Preliminaries
Before we jump into model averaging and in particular, SSD Model
Averaging, let’s backup a little and consider why we average and the
advantages and disadvantages of averaging.
The pros and cons of averaging
We’re all familiar with the process of averaging. Indeed,
averages are pervasive in everyday life - we talk of average
income; mean sea level; average global temperature; average height,
weight, age etc. etc. So what’s the obsession with averaging?
It’s simple really - it’s what statisticians call data reduction
which is just a fancy name to describe the process of summarising a lot
of raw data using a small number of (hopefully) representative
summary statistics such as the mean and the standard deviation.
Clearly, it’s a lot easier to work with just a single mean than all the
individual data values. That’s the upside. The downside is that the
process of data reduction decimates your original data - you lose
information in the process. Nevertheless, the benefits tend to outweigh
this information loss. Indeed, much of ‘conventional’ statistical theory
and practice is focused on the mean. Examples include T-tests, ANOVA,
regression, and clustering. When we talk of an ‘average’ we are usually
referring to the simple, arithmetic mean:\[\bar{X}=\frac{1}{n}\sum\limits_{i=1}^{n}{{{X}_{i}}}\]
although we recognize there are other types of mean including the
geometric mean, the harmonic mean and the weighted mean. The last of
these is particularly pertinent to model averaging.
Weighted Averages
For the simple arithmetic mean, all of the individual values receive
the same weighting - they each contribute \(\frac{1}{n}\) to the summation. While this
is appropriate in many cases, it’s not useful when the components
contribute to varying degrees. An example familiar to ecotoxicologists
is that of a time-varying concentration as shown in the figure
below.
From the figure we see there are 5 concentrations going from left to
to right: \(\left\{ 0.25,0.95,0.25,0.12,0.5
\right\}\). If we were to take the simple arithmetic mean of
these concentrations we get \(\bar{X}=0.414\). But this ignores the
different durations of these 5 concentrations. Of the 170
hours, 63 were at concentration 0.25, 25 at concentration 0.95, 23 at
concentration 0.25, 23 at concentration 0.12, and 36 at concentration
0.50. So if we were to weight these concentrations by time
have:
\[{{{\bar{X}}}_{TW}}=\frac{\left(
63\cdot 0.25+25\cdot 0.95+23\cdot 0.25+23\cdot 0.12+36\cdot 0.50
\right)}{\left( 63+25+23+23+36
\right)}=\frac{56.01}{170}=0.33\]
So, our formula for a
weighted average is:\[\bar{X}=\sum\limits_{i=1}^{n}{{{w}_{i}}{{X}_{i}}}\]
with \(0\le {{w}_{i}}\le 1\) and \(\sum\limits_{i=1}^{n}{{{w}_{i}}=1}\).
Note, the simple arithmetic mean is just a special case of the
weighted mean with \(\sum\limits_{i=1}^{n}{{{w}_{i}}=\frac{1}{n}}\)
; \(\forall i=1,\ldots ,n\)
Model Averaging
The weighted average acknowledges that the elements in the
computation are not of equal ‘importance’. In the example above,
this importance was based on the proportion of time that the
concentration was at a particular level. Bayesians are well-versed in
this concept - the elicitation of prior distributions for model
parameters provides a mechanism for weighting the degree to which the
analysis is informed by existing knowledge versus using a purely
data-driven approach. Model averaging is usually used in the context of
estimating model parameters or quantities derived from a fitted model -
for example an EC50 derived from a C-R model. Let’s motivate the
discussion using the following small dataset of toxicity estimates for
some chemical.
#> [1] 1.73 0.57 0.33 0.28 0.30 0.29 2.15 0.80 0.76 0.54 0.42 0.83 0.21 0.18 0.59
Now, suppose we have only two possibilities for fitting an SSD - both
lognormal distributions. Model 1 is the LN(-1.067,0.414) distribution
while Model 2 is the LN(-0.387,0.617) distribution. A plot of the
empirical cdf and Models 1 and 2 is shown below.
Emprirical cdf (black); Model 1(green); and
Model 2 (blue)
We see that Model 1 fits well in the lower, left region and
poorly in the upper region, while the reverse is true for Model 2. So
using either Model 1 or Model 2 is going to
result in a poor fit overall. However, the obvious thing to do is to
combine both models. We could just try using 50% of
Model 1 and 50% of Model 2, but that may be sub-optimal. It turns out
that the best fit is obtained by using 44% of Model 1 and 56% of Model
2. Redrawing the plot and adding the weighted average of Models
1 and 2 is shown below.
