Article from:
Risk Management
August 2008 – Issue No. 13
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estimator P̂ has an approximate normal distribution.
gh, and a few other technical conditions are satisfied, the
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normal distribution.
on page
38
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The distribution’s variance can be estimated from Vˆ 2
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Md. He can be reached at
[email protected].
Geoffrey Hancock, FSA, CERA,
FCIA, is a director at Oliver
Wyman in Toronto, ON. He
can be reached at geoffrey.
[email protected].
Page 37 w
Risk Management
The standard approach to this problem is to start with a ran
w August 2008
Chairman’s
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more
practical
process
for
confirming
the
asymptotic
variance
formula,
which we call
To be more specific, assume the option matures in T 10 Table
years with
1: a strike price of X 110 .
4
CTE(95%)
10-year
(1000 Trials), X=$110, S=$100
the
stock
price
at
isisagiven
by: European
15% . That
maturity
is given
by:
Theis,
current
stock price
isVerification,
assumed
to follow
a lognormalPut
returnOption
process with
S maturity
100 andfor
Variance
as
follows:
SPuppose
you
have
aThat
model
Sthe
uppose
that
takes
you
all
have
night
to
run
that
n
=
takes
1,000
all
scenarios.
night
to
It
run
would
n
=
1,000
be
impractical
scenarios.
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would
be aimpractical
Monte
Carlo
Simulation
without
Variance
Reduction
CTE(95%)
for
10-year
>
PT Va Tmodel
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@
8
%
and
.
is,
stock
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at
maturity
is
given
by:
V
15
%
T S ˜ e Variance Reduction
Table 1: Monte Carlo Simulation
>PT tohundreds
V repeat
T ˜Z S
@without
European
Put
Option
(1000
Trials),
X=$110,
S=$100
to repeatSthe
run S
process
the
of
time
run
process
in
order
hundreds
to
test
the
of
validity
time
in
of
order
the
to
variance
test
the
formulas
validity
of
as
the
variance
formulas
as
T
˜
e
>PT V T ˜Z @ Put Option (1000 Trials), X=$110, S=$100
CTE(95%)
for aS10-year
T S ˜ eEuropean
>
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described above.
described
above.
5
S T mean
˜ e zero and unit variance.
where
with
whereZZisisaastandard
standardNormal
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variate
with Smean
4
FSE VaˆR CoˆvCTE ,VaR CTˆE (95%) FSE CTˆE fˆ VaR VaˆR
zero
and
unit
variance.
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A
more
process
for
more
confirming
practical
the
process
asymptotic
for
confirming
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the asymptotic
which
variance
we call formula, which we call
dard Normal
variate
with
mean
zero
and
variance.
where
is practical
a standard
Normal
variate
with
and
unit variance.
ormal
variate
mean
zero
and
unit
, the
random
variable
whose
CTE
we
wish
to
Using
aZ with
continuous
discount
rate
ofunit
Gmean
6variance.
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n/a
4.39
n/a
2.37
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as follows:
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is as Form
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4
calculate is then the present value payoff function:
13.67
1
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1.65
0.49%
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Using
discountrate
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of G 6% , the
the random variable whose CTE we wish to
Using aacontinuous
variable
whoseLast
CTE
we wish
to
ous discount
ratecontinuous
of G 6discount
% , the random
14.93we wish
1.95 to
3.33
3.07
4.91
0.22%
Trial CTE
calculate
is then
the
present
valuewe
payoff
function:
, CTE
the
random
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count rate
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G
6
%
random
whose
wish
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calcu
G ˜T
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7.72
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late payoff
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ent value
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G ˜T
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V T ˜Z
Using C
spreadsheet
software,
we
can
samples
of
this
variable.
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this
n
1000
18.89
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M
aximum
e
˜ max 0, X S ˜ e
sample,
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estimators for the
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usingFrom
the 0.18
formulas
Using spreadsheet
software,theweplug-in
can generate
this VaR
variable.
this
1.63
1.76
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G ˜T
T V T ˜Zn 1000 samples
developed
To0estimate
density
useVaR
the estimator:
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e weearlier.
˜canmax
, Xthe plug-in
Sthe˜ probability
e Pestimators
calculate
for thef (VaR
CTE) and
using the formulas
software,
we can
generate
samples
of this
variable.
From this
n the
1000
developed
earlier.
To estimate
probability
density
estimator:
f (VaR
) use
Some
keythe
takeaways
from this table are:
>
G ˜T
>
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@
1.91
2.42
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7.31
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0.77
1.06
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et
calculate the plug-in estimators forfˆ (VaR
the) CTE and [VaR using the formulas
continued on page 40
1
1
. To estimate
the probability
f (VaR
)Fˆnuse
ware,
we can
generate density
of
this
variable.
