Here I show how to take answers from one expert that came in the form of a Beta distribution via the shiny app (example interface here). There they highlighted how many whales make a transition, as well as their certainty around that answer. In the link provided above, the experts were asked to provide answers for three transitions:

  1. Animals remaining in the MIDA
  2. Animals moving south from MIDA to SEUS
  3. Animals moving from “Northern” regions into MIDA

In this last question we lump 7 different regions north of the MIDA into one. This was done to reduce the congitive load on the experts. However, because the model from Schick et al. (2013) operates with 9 geographic regions, we need to convert from the single Beta back to 7 Beta distributions, which we will then use in the Dirichlet distribution that serves as the prior for the movement transitions. This vignette describes this process.

Necessary Background

Let’s say we three possible areas and we ask for all possible transitions between and among these areas. If we do this with a slider, then we can easily generate a full Dirichlet distribution. For example, from area 1 to area 2, 20 out of 100 whales move. From area 1 to area 1, 25 out of 100 whales move. Necessarily, this means that 55 whales move from area 1 to area 3.

The weight a user assigns to these parameters controls the variance in the individual Dirichlet parameters, i.e. \(Dir(0.2, 0.25, 0.55)\) is a very different distribution from \(Dir(20, 25, 55)\), that is while their means are the same, the variance is very different:

When we set up the shiny interface to ask the questions about the transitions, we combined several possible transitions into one. Now that we’ve done that, we need to reverse engineer it, and make the values from the one distribution match up to the values from many Dirichlet’s.

We’ll start with a toy example of a Beta distribution with parameters Beta(10, 30) which we want to turn into 5 Beta distributions with parameters Be(a, b).

We start with factoring them down into the 5 distributions as follows:

\[ \mu = \frac{a}{a+b} = \frac{10}{10+30} \cdot \frac{1}{5} = 0.05 \] and

\[ a+b = \frac{40}{5} \] which yields

\[ b = 8 - a \]

Finally, \[ a = 2, b = 6. \]

Our toy example, then included 3 transitions to be turned into 7 transitions - the last 5 of which are identical. Say we have three distributions: Be(4, 20), Be(15, 10), and Be(10, 30); after dividing, those will translate to:

  1. Be(4, 20)
  2. Be(15, 10)
  3. Be(2, 6)
  4. Be(2, 6)
  5. Be(2, 6)
  6. Be(2, 6)
  7. Be(2, 6)

Real Worked Example

Partitioning up the Beta Distributions

Now let’s look at taking some real answers for one of the questions above, and turning it into priors for the movement analysis. Looking at one person’s answers:

Table 1. Example answers from one expert for one question in the expert elicitation conducted in June 2015.
question gender rawConf scaleConf mida2mida mida2seus mida2northern
13 MALE 4 19.8181 40 10 50

Let’s explain these variables and values just a bit for clarification:

How these values then translate into parameters for beta distribution is as follows. We used the following parameterisation: \(\alpha = n \times m_1\) and \(\beta = n - n \times m_1\), where \(n\) = scaleConf, and \(m_1\) = mida2mida / 100. For the first transtion this yields, \(\alpha = 19.8181 \times 40 / 100\), or 7.92724, and \(\beta = 19.8181 - 19.8181 \times 40 / 100\), or 11.89086. So for this Beta distribution we would have Be(7.93, 11.89).

For the second transition, we’ll repeat with the appropriate substitution for the new answers: \(\alpha = 19.8181 \times 10 / 100\), or 1.98181, and \(\beta = 19.8181 - 19.8181 \times 10 / 100\), or 17.83629. So for this Beta distribution we would have Be(1.98, 17.84).

For the last transition, we need to calculate the Beta distribution parameters, and then divide by 7. The distribution for the transition is: \(\alpha = 19.8181 \times 50 / 100\), or 9.90905, and \(\beta = 19.8181 - 19.8181 \times 50 / 100\), or 9.90905. So for this Beta distribution we would have Be(9.91, 9.91). However, we need to then divide this by 7 to get the final distributions. This would yield Be(1.41, 1.41) for each of these 7 transitions.

The summary table of all these transitions for the Beta parameters is now:

Table 2. Marginal Beta distributions for 9 possible transitions from MIDA to other regions from January to February.
transition alpha beta
mida2mida 7.93 11.89
mida2seus 1.98 17.84
mida2bof 1.41 1.41
mida2gom 1.41 1.41
mida2gsc 1.41 1.41
mida2jl 1.41 1.41
mida2ne 1.41 1.41
mida2nrth 1.41 1.41
mida2rb 1.41 1.41

This table shows the marginal distributions for each of the \(\pi_i\) parameters of the Dirichlet distribution. In the last section, we go from the marginals to a final prior distribution.

Assembling the Dirichlet Prior

In the final analysis of the movement data, we are going to use a mixture of Dirichlet distributions that is comprised of each experts’ unique Dirichlet prior distribution for a particular transition. This section describes how to take the marginal distributions given above, and turn them into the final expert-specific prior.

We need to assemble the Dirichlet comprised of 9 possible transitions from the 9 Beta distributions outlined above, e.g. \(Dir(d_1, d_2, \ldots, d_9)\). In this instance \(n = d_1 + d_2 + \ldots + d_9\). Each Beta distribution, \(Be(a_{11}, a_{12})\) can be expressed as \(Be(d_1, n - d_1)\); therefore \(d_1 = a_{11}, d_2 = a_{21}\), etc. We use these definitions to construct the distribution as follows.

\[ n = 7.93 + 1.98 + 1.41 + 1.41 + 1.41 + 1.41 + 1.41 + 1.41 + 1.41 = 39.6 \]

This allows us to express this experts’ Dirichelt as:

\[Dir(7.93, 1.98, 1.41, 1.41, 1.41, 1.41, 1.41, 1.41, 1.41).\]