Empirical cdf (black); Model 1(green); Model 2
(blue); and averaged Model (red)
Clearly the strategy has worked - we now have an excellent
fitting SSD. What about estimation of an HC20? It’s a simple
matter to work out the individual HC20 values for
Models 1&2 using the appropriate qlnorm() function in
R. Thus we have:
# Model 1 HC20
cat("Model 1 HC20 =", qlnorm(0.2, -1.067, 0.414))
#> Model 1 HC20 = 0.2428209
# Model 2 HC20
cat("Model 2 HC20 =", qlnorm(0.2, -0.387, 0.617))
#> Model 2 HC20 = 0.4040243
What about the averaged distribution? An intuitively appealing
approach would be to apply the same weights to the individual
HC20 values as was applied to the respective models. That is
0.44*0.2428209 + 0.56*0.4040243 = 0.33.
So our model-averaged HC20 estimate is 0.33. As a check, we
can determine the fraction affected at concentration = 0.33 -
it should of course be 20%. Let’s take a look at the plot.
Something’s wrong - the fraction affected at concentration 0.33 is
30% - not the required 20%. This issue is taken up in
the next section
Model Averaged SSDs
As we’ve just seen, applying the model weights to component
HCx values and summing does not produce the
correct result. The reason for this can be explained mathematically as
follows (if your not interested in the mathematical explanation -
skip ahead to the next section).
The fallacy of weighting individual HCx values
The correct expression for a model-averaged SSD is: \[G\left( x \right) = \sum\limits_{i = 1}^k {{w_i}}
{F_i}\left( x \right)\] where \({F_i}\left( \cdot \right)\) is the
ith component SSD (i.e. cdf) and
wi is the weight assigned to \({F_i}\left( \cdot \right)\).
Notice
that the function \(G\left( x \right)\)
is a proper cumulative distribution function (cdf)
which means for a given quantile, x, \(G\left( x \right)\) returns the
cumulative probability: \[P\left[ {X
\leqslant x} \right]\]
Now, the incorrect approach takes a weighted sum of the
component inverse cdf’s, that is:
\[H\left( p \right) = \sum\limits_{i = 1}^k
{{w_i}} {F_i}^{ - 1}\left( p \right)\] where
\({F_i}^{ - 1}\left( \cdot \right)\) is the
ith inverse cdf. Notice that
\({G_i}\left( \cdot \right)\) is a function
of a
quantile and returns a
probability while
\({H_i}\left( \cdot \right)\) is a
function of a
probability and returns an
quantile.
Now, the correct method of determining the HCx is to
work with the proper model-averaged cdf \(G\left( x \right)\). This means finding the
inverse function \({G^{ -
1}}\left( p \right)\). We’ll address how we do this in a moment.
The reason why
\(H\left( p \right)\)
does
not return the correct result is because of
the
implicit assumption that the inverse of \(G\left( x \right)\) is equivalent to \(H\left( p \right)\). This is akin to
stating the inverse of a
sum is equal to the sum of the
inverses i.e.
\[\sum\limits_{i = 1}^n
{\frac{1}{{{X_i}}}} = \frac{1}{{\sum\limits_{i = 1}^n {{X_i}}
}}{\text{ ???}}\]
For the mathematical nerds: There are some very special
cases where the above identity does in fact hold, but for that you need
to use
complex numbers.
For example, consider two complex numbers \[{\text{a = }}\frac{{\left( {5 - i}
\right)}}{2}{\text{ and }}b = - 1.683 - 1.915i\] It can be
shown that \[\frac{1}{{a + b}} = \frac{1}{a}
+ \frac{1}{b} = 0.126 + 0.372i\]
Back to the issue at hand, and since we’re not dealing with complex
numbers, it’s safe to say:\[{G^{ - 1}}\left(
p \right) \ne H\left( p \right)\]
If you need a visual demonstration, we can plot \(G\left( x \right)\) and the
inverse of \(H\left( p
\right)\) both as functions of x (a quantile) for our
two-component lognormal distribution above.