From
this estimate of what would happen if the simulation were
n 1000
(D the
)[ Fˆestimator:
[ ) given
Any
trial provides
a reasonable
n ƒ(D
fˆ (VaR
) samples
1
1
Fˆn (D ) Fˆn (D [ )
repeated,
but
with adate
of n strike
there is considerable variability, especially for
1000price.
European putestimators
option gives the holder
right toCTE
sell the underlying
asset onusing
the
maturity
for thesize
specified
ate the Aplug-in
forthethe
and VaR
thesample
formulas
ˆR
V
a
.
with [ ˆ1
.
[ f (VaR) use the estimator:
stimate the
density
f (1100
VaR.)
with [probability
1
1
1
100
Fˆn (D ) Fˆn (D [ )
The CTE plug-in estimator is biased below the true closed form value of 13.80 (i.e., average
CTˆE estimator
is 13.70). as
However the bias is much smaller than the sampling error.
We can then calculate the Formula Standard Error (FSE) of each
We can then calculate the Formula Standard Error (FSE) of each estimator as
ƒ The asymptotic variance formula for the CTE estimator performs quite well on average (i.e.,
fˆ (VaR)
Page 39 w
ˆ FSE
1
1
VAR
X
X
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aximum
1.63
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0.19%
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14.93
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0.18
1.76
0.77
Last
Trial
key
takeaways
from
this
table
are:0.18 1.65
more 3.33
practical
process
confirming
the
1.63 3.077.72
0.18
1.76asymptotic0.22%
0.77
0.19%
14.93
1.95
4.91
Std
Deviation 3.33
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which we0.22
call
1.01
0variance formula,
0.19 1.06
Minimum
Some
keya takeaways
from
this
table are:
Some
key
takeaways
from
this
table
are:
Some
key
takeaways
from
this
table
are:
•
A
ny
given
trial
provides
reasonable
esVariance
Verification,
is
as
follows:
7.72
1.01
0
0.19
0.22
0.09%
Minimum
13.70
1.63
4.50
1.91
2.42
Average
timate
of what would
happen
if the
simuSome
key
takeaways
from
this
table
are:
13.70
1.63
4.50
1.91
2.42
0.40%
Average
18.89trial provides
2.27 a reasonable
9.17
7.31
13.05
Maximum
ƒ ofAny
estimate
of what ifwould
happen
if we
th
ƒ Any given trial provides a reasonable
estimate
what
would
the1. simulation
were
were
repeated,
butgiven
with
a happen
sample if estimate
As part
model
development,
orthe
in an
off
ƒlation
Any
given
trial
provides
a reasonable
of of
what
would
happen
simulation
18.89
2.27
9.17
7.31
13.05
3.65%
Maximum
1.63
0.18
1.76
0.77
1.06
Std
Deviation
repeated,
but size
with
aestimate
sample
size
of is
thereaifis
considerable
nwould
1000
repeated, but with a sample
sizerepeated,
of
there
considerable
variability,
especially
for
n 1000
size
isis
varipeak
time,
generate
(once)
larger
sampleespecially
of variabil
butprovides
with
a considerable
sample
of
there
considerable
variability,
n 1000
ƒ Any
given
trial
a reasonable
of
what
happen
the
simulation
were f
1.63
0.18
1.76
0.77
1.06
0.19%
Std Deviation
ˆ
VaˆR.
V
a
R
.
ability,
especially
for
say
N
=
5,000
scenarios.
Let
CTE
be
the
esˆ
V
a
R
.
N
repeated, but with a sample size of n 1000 there is considerable variability, especially for
Some
keyestimator
takeaways
from this
table are:timated CTE based on this large sample and
• The
plug-in
is biased
below
ˆR.CTE
ƒ The CTE plug-in estimator isVa
biased
belowplug-in
the
true
closed
form
value
of
13.80
(i.e.,
average
ƒ value
The
CTE
plug-in
estimator
isthe
biased
below standard
the true
closed
form
valueavera
of 1
ƒ
The
CTE
estimator
is
biased
below
true
closed
form
value
offor13.80
(i.e.,
the
true
closed
form
of
13.80
(i.e.,
averFSE
the formula
error
a given
Some key ˆtakeaways from this table are:
N
ˆ
CTE is 13.70). However theage
bias isˆ much
smaller
than
the
sampling
error.
C
Ttrial
E isthe
13.70).
However
the
bias
isthan
much
smaller
than
the sampling
error.
13.70).
However
is much
confidence
level
α.
TƒE isis
13.70).
However
the
bias
is
much
smaller
the
sampling
ƒ The CCTE
plug-in
estimator
isbias
biased
the
true
closed
form
value
oferror.
