Public perception of climate change: the `icons` dataset in the `hyper2` package
In hyper2: The Hyperdirichlet Distribution, Mark 2

library("hyper2",quietly=TRUE)
set.seed(0)
knitr::opts_chunk$set(echo = TRUE)

knitr::include_graphics(system.file("help/figures/hyper2.png", package = "hyper2"))

To cite the hyper2 package in publications, please use @hankin2017_rmd. This short document analyses the climate change dataset introduced by @oneill2008 and discussed in the hyperdirichlet R package [@hankin2010], but using the improved hyper2 package instead. @oneill2009 observe that lay perception of climate change is a complex and interesting process, and here we assess the engagement of non-experts by the use of "icons" (this word is standard in this context. An icon is a "representative symbol") that illustrate different impacts of climate change.

Here we analyse results of one experiment [@oneill2008], in which subjects were presented with a set of icons of climate change and asked to identify which of them they found most concerning. Six icons were used: PB [polar bears, which face extinction through loss of ice floe hunting grounds], NB [the Norfolk Broads, which flood due to intense rainfall events], L [London flooding, as a result of sea level rise], THC [the thermo-haline circulation, which may slow or stop as a result of anthropogenic modification of the water cycle], OA [oceanic acidification as a result of anthropogenic emissions of ${\mathrm C}{\mathrm O}_2$], and WAIS [the West Antarctic Ice Sheet, which is rapidly calving as a result of climate change]. Methodological constraints dictated that each respondent could be presented with a maximum of four icons. The R idiom below (dataset icons in the package) shows the experimental results.

library("hyper2",quietly=TRUE)
M <- icons_table # saves typing
M

(M is called icons_table in the package). Each row of M corresponds to a particular cue given to respondents. The first row, for example, means that a total of $5+3+4+3=15$ people were shown icons NB, L, THC, WAIS [column names of the the non-NA entries]; 5 people chose NB as most concerning, 3 chose L, and so on. The dataset is more fully described in the package. The builtin icons likelihood function in the hyper2 package may be created by using the saffy() function:

icons
icons == saffy(icons_table)  # should be TRUE

At this point, the icons object as created above is mathematically identical to the icons object in the hyperdirichlet package (and indeed the hyper2 package), but the terms might appear in a different order due to disordR discipline.

Analysis of the icons dataset

The first step is to find the maximum likelihood estimate for the icons likelihood:

options("digits" = 4)
(mic <- maxp(icons))
dotchart(mic,pch=16)

We also need the log-likelihood at an unconstrained maximum:

L1 <- loglik(indep(mic),icons)
options(digits=9)
L1

We see agreement to 4 decimal places with the value given in the hyperdirichlet package. The next step is to assess a number of hypotheses concerning the relative sizes of $p_1$ through $p_6$.

Hypothesis testing.

Below, I investigate a number of inequality hypotheses about $p_1,\ldots, p_6$. The general analysis proceeds as follows; we can use $H_0\colon p_1=1/6$ as an example but the method applies to any observation. I consider the general case first, then modifications for one-sided hypotheses.

We have an observation ${\mathcal O}$ and wish to quantify the strength of evidence to support this.
Translate our observation into a hypothesis phrased in terms as a restriction on parameter space.
State a null hypothesis and alternative. For example, we might have $H_0\colon p_1=1/6$ and $H_A\colon p_1\neq 1/6$, and $H_A$ is the complement of $H_0$. We sometimes write $H_A\colon\sum p_i=1$, ignoring the (measure zero) $H_0$.
We then attempt to reject $H_0$ and to do this we perform two likelihood maximizations: one unconstrained, corresponding to $H_A$, and one constrained, corresponding to $H_0$.
We calculate ${\mathcal L}{H_0} :=\operatorname{argmax}{p\in H_0}{\mathcal L}(p)$ and ${\mathcal L}{H_A} := \operatorname{argmax}{p\in H_A}{\mathcal L}(p)$ and observe that, in general, ${\mathcal O}$ implies the strict inequality ${\mathcal L}{H_A} > {\mathcal L}{H_0}$.
Calculate the likelihood ratio $R={\mathcal L}{H_A}/{\mathcal L}{H_0}>1$ or equivalently the support for $H_0$, $\Lambda = \log R>0$.
We may either use Edwards's criterion of two units of support per degree of freedom, or Wilks's theorem: $H_0\longrightarrow 2\Lambda\sim\chi^2_d$. If we are considering inequality hypotheses, we have $d=1$.
If the criterion is met we may reject $H_0$ and infer that $H_A$ is the case.