Clearly, the two functions are
not the same and thus
HCx values derived from each will nearly always be different
(as indicated by the positions of the vertical red and green dashed
lines in the Figure above corresponding to the 2 values of the
HC20). (Note: The two curves do cross over at a concentration
of about 1.12 corresponding to the 90
th percentile, but in
the region of ecotoxicological interest, there is no such cross-over and
so the two approaches will
always yield different
HCx values with this difference → 0 as x → 0).
We next discuss the use of a model-averaged SSD to obtain the
correct model-averaged HCx.
Computing a model-averaged HCx
A proper
HCx needs to satisfy what David Fox refers to as
the inversion principle.
More formally, the inversion principle states that an HCx
(denoted as \({\varphi _x}\))
must satisfy the following:
\[df\left( {{\varphi _x}} \right) = x\quad
\quad and\quad \quad qf\left( x \right) = {\varphi _x}\]
where \(df\left( \cdot \right)\) is a
model-averaged distribution function (i.e. SSD) and \(qf\left( \cdot \right)\) is a
model-averaged quantile function. For this equality to hold, it
is necessary that \(qf\left( p \right) = d{f^{
- 1}}\left( p \right)\).
So, in our example above, the green curve was taken to be
\(qf\left( x \right)\) and this was used to
derive
\({\varphi _x}\) but the
fraction affected \(\left\{ { =
df\left( {{\varphi _x}} \right)} \right\}\) at
\({\varphi _x}\) is computed using the red
curve.
In ssdtools the following is a check that the inversion
principle holds:
# Obtain a model-averaged HCx using the ssd_hc() function
hcp<-ssd_hc(x, p = p)
# Check that the inversion principle holds
ssd_hp(x, hcp, multi_est = TRUE) == p # this should result in logical `TRUE`
Note: if the multi_est argument is set to
FALSE the test will fail.
The inversion principle ensures that we only use a
single distribution function to compute both the
HCx and the fraction affected. Referring to the figure
below, the HCx is obtained from the MA-SSD (red curve) by
following the → arrows while the fraction affected is obtained by
following the ← arrows.
Finally, we’ll briefly discuss how the HCx is computed in
R using the same method as has been implemented in
ssdtools.
Where do the model-averaged weights come from?
This is a little more complex, although we’ll try to provide a
non-mathematical explanation. For those interested in going deeper, a
more comprehensive treatment can be found in (Burnham and Anderson 2002) and (Fletcher 2018).
This time, we’ll look at fitting a gamma, lognormal, and pareto
distribution to our sample data:
#> [1] 1.73 0.57 0.33 0.28 0.30 0.29 2.15 0.80 0.76 0.54 0.42 0.83 0.21 0.18 0.59
The adequacy (or otherwise) of a fitted model can be assessed
using a variety of numerical measures known as
goodness-of-fit or GoF statistics. These are invariably
based on a measure of discrepancy between the emprical data and the
hypothesized model. Common GoF statistics used to test whether the
hypothesis of some specified theoretical probability distribution is
plausible for a given data set include: Kolmogorov-Smirnov test;
Anderson-Darling test; Shapiro-Wilk test;and Cramer-von Mises test.
The
Cramer-von Mises test is a good choice and is readily performed
using the cvm.test() function in the goftest
package in R as follows:
dat <- data.frame(Conc = c(1.73, 0.57, 0.33, 0.28, 0.3, 0.29, 2.15, 0.8, 0.76, 0.54, 0.42, 0.83, 0.21, 0.18, 0.59))
library(goftest)
library(EnvStats) # this is required for the Pareto cdf (ppareto)
# Examine the fit for the gamma distribution (NB: parameters estimated from the data)
cvm.test(dat$Conc, null = "pgamma", shape = 2.0591977, scale = 0.3231032, estimated = TRUE)
# Examine the fit for the lognormal distribution (NB: parameters estimated from the data)
cvm.test(dat$Conc, null = "plnorm", meanlog = -0.6695120, sd = 0.7199573, estimated = TRUE)
# Examine the fit for the Pareto distribution (NB: parameters estimated from the data)
cvm.test(dat$Conc, null = "ppareto", location = 0.1800000, shape = 0.9566756, estimated = TRUE)
Cramer-von Mises test of goodness-of-fit
Braun's adjustment using 4 groups
Null hypothesis: Gamma distribution
with parameters shape = 2.0591977, scale = 0.3231032
Parameters assumed to have been estimated from data
data: dat$Conc
omega2max = 0.34389, p-value = 0.3404
Cramer-von Mises test of goodness-of-fit
Braun's adjustment using 4 groups
Null hypothesis: log-normal distribution
with parameter meanlog = -0.669512
Parameters assumed to have been estimated from data
data: dat$Conc
omega2max = 0.32845, p-value = 0.3719
Cramer-von Mises test of goodness-of-fit
Braun's adjustment using 4 groups
Null hypothesis: distribution ‘ppareto’
with parameters location = 0.18, shape = 0.9566756
Parameters assumed to have been estimated from data
data: dat$Conc
omega2max = 0.31391, p-value = 0.4015
From this output and using a level of significance of \(p = 0.05\), we see that none of the
distributions is implausible. However, if forced to choose just
one distribution, we would choose the Pareto distribution
(smaller values of the omega2max statistic are better).
However, this does not mean that the gamma and lognormal distributions
are of no value in describing the data. We can see from the plot below,
that in fact both the gamma and lognormal distributions do a reasonable
job over the range of toxicity values. The use of the Pareto may be a
questionable choice given it is truncated at 0.18 (which is the minimum
value of our toxicity data).
Emprirical cdf (black); lognormal (green); gamma
(blue); and Pareeto (red)
As in the earlier example, we might expect to find a better fitting
distribution by combining all three distributions using a
weighted SSD. The issue we face now is how do we choose the
weights to reflect the relative fits of the three distributions?
Like all tests of statistical significance, a p-value is
computed from the value of the relevant test statistic - in
this case, the value of the omega2max test statistic. For
this particular test, it’s a case of the smaller the better. From
the output above we see that the omega2max values are \(0.344\) for the gamma distribution, \(0.328\) for the lognormal distribution, and
\(0.314\) for the Pareto distribution.
We might somewhat naively compute the relative weights as:
\({w_1} = \frac{{{{0.344}^{ - 1}}}}{{\left(
{{{0.344}^{ - 1}} + {{0.328}^{ - 1}} + {{0.314}^{ - 1}}} \right)}} =
0.318\) \({w_2} =
\frac{{{{0.328}^{ - 1}}}}{{\left( {{{0.344}^{ - 1}} + {{0.328}^{ - 1}} +
{{0.314}^{ - 1}}} \right)}} = 0.333\) and \({w_3} = \frac{{{{0.314}^{ - 1}}}}{{\left(
{{{0.344}^{ - 1}} + {{0.328}^{ - 1}} + {{0.314}^{ - 1}}} \right)}} =
0.349\)
(we use reciprocals since smaller values
of omega2max represent better fits). As will be seen
shortly - these are incorrect.
However, being based on a simplistic measure of discrepancy between the
observed and hypothesized distributions, the
omega2max statistic is a fairly ‘blunt instrument’ and has
no grounding in information theory which is the basis for
determining the weights that we seek.
A discussion of
information theoretic methods for assessing
goodness-of-fit is beyond the scope of this vignette. Interested readers
should consult
(Burnham and Anderson
2002). A commonly used metric to determine the model-average
weights is the
Akaike Information Criterion or
AIC.
The formula for the
\(AIC\) is:
\[AIC = 2k - 2\ln \left( \ell \right)\]
where
\(k\) is the number of model
parameters and
\(\ell\) is the
likelihood for that model. Again, a full discussion of
statistical likelihood is beyond the present scope. A relatively gentle
introduction can be found
here.
The likelihood for our three distributions can be computed in
R as follows:
sum(log(dgamma(dat$Conc, shape = 2.0591977, scale = 0.3231032)))
#> [1] -7.020597
sum(log(dlnorm(dat$Conc, meanlog = -0.6695120, sdlog = 0.7199573)))
#> [1] -5.812947
sum(log(EnvStats::dpareto(dat$Conc, location = 0.1800000, shape = 0.9566756)))
#> [1] -5.621683
From which the AIC values readily follow:
#> AIC for gamma distribution = 18.04119
#> AIC for lognormal distribution = 15.62589
#> AIC for Pareto distribution = 15.24337
As with the omega2max statistic,
smaller values of AIC are better. Thus, a
comparison of the AIC values above gives the ranking of distributional
fits (best to worst) as: Pareto > lognormal > gamma
Computing model weights from the AIC
We will simply provide a formula for computing the model weights from
the AIC values. More detailed information can be found here.