13.80
(i.e.,
average
Any
given
provides
abelow
reasonable
estimate
of
what
would
happen
if the
simula
smaller
the
sampling
2.on
Fnrom
the large
sample
of size
N, drawvariability,
m=100
buterror.
with
a sample
size
of
there
is considerable
1000
CTˆE for
isthan
13.70).
the
bias
isvariance
much
smaller
than
sampling
error.
ƒ Any
given
trial provides
a reasonable
estimate
of
what
would
happen
ifwell
the
simulation
were
ƒ The
asymptotic
variance
formula
therepeated,
CTE
estimator
performs
quite
average
(i.e.,
ƒHowever
The
asymptotic
formula
forthe
the
CTE
estimator
performs
quiteespe
we
ƒTheThe
asymptotic
variance
formula
for
the
CTE
estimator
performs
quite
well
on average
(i.
•
asymptotic
variance
formula
for
the
CTE
random
sub-samples
of
size
n=1,000
without
ˆE ).
Vdeviation
aˆR.there
repeated,
butFSE
with
size of
is
considerable
variability,
especially
for
n 1000
CTaˆEsample
empirical
standard
of
C
T
average
ˆ
ˆ
FSE
C
T
E
empirical
standard
deviation
of
C
T
E
).
average
average
FSEvariance
CTˆE wellformula
empirical
standard
deviation
of performs
CTˆE ). quite well on average (i.e.,
ƒ The
asymptotic
for
the CTE
estimator
estimator
performs
quite
on average
(i.e.,
replacement.
VaˆR.
ˆ
ˆ
The
plug-in
estimator
is
biased
below
the
true
closed
form value
of
13.80
(i.e
3. or
F
ofEpeak
the
mtime,
sub-samples
calculate
a
ƒ The VaR plug-in estimator isaverage
biasedƒhigh
(average
isempirical
4.50),
butstandard
again
thedeviation
bias
isor
much
smaller
FSE
TAs
E part
of
T
). (average
1.CCTE
of model
development,
ineach
an C
off
generate
(once)
a larger
samp
ƒ The
VaR
plug-in
estimator
is biased
high
is 4.50),
butisagain
the
bia
ƒ
The
VaR
plug-in
estimator
is
biased
high
(average
is
4.50),
but
again
the
bias
much
smal
ˆ
say
N
=
5,000
scenarios.
Let
CTE
be
the
estimated
CTE
based
on
this
large
samp
deviation
of
).
CTE
estimate
and
an
FSE
estimate.
Also
ƒ Thethan
CTEthe
plug-in
estimator
true
closed
form value
of 13.80
average
sampling
error. is biased below
CTthe
E is
13.70).
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the bias
is much
smaller
than the sampling error.
N(i.e.,
than
the
sampling
error.
than
the
sampling
error.
ƒ bias
VaR
plug-in
estimator
biased
high
(average
4.50),
but again
the
is much
formula
standard
errorcheck
for is
a given
D.
FSE
• The
Theis
VaR
plug-in
estimator
isissampling
biased
high
to
see confidence
whether
ourlevel
bestbias
answer
CTEN smaller
N thethe
CTˆE is 13.70). However the
much
smaller
than
error.
1000
pairs
of
estim
ƒ The sample covariance for all
ators
is
2.37,
which
is
higher
(lower)
than
ƒ
The
asymptotic
variance
formula
for
the
CTE
estimator
performs
quite
well on ave
than
the is
sampling
(average
4.50),ƒbuterror.
againsample
the biascovariance
is much for lies
in thepairs
approximate
95% is
The
1000
of
estim
ators
2.37, which
is th
h
Thefirst
sample
covariance
for sample
allto
1000
pairs
ofall
estim
ƒ the
ators
is
2.37,
which
isconfidence
higher
(lower)
the
estimated
covariance
from
(last)
trials,
but
close
the
mean
of
all
covariance
ˆ
2.
s
of
size
n=1,000
From
the
large
of
size
N,
draw
m=100
random
sub-sample
ˆ
ƒ The asymptotic variance formula
for the
estimator
quite well
on
average
(i.e.,
FSEestimated
Cperforms
TE empirical
standard
deviation
of Ctrials,
TIfE it). does
average
smaller
thanCTE
the sampling
error.
interval
CTE+2xFSE.
we set to
thethe
the
covariance
from
the
first
(last)
but
close
mean
estimated
covariance
from
firstof(last)
close
to the
all covarian
without
replacement.
sample
covariance
for
1000the
pairs
estimtrials,
ƒ Thethe
ators isbut
2.37,
which
is mean
higherof(lower)
than
estimators.
FSE CTