X = a*k

  (k+n+1) ((k+n)! / ((X+x)! (k+n-X-x)!)

  (k+n+1) ((k+n)! / ((ak+x)! (k+n-ak-x)!)

                (n+k) * (n + 1) * (k! / (ak! (k-ak)!) * (n! / x! (n-x)!)

pdf?(p=a) = k->inf --------------------------------------------------------- (k+n+1) ((k+n)! / ((ak+x)! (k+n-ak-x)!)

f(p=a) = (n x)p^x*(1-p)^(n-x)

pdf(p=a) == f(p=a)

In this section we attempt to improve the confidence calculations in PLN inference rule formulas. The goal is to get calculations which are efficient yet justified by probability theory. It is our belief that the ideal calculations can only be obtained via uncomputable integration over multiple worlds (and to that end we intend to eventually introduce a Universal Truth Value, or UTV for short). In this work we suggest localized and efficient forms of such integration, that we hope can later be formally justified as approximations of such universal integration.

We start with the conjunction introduction rule for its simplicity. All the code involved in these experiments can be found in conjunction-xp.mac.

A question that has been in the back my mind for a while is:

What is the probability distribution obtained from applying a rule formula to premises which are themselves first order probabilities?

Let us for instance consider the conjunction introduction rule

A ≞ TVA
B ≞ TVB
⊢
(A ∧ B) ≞ TVAB

In order to obtain an accurate estimate of TVAB one would ideally

1. sample TVA and TVB to obtain first order probabilities, say pA and pB for a sample size of 1,
2. multiply pA and pB to obtain pAB, i.e. pAB = pA*pB,
3. repeat to 1 to get enough a large sample of pAB and fit whatever underlying distribution we choose to represent TVAB (like a beta distribution, as unfit as it may generally be).

The question in step 2 is: should we casually multiple pA with pB to obtain pAB? Meaning should we assume that pA and pB are independent? Or do we need to worry about the fact that we don't know if pA and pB are independent and therefore we should instead integrate over all possible ways that A and B could be distributed? Indeed, A and B could have true probabilities pA and pB respectively, yet perfectly overlap i.e. pAB = min(pA, pB), or not overlap at all, that is pAB = 0, etc. If we integrate over all possible ways that A and B could be distributed, maybe we get a variance that we need to account for so that instead of obtaining pAB = pA*pB we obtain a distribution over pAB? If this is the case then the concluding TV TVAB would take into account the variances of these "distrolets" obtained during step 2.

Given these specific premises the answer to this question is no, we do not need to worry about integrating over all possible ways that A and B could be distributed. We merely can assume that they are independent because the distrolet obtained from such integration when the size of the universe tends to infinity tends to a Dirac delta distribution with mean pA*pB. In order words, considering all possible ways that A and B could be distributed without any additional knowledge about them other than the fact that they are uniformly randomly distributed is equivalent to assuming that they are independent! I know it sounds obvious when phrased in that way, but keep in mind that this is only true when the size of the universe is infinit.

To reach this conclusion we have conducted the following experiment using Maxima.

We begin by defining how many ways A and B could intersect with intersection |A ∩ B| = k. For that we use combinations to obtain the following

Pr(|A ∩ B| = k) = binomial(a, k) * binomial(n-a, b-k) / binomial(n, b)

where |A|=a, |B|=b and |U|=n (U is the universe).

Let us replace k by n*x, a by n*pa and b by n*pb, where pa (resp. pb) is the marginal probability of A (resp. B). Using the Stirling approximation of factorial we can derive that

Pr(pAB = x) = (sqrt(pa*pb*(1-pa)*(1-pb)) / (sqrt(2*%pi*n*(pa-x)*(pb-x)*x*(x-pb-pa+1))))
            * ((pa**pa * pb**pb * (1-pa)**(1-pa) * (1-pb)**(1-pb))
               /
               ((pa-x)**(pa-x) * (pb-x)**(pb-x) * x**x * (x-pb-pa+1)**(x-pb-pa+1)))**n

Or as rendered by Maxima

This allows us to calculate Pr(pAB = x) for very high values of n.

Let us start by plotting distrolets for pA=pB=0.5 using combinations varying n from 50 to 400.

Already we can see a clear trend where the distrolet is shrinking as n goes up.

Using Stirling approximation we can push n much larger and see that the distrolet tends to a Dirac delta distribution.

I know a rigorous proof would be better, but I think it is already convincing enough that it converges to a Dirac delta distribution of mean pAB=pA*pB.

Obviously the reason we obtain Dirac delta distributions is because A and B are uniformly independently sampled, which is equivalent to assuming that they are independent. Indeed, as their sizes tend to infinity all random fluctuations that may have introduce dependencies vanish.

In this second attempt we introduce an extra premise that explicitly states dependencies. The conjunction introduction rule is thus reformulated as

A ≞ TVa
B ≞ TVb
A → B ≞ TVab
⊢
A ∧ B ≞ TVAB

This formulation generalizes the one in the section above, which can be recovered by setting TVab = <1, 0>, representing the absence of knowledge about A → B.

If TVab = <1, 0>, the sampling process to estimate the second order distribution associated to TVAB will still account for A → B but consider a beta distribution prior, such as the Bayes' prior, to sample from.

Like for the deduction rule, probability theory imposes some constraints over what probabilities are possible. Specifically, given first order probabilities pa sampled from TVa and pb sampled from TVb, probability pab sampled from TVab must be within the following bounds

max(1+(pb-1)/pa, 0) <= pab <= min(1, pb/pa)

assuming that 0 < pa, otherwise pAB = 0 anyway.

This inequality is justified by considering the extremes within which Pr(A,B) must be bound, corresponding to two situtations

1. A and B intersect the least, corresponding to max(pa+pb-1, 0),
2. A and B intersect the most, corresponding to min(pa, pb).

That is

max(pa+pb-1, 0) <= pAB <= min(pa, pb)

since

pab = Pr(A|B) = Pr(A,B)/Pr(B) = pAB/pa

the inequality of pab is obtained by dividing everything by pa.

Interestingly, given the localized nature introduced by these bounds, we can resurrect the idea of replacing concluding first order probabilities by distrolets. Below is a plot showing such distrolets for various values of pa and pb.

Let us now put all this together and build second order distributions obtained from the sampling method described earlier, which, to sum up, consists of sampling first order probabilities from premises, then either apply the product or sample from distrolets to obtain first order probability samples to build the concluding second order distribution.

What we would like to see is how much variance in the concluding distribution the distrolets bring vs using the independence assumption. To that end we produce for each pair of premises, ranging from high to low confidence, three plots, one with independence assumption (Dirac delta distrolet), a second derived from the extra premise A → B with null confidence, i.e. a uniform distrolet with the admissible range constrained by probability theory, a third obtained from replacing the uniform distrolet by a bimodal distribution with two Dirac deltas at its lower and upper admissible extremities. The later is hoped to emulate the most conservative conjunction introduction, producing conclusions with the maximum variance.

In all plots, the blue and green curves indicate the PDFs of the premises, while the red histogram indicates the PDF of the conclusion.

We start with premises with high confidences (α=6000 and β=3000 for the first premise, α=1000 and β=100 for the second premise).

One can observe that for such high confidences the desired effect is obtained, meaning that the variance of the conclusion using uniform distrolets, 0.00073803, is substantially greater than that of both its premises, 0.00002453 and 0.00007300. It is also substantially greater than the one obtained by making the independence assumption, 0.00005292. The variance obtained from the extreme distrolets is even greater, 0.00212295, but result in a bimodal distribution which is probably not desirable.

We continue with similar premises but with reduced confidences (α=500 and β=300 for the first premise, α=100 and β=10 for the second premise).

The same phenomenon, albeit less acute, can be observed, with variances 0.00122727 for the uniform distrolets and 0.00266664 for the extreme distrolets, greater than the variances of their premises.

In the following plots the confidence of the first premise is substantially reduced (α=50 and β=30 for the first premise).

The same phenomenon is still observed albeit less acutely.

The next plots consider premises with 4 positive and 4 negative pieces of evidence for both premises.

As before, using the uniform distrolet provides a larger variance, how the resulting concluding distribution is now quite far from a beta distribution as it has a bell like shape that, unlike with beta distribution, is cut a third of the way left. This would badly fit a beta distribution, which is problematic if we wish to keep simple truth values all the time. This will have to be studied in the future.

The next plots consider premises with only positive evidence and very low confidences.

Interestingly here even the concluding variance obtained from the independence assumption, 0.004037894, is greater than the variances of its premises, 0.033756043 and 0.02785076. In fact, the concluding variances in all three cases are not far apart, with 0.04628453 for the uniform distrolets and 0.04888252 for the extreme distrolets. For the latter, the extreme distrolets, one may notice a spike at 0. This is because when the first order sampled probabilities sum up to less than one, zero is selected as concluding probability half of the time.

In the next plots, the premises have respectively one positive and one negative evidence.

Contrary to the previous case, the concluding variances, 0.03552852 for the independence assumption, 0.03872038 for the uniform distrolets and 0.04822294 for the extreme distrolets, are all below that of their premises, which are approximately 0.05. All concluding second order distributions are shifted to the left.

These next plots show the concluding distributions when the premises have null confidence.

Like above, all concluding variances are less than the variances of the premises, and all concluding second order distributions are shifted to the left. Clearly these second order distributions would corresponds to simple truth values with a positive count of positive and total evidence, while no observation has ever been made.

Let us push the experiment even further by considering other priors, start with Jeffreys'.

For all three cases (independence, uniform and extreme), the resulting variances, around 0.07, are smaller than that the variances of the premises, even more so than with a Bayes' prior. Note that the resulting mean, around 0.23, remains close to 0.25. Overall the results make sense.

Then let us consider a prior somewhere between Jeffreys and Haldane with α=β=0.1.

The resulting means, around 0.22, are drifting a bit away from 0.25. One may notice that the means of premises are around 0.47 instead of 0.5, likely due to some floating point number calculation imprecisions, so that might explain it. But overall the results still make sense.

Finally let us consider a prior that is getting even closer to the Haldane prior with α=β=0.01.

The concluding distributions, in all three cases, is now almost completely equivalent to Bernoulli distributions, where the densities are split between completely zero or completely one with means around 0.25.

To conclude, it interestingly seems that the structure of a composite predicate, like a conjunction of two predicates, by itself provides some forms of evidence. In other words, evidence can be obtained not just from direct observations but from structures as well.

Let us explore a variation where the extra premise to capture dependence is formulated using Equivalence rather that Implication (as suggested in Ben's document). In this case the inference rule for conjunction becomes

A ≞ TVa
B ≞ TVb
A↔B ≞ TVab
⊢
A∧B ≞ TVAB

The semantics of A↔B corresponds, in probabilistic terms, to

P(A∧B|A∨B)

Let us assume that the first order probabilities are sampled as follows

pa ~ TVa
pb ~ TVb
pab ~ TVab

and let us examine the question of how to obtain pAB=P(A∧B).

P(A∧B|A∨B) = P(A∧B) / P(A∨B)
P(A∧B|A∨B) = P(A∧B) / (P(A) + P(B) - P(A∧B))
pab = pAB / (pa + pb - pAB)

Let us express pAB w.r.t. to pa, pb and pab

pab = pAB / (pa + pb - pAB)
(pa + pb - pAB) * pab = pAB
pa*pab + pb*pab - pAB*pab = pAB
pa*pab + pb*pab = pAB + pAB*pab
pa*pab + pb*pab = pAB * (1 + pab)
pAB = (pa*pab + pb*pab) / (1 + pab)

The procedure to obtain pAB is therefore going to be

1. Sample pa ~ TVa
2. Sample pb ~ TVb
3. Sample pab ~ TVab (within valid bounds)
4. Calculate pAB = (pa*pab + pb*pab) / (1 + pab)