The AIC for the ith distribution fitted
to the data is \[AI{C_i} = 2{k_i} - 2\ln
\left( {{L_i}} \right)\] where \({L_i}\) is the ith
likelihood and \({k_i}\) is the
number of parameters for the ith
distribution. Next, we form the differences: \[{\Delta _i} = AI{C_i} - AI{C_0}\] where
\(AI{C_0}\) is the AIC for the
best-fitting model (i.e.\(AI{C_0} = \mathop {\min }\limits_i \left\{
{AI{C_i}} \right\}\) ). The model-averaged weights \({w_i}\) are then computed as:
AIC Model averaging weights\[{{w}_{i}}=\frac{\exp \left\{ -\frac{1}{2}{{\Delta }_{i}} \right\}}{\sum{\exp \left\{ -\frac{1}{2}{{\Delta }_{i}} \right\}}}\]
The model-averaged weights for the gamma, lognormal, and Pareto
distributions used in the previous example can be computed ‘manually’ in
R as follows:
dat <- c(1.73, 0.57, 0.33, 0.28, 0.3, 0.29, 2.15, 0.8, 0.76, 0.54, 0.42, 0.83, 0.21, 0.18, 0.59)
aic <- NULL
k <- 2 # number of parameters for each of the distributions
aic[1] <- 2 * k - 2 * sum(log(dgamma(dat, shape = 2.0591977, scale = 0.3231032))) # Gamma distribution
aic[2] <- 2 * k - 2 * sum(log(dlnorm(dat, meanlog = -0.6695120, sdlog = 0.7199573))) # lognormal distribution
aic[3] <- 2 * k - 2 * sum(log(EnvStats::dpareto(dat, location = 0.1800000, shape = 0.9566756))) # Pareto distribution
delta <- aic - min(aic) # compute the delta values
aic.w <- exp(-0.5 * delta)
aic.w <- round(aic.w / sum(aic.w), 4)
cat(
" AIC weight for gamma distribution =", aic.w[1], "\n",
"AIC weight for lognormal distribution =", aic.w[2], "\n",
"AIC weight for pareto distribution =", aic.w[3], "\n"
)
AIC weight for gamma distribution = 0.1191
AIC weight for lognormal distribution = 0.3985
AIC weight for pareto distribution = 0.4824
Finally, let’s look at the fitted model-averaged SSD:
Empirical cdf (black) and model-averaged fit
(magenta)
As can be seen from the figure above, the model-averaged fit provides
a very good fit to the empirical data.
Correcting for distributions having differing numbers of
parameters
In deriving the AIC, Akaike had to make certain, strong assumptions.
In addition, the bias factor (the \(2k\) term) was derived from theoretical
considerations (such as mathematical expectation) that relate
to infinite sample sizes. For small sample sizes, the AIC is
likely to select models having too many parameters (i.e models which
over-fit). In 1978, Sugiura proposed a modification to the AIC
to address this problem, although it too relied on a number of
assumptions. This ‘correction’ to the AIC for small samples (referred to
as \(AI{{C}_{c}}\)) is:
Corrected Akaike Information Criterion (AICc)\[AI{{C}_{c}}=AIC+\frac{2{{k}^{2}}+2k}{n-k-1}\]where n is the sample size and k is the number of parameters.
It is clear from the formula for \(AI{{C}_{c}}\) that for \(n\gg k\), \(AI{{C}_{c}}\simeq AIC\). The issue of
sample size is ubiquitous in statistics, but even more so in
ecotoxicology where logistical and practical limitations invariably mean
we are dealing with (pathologically) small sample sizes. There are no
hard and fast rules as to what constitutes an appropriate
sample size for SSD modelling. However, Professor David Fox’s personal
rule of thumb which works quite well is:
Sample size rule-of-thumb for SSD modelling \[n\ge 5k+1\]where n is the sample size and k is the number of parameters.
Since most of the common SSD models are 2-parameter, we should be
aiming to have a sample size of at least 11. For 3-parameter models
(like the Burr III), the minimum sample size is 16 and if we wanted to
fit a mixture of two, 2-parameter models (eg.
logNormal-logNormal or logLogistic-logLogistic) the
sample size should be at least 26. Sadly, this is rarely the
case in practice!
