The scourge of statistical modelling
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Definition
Examples (many)
Possible solutions
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Vague def: A model is identifiable if all elements
of it can be estimated from the data available.
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Statistical def: A statistical model is identifiable if
each set of parameter values gives a unique
probability distribution for the possible outcomes
(the likelihood).
f (D| ) f (D| ') D

'
Lack of identifiability You have a problem.
Pops up ever so often!
We want to estimate the
length of a specific lizard.
We’ve got multiple
measurements.
We model the measurements
of total length as “body
length” + “tail length” + noise,
but we only measure total
length!
Total length, L
Body length,
Lb
Tail
length,
Lt
Li Lb Lt i
The parameters (Lb,Lt) are not where i ~ N(0, 2 ) i.i.d.
identifiable.
Li Lb Lt i
where i ~ N(0, 2 ) i.i.d.
Plotting Lb on the x axis and Lt
on the y axis, the likelihood
surface has a “ridge” on a
diagonal, going from (0,mean)
to (mean,0).
We want to study the variation in the length of
a species of lizard.
We’ve got multiple measurements, but only
one per organism. We model the
measurements as normally distributed
variation in individual length plus normally
distributed measurement noise.
Li L t i
where i ~ N(0, 2) is i.i.d.variation
in length
and i ~ N(0, 2) is i.i.d.measuremen
t noise
The variation parameters ( , ) are also not identifiable.
Total length, L
Occupancy analysis consists of trying to establish
how many areas are populated with a given
organism.
 It is performed by doing multiple transects in
multiple areas. One is interested in the
occupancy frequency, po . If the area is occupied,
there is a detection frequency, pd , associated
with each transect.
 What if we only have one transect per area? The
encounters are then binomially distributed with
frequency po* pd, so the detection and occupancy
frequencies are not identifiable.
 If po* pd =50%, that can either be because all the
sites are occupied but the chance of detecting
them are only 50%, or it could be because the
chance of detecting them is 100% and 50% of the
areas are occupied. (Or something in between).
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0
1
1
0
1
1
0
0
0
1
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1
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Newton’s law of gravity gave a
mechanistic description of the elliptical
orbits of the planets around the sun,
the deviances from perfect ellipses due
to the planets effect on each other,
Earth’s effect on us, the moons orbit
around Earth, the tides, the orbits of
comets and the orbits of the moons of
Jupiter. In short, a spectacular success!

It did however contain an identifiability
problem. Each orbit could be described
by the acceleration caused by other
bodies. But there was a gravitational
constant involved; constant times
mass. The masses could thus not be
identified, only their ratio compared to
each other!
a
GMe
2 R
R
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A standard (one-way) ANOVA model is defined
as: yij
i
ij where i denotes the group
and j denotes the measurement index of that
group. But and are not identifiable!

If you do linear regression on linearly
dependent covariates, you get the same
problem. This is bound to happen if you have
more covariates than measurements.
Sometimes a complicated distribution is
needed for describing variation in
nature. One way of modeling this is by
mixture distributions. Think of this as
first drawing a parametric distribution
and then drawing from that distribution.
f (x) pfN (x| 1, 1) (1 p) fN (x| 2, 2)
where fN is the normal distribution with
expectation and standard deviation .
What constitutes peak number one and
two is not clear though.
However if we could determine that, the
ratio, p, and the individual expectations,
, and standard deviations, , are
possible to estimate.
Peak 1 or 2?
Peak 2 or 1?
The hare/lynx capture data from
Hudson Bay company is often used
for studying fluctuations in
hare/lynx abundance.
But the abundances themselves
are not possible to estimate from
this data. The data can be modeled
as capture=abundance*hunting intensity*hunting success.
By assuming hunting intensity and hunting success to be constant,
one can at least get relative abundance from year to year, but the
absolute abundance is not identifiable.
Additional processes like the abundance of the plants that the
hares are eating, will also be subject to identification problems.
Phenotypic evolution of a single trait (fixed optimum)
modelled as Ornstein-Uhlenbeck process. Aside from
location and spread, the dynamics is given by a
characteristic time, t.
The optimum can also change (OU?). Phenotypic
evolution would then be linear tracking (natural
selection) of this underlying process. The presence and
the characteristic times of both the two dynamics can
be found given enough fossil data.
Which dynamics belong to which process can however
not be identified. If we find one slow-moving and one
rapidly moving component in our phenotypic data then;
a) Natural selection is rapidly tracking slow-moving
changes in the optimum.
b) Natural selection is slowly tracking fast-moving
changes in the optimum!
1.96 s
t
-1.96 s
Slow optimum
Fast selection
Fast optimum
Slow selection
If you really are interested in all the parameters in your model and
can’t identify them, then get better data!
Examples:
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(1) If you want to know a lizards body and tail lengths
then measure that!
(2) If you want to separate the variation of a species
from measurement noise, do multiple measurements
on each organism.
(3) If you want to estimate the occupancy rate, do
multiple transects in each area.
(6) If you have too few measurements to estimate all
regression parameters, get more data!
(4) If you really are interested in the gravitational
constant and the mass of planets and the sun, measure
the gravity between objects of known mass at a known
distance. Cavendish did, 110 years after Newton made
his law of gravity!
Cavendish experiment
Note: For ANOVA (5), it’s the definition of the parameters which is the problem. For hare-lynx
(8), the dynamics, not the absolute abundances are the issue. For separating selection from
optimum dynamics (9), getting data from the optimum in the past is not feasible!
If you have no interest in the parameters that are
causing problems, try to get rid of them!
Examples:

(1) If we are only interested in the total length of a lizard,
modelling each measurement as Li Lb Lt i (body
length+tail length+noise) makes no sense. Replace these two
parameters with a parameter representing total body length,
LT.

(3) If the probability of encountering a species is of interest,
don’t model that probability as po* pd . Use one encountering
probability, p , instead.
If you aren’t interested in all the problematic parameters,
but can’t get rid of them, fixate them.
Examples:
(5) In ANOVA the identifiability problem in yij
i
ij can be
fixed either by setting
i 0 (sum contrast) or by comparing all
groups to a single default group, 1 0 (treatment contrast).
 (4) Since any value of the gravitational constant could be used (preCavendish), one can fixate it so that it yields a “reasonable” estimate
for the mass of the Earth.
 (8) If it’s the dynamics of the ecology which is of interest, not absolute
abundances, fixate the hunting success parameters to something
reasonable.

Note: This will not work in case (7) and (9) since the identifiability
problem there is discrete. Nor will it work in case (6), since the
regression parameters themselves are of interest.
If multiple but discrete places in parameter space yields the same
likelihood, try putting restrictions on the parameters.
Examples:
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(7) In order to identify distribution one and two in a mixture, put restriction on
the mixture probability, p>0.5 , or the peak of the distributions, 1< 2 , or on
the spread of the distributions, 1> 2.
(9) Assume natural selection to move faster than the optimum, t1< t2 .
Note: In Bayesian analysis, you might be tempted
to give the same prior distribution for each
characteristic time as you would for a one layer
model. But when you then impose the restriction,
each prior gets more narrow and the two layer
model gets an unfair advantage in model
comparison!
Characteristic time
priors:
•One layer prior
•Lower two layer prior
•Upper two layer prior
So what if there are multiple solutions? They yield the same predictions. So if
you do ML optimization, just start from somewhere and seek an optimum
until you can’t get any higher. You’ll get different estimates when starting
from different places, but predictions-wise this doesn’t matter.
Examples:
 (1) You will get an estimate for body length between 0 and your mean total
length. The tail length will be the mean minus that. Total length will be
estimated as the mean.
 (9) Depending on where you start the optimization, you will get t1=ta and
t2=tb and or vice versa t1=tb and t2=ta .
Notes:
a) You probably will not get an estimate of the parameter uncertainty.
b) Other people using your model will get different answers and wonder why.
c)
The actual values you get will be a result of your optimization scheme.
An informative prior will give different probabilities to different parameter sets even when the
likelihood is the same.
Examples:
 (1) As with ML, you will get a distribution for body length between 0 and your mean total
length (maybe a little further). If you make your prior so that body length and tail length are
approximately the same, that will be the result for the posterior distribution also.
 (2) A strong prior on the size of the measurement noise may yield reasonable results for the
species variation, while taking the uncertainty due to the size of the measurement noise
into account.
 (6) A sensible prior can give regression estimates even when there are less measurements
than covariates! This can be viewed as the principle behind ridge regression and lasso
regression.
 (9) You will get a distribution for each characteristic time having two peaks, though they
may be of different height due to the prior distribution. (Same with 7.)
Notes:
a)
In the direction in parameter space dominated by the identifiability problem, only the
prior will have any real say in the outcome!
b)
This method is however much easier than the restriction method for model comparison!