As another example, consider $H_0\colon p_1=p_2=p_3$. The vector ${\mathbf p}\in \left{\left.\left(p_1,\ldots,p_6\right)\right| \sum p_i=1\right}$ is a point in a 5-dimensional manifold, and $H_0$ asserts that ${\mathbf p}\in \left{\left.\left(p_1,p_1,p_1,p_4,p_5,p_6\right)\right| 3p_1+p_4+p_5+p_6=1\right}$, that is, a 3-dimensional manifold. The null imposes a loss of two degrees of freedom.

The logic above operates for one-sided tests. We might observe that $\hat{p_1}$ is the largest and wish to test a null of $p_1\leqslant 1/6$ against an alternative of $p_1>1/6$. Note that the one-sided p-value and likelihood ratio statistic are the same as the two-sided values.

Below I will rework some of the hypotheses tested in the hyperdirichlet package, with consistent labelling of null and alternative hypotheses, and renumbering

Equality of strengths

The most straightforward null would be the hypothesis of player equality, specifically:

[ H_E\colon p_1=p_2=\cdots=p_n=\frac{1}{n}. ]

("E" for equal). This was not carried out in @hankin2010, because I had not thought of it. Testing $H_E$ is implemented in the package as test.equalp(). Note that because $H_E$ is a point hypothesis we would have $n-1$ degrees of freedom (because $H_0$ has $n-1$ df).

equalp.test(icons)

Hypothesis 1: $p_1 = 1/6$

Following the analysis in @hankin2010, and restated above, we first observe that NB [the Norfolk Broads] is the icon with the largest estimated probability with $\hat{p_1}\simeq 0.25$ (O'Neill gives a number of theoretical reasons to expect $p_1$ to be large). This would suggest that $p_1$ is in fact large, in some sense, and here I show how to assess this statement statistically. Consider the hypothesis $H_0\colon p_1=1/6$, and as per the protocol above we will try to reject it.

To that end, we perform a constrained optimization, with (active) constraint that $p_1\leqslant 1/6$ (note that the inequality constraint allows us to use fast maxp(), which has access to derivatives). We note the support at the evaluate and then compare this support with the support at the unconstrained evaluate; if the difference in support is large then this constitutes strong evidence for $H_A$ and then would conclude that $p_1 > 1/6$. In package idiom, the optimization is implemented by the specificp.test() suite of functions; these work by imposing additional constraints to the maxp() function via the fcm and fcv arguments. Using the defaults we have:

specificp.test(icons,1)

which tests a null of $p_1\leqslant\frac{1}{6}$; see how the evaluate under the null is on the boundary and we have $\hat{p_1}=\frac{1}{6}$. Compare the support of 2.607 with 2.608181 from the hyperdirichlet package. This exceeds Edwards's two-units-of-support criterion; the $p$-value is obtained by applying Wilks's theorem on the asymptotic distribution of 2\log\Lambda.

Both these criteria indicate that we may reject that hypothesis that $p_1\leqslant 1/6$ and thereby infer $p_1>\frac{1}{6}$.

