What’s Reasonable About a Range? Casualty Loss Reserve Seminar September 11-12, 2006

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What’s Reasonable About a Range?
Casualty Loss Reserve Seminar
September 11-12, 2006
Roger M. Hayne, FCAS, MAAA
How Did We Get Here?
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Traditional actuarial methods are
deterministic
At their best they say if certain specific things
happen then the ultimate payments (incurred
losses, reserves) will be such and such
No models, so no statement as to how likely
those “certain specific things” are to happen,
nor how likely the final payments will be
“close” to “such and such”
Some Estimates are Better than Others
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Actuaries have long known estimates for
certain lines are “better” than others, but
how?
Traditionally apply many “methods” (i.e.
many “certain specific things”) and review
resulting “such and such” outcomes
If the “such and such” “bunch” the actuary
“feels good” about the final estimates
If the “such and such” are spread out
considerably, the actuary “feels
uncomfortable”
Enter Reasonable Estimates
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How can one tell a “good” “such and such”
from a not-so-good “such and such”?
Without underlying models we cannot tell
which is “more likely”
We do the next best thing and ask whether
the “certain specific things” are “reasonable”
or not
If they are “reasonable” we say the
“estimate” (whatever that is) is “reasonable”
Range of Reasonable Estimates
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Various “certain specific things,” seem
“reasonable” to a “reasonable person” and
others do not
A “Range of Reasonable Estimates” is a
collection of “such and such” that follow from
“reasonable” “certain specific things”
As with beauty, “reasonable” is in the eye of
the beholder
Honest “reasonable” people can disagree
Can be the fodder for disputes
Enter the Accountants
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Problem
– There can be many “reasonable” estimates
– Methods do not quantify uncertainty
– Accounting statements require single numbers
that are treated as “fact”
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Solution, book a “reasonable estimate” of the
unpaid claim costs
Implication – book the number that will
happen
But, no single number is very likely
Let’s Put Some Numbers Around This
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Begin by assuming the goal is to estimate the
amount that will ultimately happen
In addition assume that you know all possible
outcomes (Z) and the probabilities associated with
those outcomes, i.e. you know the “distribution of
outcomes” with certainty
What would be “reasonable?”
– X with 1/3≤Pr(Z≤X)≤2/3
– Between X and Y where X and Y are values
that minimize Y-X subject to Pr(X≤Z≤Y)=1/3
– Between E(Z)-T and E(Z)+T where T is
selected such that Pr(E(Z)-T ≤Z≤ E(Z)+T)=1/3
– Something else?
The Infamous Die
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Running the risk of Bob’s ire suppose the ultimate
payment is determined by the roll of a fair die (rolled
and hidden when the policy is written)
Here we know the distribution – each number 1
through 6 equally likely
Symmetric so the “reasonable range” of between 2.5
and 4.5 satisfies all 3 of the conditions but not
necessarily unique
Comes down to what “feels right”
Focus is the outcome not the estimate
Even with perfect knowledge outcome is uncertain
Narrowing the Gap
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Instead of focusing on the (impossible to
forecast) outcome why not some summary
statistic of the distribution?
In our ideal world statistics of the variable will
have known values
For example, even though the outcome is
unknown, its mean or “expected value” is
known
If we define the mean as a reasonable
estimate then the “range of reasonable
estimates” is a single point
What Is It Going To Be
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We have a fundamental choice to make
– Do we pick the reserve booked as a point on
the distribution (“what will happen”)?
– Do we pick the reserve booked as some
statistic or descriptor of the distribution?
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If a descriptor which one?
–
–
–
–
Mean
Mode
Median
Other?
My Favorite Descriptor
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What is the “best” amount to book (Z)?
For each potential outcome X calculate the decrease
in value of the company if the company booked Z but
actual losses are X, call this amount g(X,Z)
A rational amount to book as reserves is that value of
Z for which the expected value of g(X,Z), expected
over all values of X is as small as possible (the least
pain)
It turns out the mean solves this problem if
g(X,Z)=(X-Z)2
Is the mean really “rational” or even “reasonable”???
Which is “Better?”