Let us assign a Bayes' prior over the second order distribution over TVab when A↔B is completely unknown, that is a uniform second order distribution. We still need to sample TVab so that pAB is within valid bounds. To that end we attempt to derive the bounds of pab given the bounds of pAB recalled below

max(pa+pb-1, 0) <= pAB <= min(pa, pb)

According to DeepSeek, we get

max(pa+pb-1, 0) <= pab <= min(pa/pb, pb/pa)

We did not go through its log to verify that its reasoning is correct. Let us however plot pAB with respect to pab and vice versa for various values of pa and pb while maintaining pAB and pab within their supposedly valid bounds.

It is clear that the plots of pAB with respect to pab perfectly mirror the plots of pab with respect to pAB, bounds included, thus the lower and upper bounds of pab inferred by DeepSeek are likely correct.

Note that since the function mapping pab to pAB is not linear, building the second order distribution of A∧B via sampling pab (then mapping pab to pAB) according to A↔B is going to yield different results than sampling according to A→B. Indeed, the relationship between A→B and A∧B is linear while the relationship between A↔B and A∧B is not. To see how they differ let us plot the distrolets associated to A↔B and A∧B for various first order probabilities of A and B.

Across all plots, the distrolets associated to A∧B, in red, look like they are quadratically increasing. We anticipate that this will have the effect of shifting the means of the concluding distributions to the right. Let us plot various concluding distributions for difference simple truth values to examin that. For better comparison we also replot the concluding distribution obtained with A→B as extra premise.

Already one can see a slight increase in the means, with 0.6210 when using A→B and 0.6227 when using A↔B. The resulting distribution when using A↔B also retain the quadratic increase. The variance of the distribution obtain using A↔B is also a tiny bit greater, 0.00073951 with A↔B vs 0.00073803 with A→B.

In these two plots above the concluding distributions look visually identical but again the mean obtained with A↔B, 0.5819, is greater than the mean obtained with A→B, 0.5793. This could be however due to the fact that the samples picked in the A↔B experiment for A and B also have larger means, respectively 0.6256 and 0.9093 with A↔B and respectively 0.6248 and 0.9090 with A→B. The variances however follow another trend, being greater with A→B, 0.00122727, than with A↔B, 0.00119035.

In the two plots above the mean obtained with A↔B, 0.5813, is actually less than that obtained with A→B, 0.5813. That could be explained however by differences in means of the samples obtained from the premises A and B as above. It is unclear at this point. The variance obtained with A↔B is also less than the one obtained with A→B.

The plots above involve premises with lower counts, 4 positive and 4 negative counts each. The difference in means is more significant, being 0.2776 with A↔B and 0.2472 wth A→B. The variance with A↔B, 0.02338096, is also greater than the variance with A→B, 0.02266180, but could be from a difference of variances in the premises.

Once again the mean obtained with A↔B is greater than the one obtained with A→B. The variance obtained with A↔B is less than the one obtained with A→B.

Once again we observe that the mean obtained with A↔B is greater than the one obtained with A→B. This time the variance obtained with A↔B is greater than one obtained with A→B.

Same thing again, greater mean with A↔B than A→B. The variance obtained with A↔B is less than the variance obtained with A→B. At this point there seems a clear pattern for the mean but no pattern for the variance.

Again, greater mean with A↔B than A→B. The variance with A↔B is greater than the variance with A→B this time.

Finally, the mean with A↔B is greater than the one with A→B and the variance with A↔B is also greater than the one with A→B.

Over the 9 pairs of plots above, the mean obtained with A↔B is greater than the one obtained with A→B 8 times, which is unlikely due to chance. However the variance obtained with A↔B is greater than the one obtained with A→B only 5 times, which could be due to chance.

Thus we can conclude that using equivalence instead of implication is unlikely to have an impact on the variance of the concluding distribution. However the means is likely a bit greater when using equivalence. It is unclear from these experiments which rule is preferable.