Hypothesis 2: $p_1\geqslant\max\left(p_2,\ldots,p_6\right)$

We observe that NB (the Norfolk Broads) has large strength, and hypothesise that it is in fact stronger than any other icon. This is another constrained likelihood maximization, although this one is not possible with convex constraints. In the language of the generic procedure given above, we would have

[ H_0\colon (p_1\leqslant p_2)\cup (p_1\leqslant p_3)\cup (p_1\leqslant p_4)\cup (p_1\leqslant p_5)\cup (p_1\leqslant p_6) ]

[ \overline{H_0}\colon (p_1 > p_2)\cap (p_1 > p_3)\cap (p_1 > p_4)\cap (p_1 > p_5)\cap (p_1 > p_6) ]

(although note that $H_0$ has nonzero measure). In words, $H_0$ states that $p_1$ is smaller than $p_2$, or $p_1$ is smaller than $p_3$, etc; while $\overline{H_0}$ states that $p_1$ is larger than $p_2$, and is larger than $p_3$, and so on. Considering the evaluate, we see that ${\mathcal L}{H_0} < {\mathcal L}{\overline{H_0}}$, so optimizing over $\overline{H_0}$ is equivalent to unconstrained optimization [of course, the intrinsic constraints $p_i\geqslant 0, \sum p_i\leqslant 1$ have to be respected]. Here, $\overline{H_0}$ is regions of $(p_1,\ldots,p_6)$ with $p_1$ being greater than at least one of $p_2,\ldots,p_6$. The union of convex sets is not necessarily convex (e.g. a two-way Venn diagram). As far as I can see, the only way to do it is to perform a sequence of five constrained optimizations: $p_1\leqslant p_2, p_1\leqslant p_3, p_1\leqslant p_4, p_1\leqslant p_5$. The fillup constraint would be $p_1\leqslant p_6\longrightarrow 2p_1+p_2+\cdots +p_5\leqslant 1$. We then choose the largest likelihood from the five.

o <- function(Ul,Cl,startp,give=FALSE){
    small <- 1e-4  #  ensure start at an interior point
    if(missing(startp)){startp <- small*(1:5)+rep(0.1,5)}           
    out <- maxp(icons, startp=small*(1:5)+rep(0.1,5), give=TRUE, fcm=Ul,fcv=Cl)
    if(give){
        return(out)
    }else{
        return(out$value)
    }
}

p2max <- o(c(-1, 1, 0, 0, 0), 0)  # p1 <= p2
p3max <- o(c(-1, 0, 1, 0, 0), 0)  # p1 <= p3
p4max <- o(c(-1, 0, 0, 1, 0), 0)  # p1 <= p4
p5max <- o(c(-1, 0, 0, 0, 1), 0)  # p1 <= p5
p6max <- o(c(-2,-1,-1,-1,-1),-1)  # p1 <= p6 (fillup)

(the final line is different because $p_6$ is the fillup value).

likes <- c(p2max,p3max,p4max,p5max,p6max)
likes
ml <- max(likes) 
ml

Thus the first element of likes corresponds to the maximum likelihood, constrained so that $p_1\leqslant p_2$; the second element corresponds to the constraint that $p_1\leqslant p_3$, and so on. The largest likelihood is the easiest constraint to break, in this case $p_1\leqslant p_3$: this makes sense because $p_3$ has the second highest MLE after $p_1$. The extra likelihood is given by

L1-ml

(the hyperdirichlet package gives 0.0853 here, a surprisingly small discrepancy given the difficulties of optimizing over a nonconvex region). We conclude that there is no evidence for $p_1\geqslant\max\left(p_2,\ldots,p_6\right)$.

It's worth looking at the evaluate too:

o2 <- function(Ul,Cl){
  jj <-o(Ul,Cl,give=TRUE)
  out <- c(jj[[1]],1-sum(jj[[1]]),jj[[2]])
  names(out) <- c("p1","p2","p3","p4","p5","p6","support")
  return(out)
}
rbind(
o2(c(-1, 1, 0, 0, 0), 0),  # p1 <= p2
o2(c(-1, 0, 1, 0, 0), 0),  # p1 <= p3
o2(c(-1, 0, 0, 1, 0), 0),  # p1 <= p4
o2(c(-1, 0, 0, 0, 1), 0),  # p1 <= p5
o2(c(-2,-1,-1,-1,-1),-1)   # p1 <= p6
)

In the above, the evaluate is the first five columns ($p_6$ being the fillup) and the final column is the log-likelihood at the evaluate. See how the constraint is active in each line: M[1,] == M[1:5,2:6]. Also note that the largest log-likelihood is the second row: if we were to violate any of the constraints, it would be $p_1<p_3$, consistent with the fact that $p_3$ (polar bears) has the second highest strength, after $p_1$.