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Actuaries like to think about the average
– It all evens out in the end (pluses and minuses
cancel over the long term)
– It may even be possible to predict the mean
(or at least get “close”)
– May make life “easier”
– But usually not “verifiable” by actual events
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Our publics, however, often think in terms of
“what will happen”
We are measured by how close our
estimates are to what “actually happens”
Don’t Confuse Me With Facts…
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In reality we seldom know the distribution of
outcomes
We must estimate the distribution
We can think of a distribution of distributions
If we pick a descriptor we can calculate the value of
the descriptor for each of the distributions
We thus have a distribution of the descriptors
We can have “reasonable ranges” of the descriptors
as just as we had “ranges of reasonable” outcomes
before
Very Simple Example
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Losses have lognormal distribution,
parameters m (unknown) and σ2 (known),
respectively the mean and variance of the
related normal
The parameter m itself has a normal
distribution with mean μ and variance τ2
Here we have a distribution of distributions,
one for each value of the parameter m
Distribution of Outcomes & Means
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Expected is lognormal
– Parameters μ+ σ2/2 and τ2
– c.v.2 of expected is exp(τ2)-1
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“What will happen” is lognormal
– Parameters μ and σ2+τ2
– c.v.2 is exp(σ2+τ2)-1
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c.v. = standard deviation/mean, measure of
relative dispersion
Note expected is much more certain (smaller
c.v.) than “what will happen”
Ranges
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Suppose Pr(Z<α)=1/3 for Z~N(0,1)
Take “reasonable range” 1/3≤Pr(Z≤X)≤2/3
Reasonable range of outcomes between
exp(μ-α(σ2+τ2)) and exp(μ+α(σ2+τ2))
Reasonable range of means between
exp(μ+σ2/2-ατ2) and exp(μ+σ2/2+ατ2)
As expected the reasonable range of
outcomes is wider than the reasonable range
of expected values
Evolution of Estimates
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Think of the distribution of parameters as
describing different possible “states of the
world” with assumed likelihoods
Base these estimates on
– Prior analysis of data
– Judgment
– Other?
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We can now talk about reasonable ranges of
outcomes or of any descriptor (mean,
median, mode, least pain, …)
Evolution of Estimates (Cont.)
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Now observe the “real” losses for the next
year
For each parameter value calculate the
likelihood of observing that value given the
value
Re-weight the prior distribution giving more
weight to the parameters with a higher
likelihood of observing the real value
This gives revised descriptors and ranges
An Evolutionary (Bayesian) Model
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Again take a very simple example
Use the die example
For simplicity assume we book the mean
This time there are three different dice that can be
thrown and we do not know which one it is
Currently no information favors one die over others
The dice have the following chances of outcomes:
1
1/6
1/21
6/21
2
1/6
2/21
5/21
3
1/6
3/21
4/21
4
1/6
4/21
3/21
5
1/6
5/21
2/21
6
1/6
6/21
1/21
Evolutionary Approach
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“What will happen” is the same as the first
die, equal chances of 1 through 6
The expected has equally likely chances of
being 2.67, 3.50, or 4.33
If you set your reserve at the “average” both
have the same average, 3.5, the true
average is within 0.83 of this amount with
100% confidence
There is a 1/3 chance the outcome will be
2.5 away from this pick.
We now “observe” a 2 – what do we do?
How Likely Is It?
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Likelihood of observing a 2:
– Distribution 1
– Distribution 2
– Distribution 3
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1/6
2/21
5/21
Given our distributions it seems more likely that
the true state of the world is 3 (having observed
a 2) than the others
Use Bayes Theorem to estimate posterior
likelihoods
Posterior(model|data)likelihood(data|model)prior(model)
Evolutionary Approach
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Revised prior is now:
– Distribution 1
– Distribution 2
– Distribution 3
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0.33
0.19
0.48
Revised posterior distribution is now:
1
2
3
4
5
6
0.20
0.19
0.17
0.16
0.15
0.13
Overall mean is 3.3
The expected still takes on the values 2.67, 3.50, and
4.33 but with probabilities 0.48, 0.33, and 0.19
respectively (our “range”)
Next Iteration
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Second observation of 1
Revised prior is now (based on observing a 2 and a
1):
– Distribution 1
– Distribution 2
– Distribution 3
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0.28
0.05
0.67
Revised posterior distribution is now:
1
2
3
4
5
6
0.25
0.21
0.18
0.15
0.12
0.09
Now the mean is 3.0
The expected can be 2.67, 3.50, or 4.33 with
probability 0.67, 0.28, and 0.05 respectively
Some “Take-Aways”
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Always be clear on what you are describing
– The entire distribution of future outcomes
(“what will happen”)?
– Some statistic representing the distribution?
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Our publics probably think in terms of the
former – we in terms of the latter
The reality is that the distribution of future
outcomes (“what will happen”) is quite
disperse – maybe too wide for “comfort”
Disclosure, disclosure, disclosure
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