Tag Archives: Bayesianism

Business projects, sales programmes, often go to double the time and double the cost: how would Bayes have accounted and planned for these?

I now turn to important and sometimes critical time-measures that are used in business decision-making, strategic planning and valuation, such as ‘sales cycle’ time, customer lifetime, and various ‘time-to-market’ quantities, such as time to proof of concept or time to develop a first version of a product.

Bayesian analysis enables us to make good and common sense estimates in this area, where frequency statistics fails. It allows us to use sparse, past observations of positive cases, all of our recent observations where no good result has yet happened, and a subjective knowledge, all treated together and in an objective way, using all of the above information and data and nothing but this. That is, it will be a maximum entropy treatment of the problem where we only use the data we have and nothing more, as accurately as is possible.

We assume that the model for the probability that the time taken to success, t, in ‘quarters’ of a year, is an exponential distribution e^−λt for any positive t > 0. λ will be the mean rate of success for the next case in point. We have available some similar prior data, over a period of t quarters, where we had n clients and r ≤ n successful sales (footnote 0).

Let T = rt̂ + (n−r)t

be the total number of quarters (i.e. three-month periods of time) in which we have observed lack of success in selling something, e.g. a product or service, and where $\hat{t} = \frac{1}{r} \sum t_{j}$

$\hat{t} = \frac{1}{r} \sum t_{j}$

is the mean observed time to success t_j for the jth data point.

Let the inverse θ = λ⁻¹, be the mean time to success, for the quantity we want to estimate predictively, and track or monitor carefully, ideally in real time, from as early as possible in our business development efforts, for example, the mean sales-cycle time, i.e. the time from first contact with a new client to the time of the first sale, or possibly the time between sales, or new marketing campaigns, product releases or versions and so on. We shall create an acceptance test at some given level of credibility or rational degree of belief, P, for this θ to be above a selected test value θ₀, with some degree of belief my team of executives are comfortable with or interested in.

I wish to obtain an expression telling me the predicted time-to-success in quarters is above (or below) θ₀ in terms of θ₀ and T, n and r, i.e.given all the available evidence.

By our hypothesis (model), the probability that the lifetime θ > θ₀ is given by e^−λθ₀.

The prior probability for the subjective belief in the mean time taken t_s is taken to be distributed exponentially around this value, p_s(λ) = t_se^−λt_s, which is the maximally-equivocal (footnote 1) most objective assumption.

The small probability from the test, for a given value of λT, given the evidence in the test data, and our best expert opinion, leading to T, is given by the probability ‘density’
$p (d λ ∣ T, n, t_{1}, . . ., t_{r}) = \frac{1}{r!} (λ T)^{r} e^{- λ T} d (λ T)$

$p (d λ ∣ T, n, t_{1}, . . ., t_{r}) = \frac{1}{r!} (λ T)^{r} e^{- λ T} d (λ T)$

Multiplying the probability that the time is greater than θ₀ by this probability for each value of λ, and integrating over all positive values of λ, I find that the probability that the next sales person or next case of customer lifetime or time to sale is greater than our selected lifetime for the test, θ₀ is given by

p (θ > θ_{0} ∣ n, r, T, θ_{0}) = \int_{0}^{\infty} \frac{1}{r!} (λ T)^{r} e^{- λ T - λ θ_{0}} d (λ T) = p (D, θ_{0})

Where p(D,θ₀) is the posterior probability as a function of our data D and (acceptance) case in point θ₀, and which after some straightforward algebra turns out to be a simple expression from which result one can obtain the numerical value with T having been shifted by the inclusion of the subjective expert time, t_s, T → T + t_s, which is our subjective, common sense, maximum entropy, prior belief as to the mean length of time in quarters for this quantity.

Suppose we have an acceptance probability, of P * 100% that our rational, mean sales cycle time for the next customer or time-to-market for a product or service is less than some time θ₀. I thus test whetherp(D,θ₀) < P. If this inequality is true, (we have chosen P such that) our team will accept and work with this case, because it is sufficiently unlikely for us that the time to sale or sales cycle is longer than θ₀. Alternatively, I can determine what θ₀ is, for a given limiting value of P, say, 20%. For example: taking some data, where n = 8, r = 6, the expert belief is that the sales time mean is $t_{s} = \frac{17}{4}$

$t_{s} = \frac{17}{4}$

, i.e. just over a year, and there were specific successes, say, at t_j = (3,4,4,4,4.5,6)quarters corresponding to our r = 6, and we run the new test for t = 2 quarters. We want to be 80% sure that our next impact-endeavour for sales/etc will not last more than some given θ₀ that we want to determine. I put in the values, and find that T = 33.75, continuing to determine θ₀ I find that with odds of 4:1 on, time/lifetime/time-to-X is no greater than 8.7 quarters.

Suppose that we had more data, say an average of $\bar{t_{j}} = \frac{17}{4}$

$\bar{t_{j}} = \frac{17}{4}$

quarters with r = 15 actual successes and n = 20 trials. We decide to rely on the data and set $t_{s} = \frac{17}{4}$

$t_{s} = \frac{17}{4}$

. Now T = 78. Keeping the same probability acceptance or odds requirement at 80% or 4:1 on, we find θ₀ ≤ 8.25 quarters. If we were considering customer lifetime, rather than sales cycle time or similar measures like time to proof of concept etc, we benefit when the lifetime of the customer is more than a given value of time θ₀, and so we may look at tests where P > 80% and so on.

If we omit the quantity t_s, we find that the threshold θ₀ = 7.8 quarters, only a small tightening, since the weight of one subjective ’data’ is much smaller than the effect of so many, O(n) ‘real’ data points.

Now I wish to consider the case where we run a test for a time t with n opportunities. After a time t, we obtain a first success (footnote 2), so that then r = 1 and we note the value of t̂. I then set t_s = t and I have also t̂ = t. T reduces to T = (n+1)t, and if we look at the case θ₀ = t, our probability reduces to an expression that is a function of n:

p (θ > t ∣ n, 1, t, θ_{0} = t) = {[\frac{n + 1}{n + 2}]}^{2}

Since ∞ > n ≥ 1 then $\frac{4}{9} \leq p (θ > t ∣ n, 1, t, θ_{0} = t) < 1$

$\frac{4}{9} \leq p (θ > t ∣ n, 1, t, θ_{0} = t) < 1$

, i.e. if we are only testing one case and we stop this test after time t with one success r = 1 = n, this gives us our minimal probability that the mean is θ ≥ t, all agreeing with common sense, and interesting that the only case where we can achieve a greater than 50:50 probability of θ < t = t_s = t̂ is when we only tested n = 1 to success. This is of course probing the niches of sparse data, but in business, one often wishes to move ahead with a single ‘proof of concept’. It is interesting to be able to quantify the risks in this way.

If we consider the (extreme) case where we have no data, only our subjective belief (footnote 3), quantified as t_s. Let us take $θ_{0} = m t_{s}$

$θ_{0} = m t_{s}$

, m an integer, then our probability p(∅,θ₀) of taking this time reduces to

p (θ > θ_{0} ∣ 0, 0, t_{s}, θ_{0} = m t_{s}) = [\frac{1}{1 + m}]

This means that at m = 1 the probability of being greater or less than θ₀ is a half, which is common sense. If we want to have odds, say, of 4:1 on, or a probability of only 20% of being above θ₀ quarters, then we require m = 4, and the relationship between the odds to 1 and m is simple.

Again this all meets with common sense, but shows us how to deal with a near or complete absence of data, as well as how the situation changes with more and more data. The moral is that for fairly sparse data, when we seek relatively high levels of degree of belief in our sales or time needed the next time we attempt something, the Reverend Bayes is not too forgiving, although he is more forthcoming with useful and most concise information than an equivalent frequency statistics analysis. As we accumulate more and more data, we can see the value of the data very directly, as we have quantified how our risks are reduced as it comes in.

The results seem to fit our experiences with delays and over-budget projects. We must take risks with our salespeople and our planning times, but with this analysis, we are able to quantify and understand these calculated risks and rewards and plan accordingly.

One can update this model with a two-parameter model that reduces to it, but which allows for a shape (hyper)parameter which gives flexibility around prior data, such as the general observation that immediate success, failure or general `event’ is not common, the position of the mean relative to the mode, and also around learning/unlearning since the resulting process need not be memoryless (see another blog here!)

or customer lifetimes, or types of time-to-market, or general completions/successes etc.)↩︎
highest entropy, which uncertainty measure is given by S = − ∑_sp_slog p_s.↩︎
e.g. a sale in a new segment/geography/product/service↩︎
if we neither have any data nor a subjective belief, the model finally breaks down, but that is all you can ask of the model, and a good Bayesian would not want the model to ‘work’ under such circumstances!↩︎

Generalised Game Show Problem

August 30, 2023 admin

We generalise a result in Professor David Mackay’s book on inference. Bayes theorem also plays a crucial role in decision-making:

$P (A | D) = \frac{P (D | A) P (A)}{P (D | A) P (A) + P (D | \neg A) P (\neg A)}$

Let us consider a worked example which demonstrates how unintuitive results following from this 260-year-old theorem can be.

In the Game Show example, we can simplify Bayes’ Theorem by using a form that expands out all individual doors, and then cancelling off the unconditional probability of each door in numerator and denominator as they are each $P(any door) = \frac{1}{n}$ , where there are n doors. We consider the two distinct representative cases:

$P (prize behind my chosen door ∣ m) = \frac{P (m ∣ prize behind my chosen door)}{P}$

$P (prize behind another door ∣ m) = \frac{P (m ∣ prize behind another door)}{P}$

where here

$P = 1 / (\binom{n - 1}{m}) + (n - 1) / (\binom{n - 2}{m})$

and where $∣ m$ means that any m doors were removed at the usual intermediate stage after I chose a door. When we start with n doors, and one has been chosen, and that door happens to have the prize behind it, then the Game Show Host is free to remove m doors from the set of $n - 1$ doors and so there are $(\binom{n - 1}{m})$ available ways for the host to do so.

The probability $P(any m ∣ 1)$ is therefore $1 / (\binom{n - 1}{m})$ , since we equivocate between all the host’s options. For the other case, there are only $n - 2$ doors for the host to choose m from, so the number of ways is $(\binom{n - 2}{m})$ , and the probability $P(any m ∣ not 1)$ is therefore $1 / (\binom{n - 2}{m})$ .

after some algebra I find that

$P (prize behind my chosen door ∣ m) = \frac{n - 1 - m}{n - 1 - m + (n - 1)^{2}}$

and

$P (prize behind another door ∣ m) = \frac{n - 1}{n - 1 - m + (n - 1)^{2}}$

Thus, our probability factor is given by: $\frac{P(prize behind another door ∣ m)}{P(prize behind my chosen door ∣ m)} = \frac{n - 1}{n - 1 - m}$

This expression gives us back, from the original game, our game strategy factor of 2 times better if we shift door when $n = 3$ and $m = 1$ . The factor rises to $n - 1$ better for shifting to another door for any $n \geq 2$ and $m = n - 2$ doors (all but one other than the one the player chose), and for the shift strategy a probability factor that tends to unity from above when $n \to \infty$ and $m = 1$ , i.e. only one door is removed.

An informal survey by Student-b has shown that quite often, intuition among a random sample of people asked, is lacking. Some will believe it is better to stick, and some say in the standard three-door game that it is slightly better to shift. The mathematics show that for positive n and m: it is always better to shift!

Probability as logic for decisionmaking

Homage to Bazball Strategy

August 30, 2023 admin

We are selectors for a cricket team who wish to make decisions based on data around the success or otherwise of our players’ strategies or relative performances that we trust and believe to be comparable.
If we follow Bayes’ procedure to calculate the probability that the “real” average runs scored, b, by our batsman as player B (batting as “Bazball”) is greater than that of the same batsman as player A, a (batting as “Anodyne”) for data set averages $\bar{x}$ and $\bar{y}$ respectively for the player, and with n and m data points, then setting $s = n \bar{x} + \frac{1}{2}$ and $t = m \bar{y} + \frac{1}{2}$ , we find that:

$Prob (b > a) = \frac{m^{t}}{Γ (s) Γ (t)} \int_{0}^{\infty} a^{t - 1} e^{- m a} Γ (s, n a) d a$

where we have assumed a Poisson distribution to parametrise the average numbers of runs scored in the two strategies, and the Jeffreys’ uninformative prior distribution for this model.

For example, if there are $m = 6$ innings at $\bar{y} = 30.0$ for A, and $n = 8$ innings at $\bar{x} = 32.0$ for B, then we find that the above equation $= 0.7407$ and the odds are almost 3:1 on that the strategy as B has a greater average than that of the player as A. If we have a rule that the odds on a strategy for at least 5 innings must be better than 4:1 on, then our decision in this case would be to ask the player to continue to bat as A, pending more information. If we only required 2.5:1 then we would be inclined to try strategy B for this player.

Another angle on this is to look at the average number of balls `survived’ by the player in an innings, with the two strategies. That is, I am interested in the number of balls faced at the point the player is out. To model this, I start from the Pascal distribution with an uninformative (uniform) prior, since this distribution is a discrete one which models a future number $n$ of independent trials, which contain $r$ failures. I have $r = 1$ for this case, as cricket is unforgiving! The end of the trials is on the nth ball, when the player fails, i.e. is out. This simplifies things:

$f (x) = P (X = x) = (\binom{x - 1}{r - 1}) p^{r} (1 - p)^{x - r} = p^{r} (1 - p)^{x - r}$

where $p$ is the probability of getting out on any given ball and $X$ is the random quantity realised as $x$ in a given trial (innings), the number of balls received up to and including getting out. The mean of the quantity $x$ is $\frac{1}{p}$ . This makes sense, as the number of balls faced in an innings is greater than or equal to one. Our prior distribution (density) for the unknown value of p was taken to be $B (1, 1)$ . The Beta family is conjugate with respect to the Pascal distribution (negative binomial distribution), in such a way that for prior $B (α, β)$ , the posterior is also a Beta density with parameters $α + n r$ and $β + n \bar{x} - n r$ , where $\bar{x}$ is the mean of the data $x$ and there were $n$ trials (innings). Since $r = 1$ the conditional probability density is:

$f (p ∣ x) = B^{- 1} p^{n} (1 - p)^{s}$

where $s = n (\bar{x} - 1)$ and $B = B (1 + n, 1 + s)$ . This means that if we compare two sets of innings, n and m, with average balls faced until getting out, $\bar{x} \geq 1$ and $\bar{y} \geq 1$ respectively, then, relabelling the respective unknowns we find the probability of B getting out sooner being greater than that of A is:

$Prob (β > α) = \frac{1}{C_{n} C_{m}} \int_{0}^{1} d α \int_{α}^{1} α^{m} (1 - α)^{t} β^{n} (1 - β)^{s} d β$

where $t = m (\bar{y} - 1)$ and $C_{n} = B (1 + n, 1 + s))$ and $C_{m} = B (1 + m, 1 + t)$ are constants calculated from the data, which normalise the integrals. From this probability we can deduce the odds that one strategy is longer-lasting than the other, ignoring run rates this time.

We can then also simplify by converting the discrete geometric distribution to the continuous, and computationally slightly easier exponential distribution $f (x) = p e^{- p x}$ . The resulting relative probability, corresponding to and in good agreement with the equation above, is:

$Prob (β > α) = \frac{t^{m + 1}}{Γ (n + 1) Γ (m + 1)} \int_{0}^{1} α^{m} e^{- α t} Γ (n + 1, s α, s) d α$

where $α$ and $β$ are the real probabilities of getting out on the next or any given delivery, given by the inverse of the respective real average numbers of balls faced up to the point of failure, i.e. getting out. The $Γ (a, b, c)$ in the integrand is the generalised, incomplete Gamma function arising from the first integration over all cases in the joint distribution where $β > α$ . Note that the integral this time is over all cases from zero to one since the variables $α$ (and $β$ ) are probabilities.

One can compare a set of innings in A and B modes, this time ignoring runs scored but focusing on how long the innings were and again perhaps having a rule for deciding which strategy is optimal and how to apply the rule to make the judgement.

If in the first strategy A the player stays at the crease for an average of $\bar{y} = 25$ balls, in $m = 4$ innings and in strategy B, $\bar{x} = 20$ balls, in $n = 5$ innings, I find that the probability that the unknown parameter $β$ , representing the probability of getting out next ball in strategy B is higher than the unknown parameter $α$ , representing the probability of getting out next ball in strategy A is 0.623 in the exponential distribution calculation. The Pascal calculation gave 0.626, under $0.5 %$ off.

The applications for this kind of Bayesian joint-probability A/B comparison to get simple odds for or against are miriad and go far beyond the tip of the iceberg which is sport strategy; they are numerous in business and governmental strategy.

mathematics, Probability as logic for decisionmaking

Student-b: Probability Tables

August 30, 2023 admin

Student-b

Dear Sir/Madam,

In this letter, I derive and tabulate the maximum entropy values for the probabilities of each side of biased $n$ -sided dice, for $n = 3, 4, 6, 8, 10, 12, 15, and$ $20$ . These probabilities for each of the n options (sides), are those which have the least input information beyond what we know, which is nothing more than the bias or average score on the $n$ -sided die. I generalise the “Brandeis dice” problem from E T Jaynes’ 1963 lectures, to an $n$ -sided die, from the 6-sided case. To calculate these probabilities, I obtain the solution of an $n + 1$ -order polynomial equation, derived using a power series identity, for the value of the Lagrange multiplier, $λ$ . The resulting maximally-equivocated prior probabilities at the 5th, 17th, 34th, 50th (fair), 66th, 83rd, and 95th percentiles of the range from 1 up to $n$ will aid in decision-making, where the options are the conditions we cannot influence, but across which we may have a non-linear payoff.

We use the standard variational principle in order to maximise the entropy in the system.

$\sum_{i = 1}^{n} p_{i} f_{k} (x_{i}) = F_{k}$

$\sum_{i = 1}^{n} p_{i} = 1$

where the index $k$ is not summed over in the first equation, and where the $p_{i}$ are the probabilities of the $n$ options, e.g. sides of an n-dice. $F_{k}$ are the numbers given in the problem statement (constraints or biases), and $f_{k} (x_{i})$ are functions of the Lagrange multipliers $λ_{i}$ . The second equation is just the probability axiom requiring the probabilities to sum to one. This set of constraints is solved by using Lagrange multipliers. The formal solution is

$p_{i} = \frac{1}{Z} \exp [- λ_{1} f_{1} (x_{1}) - . . . - λ_{m} f_{m} (x_{i})]$

where $Z (λ_{1}, . . ., λ_{m}) = \sum_{i = 1}^{n} \exp [- λ_{1} f_{1} (x_{1}) - . . . - λ_{m} f_{m} (x_{i})]$ is the partition function and $- λ_{k}$ are the set of multipliers, of which for a solution to the problem there need to be fewer than n, though in our current problem as we shall see, there is only one. The constraints are satisfied if:

$F_{k} = - \frac{\partial}{\partial λ_{k}} \log_{e} Z$

for k ranging from 1 to m.

Our measure of entropy is given by $S = - \sum_{i = 1}^{n} p_{i} \log_{e} p_{i}$ and in terms of our constraints, i.e. the data, this function is:

$S (F_{1}, . . ., F_{m}) = \log_{e} Z + \sum_{k = 1}^{m} λ_{k} F_{k}$

The solution for the maximum of S is:

$λ_{k} = \frac{\partial S}{\partial F_{k}}$

For k in same range up to m. For our set of n-sided dice, $m = 1$ and so I can simplify $F_{k}$ to F. The $f_{k} (x_{i})$ are simply the set of $i$ the values on the n sides of our die.

For the problem at hand of the biased die, I introduce the quantity q which I define as the tested, trusted average score on the given n-sided die in hand. That is, I set $F = q$ here, our bias constraint number, which can range from the lowest die value 1 through to the highest value, n.

$q = q_{0} := \frac{1}{2} (n + 1)$

i.e. 3.5 on a 6-sided die, then the die is fair, otherwise, it has a bias and therefore an additional constraint. I assume this is all I know and believe about the die, other than the number of sides, n.

We see that $λ_{k}$ becomes just $λ$ and the equation for $F_{k}$ reduces to

$F = - \frac{\partial}{\partial λ} \log_{e} Z$

and the equation for $S_{k}$ reduces to

$S (F) = \log_{e} Z + λ F$

and its solution is

$λ = \frac{\partial S}{\partial F}$

After a little algebra, I found that the partition function Z is given by

$Z = \frac{x (x^{n} - 1)}{(x - 1)}$

and after some further algebra, I found that in order to determine the value of $x$ , where $x = e^{- λ}$ , corresponding to the maximum entropy (least input information) set of probabilities, we must find the positive, real root of the following equation, which is not unity:

$(n - q) x^{n + 1} + (q - (n + 1)) x^{n - 1} + q x + (1 - q) = 0$

By inspection, this equation is always satisfied by the real solution x = 1, which corresponds to the fair or unbiased die, with all probabilities equal to 1/n for n sides. We need the other real root, and we obtain this by simple numerical calculation. From the solution $x = x_{q}$ for the given value of bias q, the set of probabilities corresponding to maximum entropy for each side of the relevant n-sided die are easily generated.

The following tables may be of use in decisionmaking in business and other contexts, especially where the agent (the organisation or individual making a decision) has a non-linear desirability or utility function over the outcomes (i.e. the values of the discrete set of possible options), does not have perfect intuition and does not wish to put any more information into the decision that is not within the agent’s state of knowledge.

I present tables for n = 3, 4, 6, 8, 10, 12, 15, and 20 here, each at 7 bias values of q for each n, corresponding to percentages of the range from 1 to n of 5%, 17, 34, 50, 66, 83 and 95%. There is transformation group symmetry in this problem. If $i$ represents the side with $i$ spots up, then when we reflect from $i \to n + 1 - i$ and transform $x \to \frac{1}{x}$ we obtain the same probability, e.g. the probability of a 1 on a six sided die at 5th percentile bias is the same as a 6 at 95th percentile bias. This is why in our tables we can observe the corresponding symmetry in the values of the probabilities and in the entropy, which is maximal of all biases when there is no bias and thus no constraint. Readers may wish arbitrarily to adjust any of the probabilities in the tables in the appendix and recalculate the entropy $S = - \sum_{i = 1}^{n} p_{i} \log_{e} p_{i}$ , which will be lower than the maximum entropy value in the table.

APPENDIX Student-b Maximum Entropy Probability Tables

n=3
	q05	q17	q34	q0	q66	q83	q95
q-vals	1.1	1.34	1.68	2	2.32	2.66	2.9
Score	Probabilities
1	0.9078	0.7232	0.5064	0.3333	0.1864	0.0632	0.0078
2	0.0843	0.2137	0.3072	0.3333	0.3072	0.2137	0.0843
3	0.0078	0.0632	0.1864	0.3333	0.5064	0.7232	0.9078
entropy	0.3343	0.7386	1.0203	1.0986	1.0203	0.7386	0.3343

n=4
	q05	q17	q34	q0	q66	q83	q95
q-vals	1.15	1.51	2.02	2.5	2.98	3.49	3.85
Score	Probabilities
1	0.8689	0.6425	0.4136	0.25	0.1241	0.0324	0.002
2	0.1141	0.2374	0.2769	0.25	0.1854	0.0877	0.015
3	0.015	0.0877	0.1854	0.25	0.2769	0.2374	0.1141
4	0.002	0.0324	0.1241	0.25	0.4136	0.6425	0.8689
Entropy	0.445	0.9502	1.2921	1.3863	1.2921	0.9502	0.445

n=6
	q05	q17	q34	q0	q66	q83	q95
q-vals	1.250	1.85	2.70	3.5	4.3	5.15	5.75
Score	Probabilities
1	0.7998	0.5260	0.3043	0.1666	0.072	0.0134	0.0003
2	0.1602	0.2527	0.2282	0.1667	0.0961	0.028	0.0013
3	0.0321	0.1214	0.1711	0.1667	0.1282	0.0583	0.0064
4	0.0064	0.0583	0.1282	0.1667	0.1711	0.1214	0.0321
5	0.0013	0.028	0.0961	0.1667	0.2282	0.2527	0.1602
6	0.0003	0.0135	0.0721	0.1667	0.3043	0.526	0.7998
Entropy	0.6254	1.2655	1.6794	1.7918	1.6794	1.2655	0.6254

APPENDIX Student-b Maximum Entropy Probability Tables (ctd)

n=8
	q05	q17	q34	q0	q66	q83	q95
q-vals	1.35	2.19	3.38	4.5	5.62	6.81	7.65
Score	Probabilities
1	0.7407	0.4454	0.2412	0.125	0.05	0.0076	0.0001
2	0.1921	0.2489	0.1927	0.125	0.0626	0.0136	0.0002
3	0.0498	0.1391	0.1539	0.125	0.0784	0.0243	0.0009
4	0.0129	0.0777	0.1229	0.125	0.0982	0.0434	0.0034
5	0.0034	0.0434	0.0982	0.125	0.1229	0.0777	0.0129
6	0.0009	0.0243	0.0784	0.125	0.1539	0.1391	0.0498
7	0.0002	0.0136	0.0626	0.125	0.1927	0.2489	0.1921
8	0.0001	0.0076	0.05	0.125	0.2412	0.4454	0.7407
Entropy	0.7726	1.5012	1.9569	2.0794	1.9569	1.5012	0.7726

n=10
	q05	q17	q34	q0	q66	q83	q95
q-vals	1.45	2.53	4.06	5.5	6.94	8.47	9.55
Score	Probabilities
1	0.6896	0.3862	0.2	0.1	0.0381	0.005	0.0000
2	0.214	0.2382	0.1663	0.1	0.0458	0.0081	0.0001
3	0.0664	0.147	0.1383	0.1	0.055	0.0131	0.0002
4	0.0206	0.0907	0.115	0.1	0.0662	0.0213	0.0006
5	0.0064	0.0559	0.0957	0.1	0.0796	0.0345	0.002
6	0.002	0.0345	0.0796	0.1	0.0957	0.0559	0.0064
7	0.0006	0.0213	0.0662	0.1	0.115	0.0907	0.0206
8	0.0002	0.0131	0.055	0.1	0.1383	0.147	0.0664
9	0.0001	0.0081	0.0458	0.1	0.1663	0.2382	0.214
10	0.0000	0.005	0.0381	0.1	0.2	0.3862	0.6896
Entropy	0.8981	1.6905	2.1735	2.3026	2.1735	1.6905	0.8981

APPENDIX Student-b Maximum Entropy Probability Tables (ctd)

n=12
	q05	q17	q34	q0	q66	q83	q95
q-vals	1.55	2.87	4.74	6.5	8.26	10.13	11.45
Score	Probabilities
1	0.6451	0.3408	0.1708	0.0833	0.0306	0.0036	0.0000
2	0.2289	0.2255	0.1461	0.0833	0.0358	0.0055	0.0000
3	0.0812	0.1492	0.125	0.0833	0.0419	0.0083	0.0001
4	0.0288	0.0987	0.1069	0.0833	0.049	0.0125	0.0002
5	0.0102	0.0653	0.0915	0.0833	0.0572	0.0189	0.0005
6	0.0036	0.0432	0.0782	0.0833	0.0669	0.0286	0.0013
7	0.0013	0.0286	0.0669	0.0833	0.0782	0.0432	0.0036
8	0.0005	0.0189	0.0572	0.0833	0.0915	0.0653	0.0102
9	0.0002	0.0125	0.049	0.0833	0.1069	0.0987	0.0288
10	0.0001	0.0083	0.0419	0.0833	0.125	0.1492	0.0812
11	0.0000	0.0055	0.0358	0.0833	0.1461	0.2255	0.2289
12	0.0000	0.0036	0.0306	0.0833	0.1708	0.3408	0.6451
Entropy	1.0078	1.8001	2.1252	2.4849	1.7684	1.1462	1.0081

n=15
	q05	q17	q34	q0	q66	q83	q95
q-vals	1.7	3.38	5.76	8	10.24	12.62	14.3
Score	Probabilities
1	0.5882	0.2898	0.1402	0.0667	0.0236	0.0025	0.0000
2	0.2422	0.2063	0.1235	0.0667	0.0269	0.0035	0.0000
3	0.0997	0.1469	0.1087	0.0667	0.0305	0.0049	0.0000
4	0.0411	0.1046	0.0957	0.0667	0.0346	0.0069	0.0000
5	0.0169	0.0745	0.0843	0.0667	0.0393	0.0097	0.0001
6	0.007	0.053	0.0743	0.0667	0.0447	0.0136	0.0002
7	0.0029	0.0378	0.0654	0.0667	0.0507	0.0191	0.0005
8	0.0012	0.0269	0.0576	0.0667	0.0576	0.0269	0.0012
9	0.0005	0.0191	0.0507	0.0667	0.0654	0.0378	0.0029
10	0.0002	0.0136	0.0447	0.0667	0.0743	0.053	0.007
11	0.0001	0.0097	0.0393	0.0667	0.0843	0.0745	0.0169
12	0.0000	0.0069	0.0346	0.0667	0.0957	0.1046	0.0411
13	0.0000	0.0049	0.0305	0.0667	0.1087	0.1469	0.0997
14	0.0000	0.0035	0.0269	0.0667	0.1235	0.2063	0.2422
15	0.0000	0.0025	0.0236	0.0667	0.1402	0.2898	0.5882
Entropy	1.1517	2.0471	2.5698	2.7081	2.5698	2.0471	1.1517

APPENDIX Student-b Maximum Entropy Probability Tables (ctd)

n=20
	q05	q17	q34	q0	q66	q83	q95
q-vals	1.95	4.23	7.46	10.5	13.54	16.77	19.05
Score	Probabilities
1	0.5128	0.2318	0.108	0.05	0.0171	0.0016	0.0000
2	0.2498	0.1784	0.098	0.05	0.0188	0.0021	0.0000
3	0.1217	0.1372	0.0889	0.05	0.0207	0.0027	0.0000
4	0.0593	0.1056	0.0807	0.05	0.0229	0.0035	0.0000
5	0.0289	0.0812	0.0732	0.05	0.0252	0.0045	0.0000
6	0.0141	0.0625	0.0665	0.05	0.0278	0.0059	0.0000
7	0.0069	0.0481	0.0603	0.05	0.0306	0.0077	0.0000
8	0.0033	0.037	0.0547	0.05	0.0337	0.01	0.0001
9	0.0016	0.0285	0.0497	0.05	0.0371	0.013	0.0002
10	0.0008	0.0219	0.0451	0.05	0.0409	0.0169	0.0004
11	0.0004	0.0169	0.0409	0.05	0.0451	0.0219	0.0008
12	0.0002	0.013	0.0371	0.05	0.0497	0.0285	0.0016
13	0.0001	0.01	0.0337	0.05	0.0547	0.037	0.0033
14	0.0000	0.0077	0.0306	0.05	0.0603	0.0481	0.0069
15	0.0000	0.0059	0.0278	0.05	0.0665	0.0625	0.0141
16	0.0000	0.0045	0.0252	0.05	0.0732	0.0812	0.0289
17	0.0000	0.0035	0.0229	0.05	0.0807	0.1056	0.0593
18	0.0000	0.0027	0.0207	0.05	0.0889	0.1372	0.1217
19	0.0000	0.0021	0.0188	0.05	0.098	0.1784	0.2498
20	0.0000	0.0016	0.0171	0.05	0.108	0.2318	0.5128
Entropy	1.351	2.3085	2.8526	2.9957	2.8526	2.3085	1.351

Cambridge Investment Research

Tag Archives: Bayesianism

Student-b: Realistic decision-making for time-related quantities in business

Generalised Game Show Problem

Homage to Bazball Strategy

Student-b: Probability Tables