Low frequency responses

The next hypothesis follows from the observed smallness of $\hat{p_5}$ (ocean acidification) and $\hat{p_6}$ (West Antarctic Ice Sheet) at 0.111 and 0.069 respectively. These two strengths correspond to "distant" concerns and O'Neill had reason to consider their sum (which she argued would be small). Thus we specify $H_0\colon p_5+p_6\leqslant \frac{1}{3}$.

The optimizing constraint of $p_5+p_6\geqslant\frac{1}{3}$ translates to an operational constraint of $-p_1-p_2-p_3-p_4\geqslant-\frac{2}{3}$ (because $p_5+p_6=1-p_1-p_2-p_3-p_4$):

jj <- o(c(-1,-1,-1,-1,0) , -2/3, give=TRUE,start=indep((1:6)/21))$value
jj

then the extra support is

L1-jj

(compare 7.711396 in hyperdirichlet, not sure why the discrepancy is so large).

Final example

The final example is motivated by the fact that both the distant icons $p_5$ and $p_6$ had lower strength than any of the local icons $p_1,\ldots, p_4$. Thus we would have $H_0\colon\max\left{p_5,p_6\right}\geqslant\min\left{p_1,p_2,p_3,p_4\right}$. This means the null optimization is constrained so that at least one of $\left{p_5,p_6\right}$ exceeds at least one of $\left{p_1,p_2,p_3,p_4\right}$. So we have the union of the various possibilities:

\begin{equation}\label{eq:Habar}H_0= \overline{H_A}\colon \bigcup_{j\in\left{5,6\right}\atop k\in\left{1,2,3,4\right}} \left{\left(p_1,p_2,p_3,p_4,p_5,p_6\right)\left|\sum p_i=1, p_j\geqslant p_k\right.\right} \end{equation}

The fillup value $p_6$ behaves differently in this context and $p_6\geqslant p_2$, say, translates to $-p_1-2p_2-p_3-p_4-p_5\geqslant -1$.

small <- 1e-4
start <- indep(c(small,small,small,small,0.5-2*small,0.5-2*small))
jj <- c(
   o(c(-1, 0, 0, 0, 1), 0,start=start),  # p1 >= p5
   o(c( 0,-1, 0, 0, 1), 0,start=start),  # p2 >= p5
   o(c( 0, 0,-1, 0, 1), 0,start=start),  # p3 >= p5
   o(c( 0, 0, 0,-1, 1), 0,start=start),  # p4 >= p5

   o(c(-2,-1,-1,-1,-1),-1,start=start),  # p1 >= p6
   o(c(-1,-2,-1,-1,-1),-1,start=start),  # p2 >= p6
   o(c(-1,-1,-2,-1,-1),-1,start=start),  # p3 >= p6
   o(c(-1,-1,-1,-2,-1),-1,start=start)   # p4 >= p6
   )
jj

Above, the elements of vector jj are the maximum likelihoods for each of the separate components of the parameter space allowed under $H_0$. Note that these components are not disjoint. So the maximum likelhood for the whole of allowable parameters under $H_0$ would be the maximum of these maxima:

max(jj)

This corresponds to the fourth component of $H_0$, viz $p_4=p_5$. The extra support is thus

L1-max(jj)

(compare hyperdirichlet which gives 3.16, not sure why the difference although if pressed I would point to hyper2 using multiple walkers in its optimization [defaulting to $n=10$] ). We should look at the maximum value:

o(c( 0, 0, 0,-1, 1), 0,give=TRUE,start=start)

So the evaluate is at the boundary, for $p_4=p_5$; THC and OA have the same Bradley-Terry strength. The small amount of extra support given by allowing an unconstrained optimization would suggest that there is no strong evidence for the contention investigated, viz "both the distant icons $p_5$ and $p_6$ have lower strength than any of the local icons $p_1,\ldots, p_4$".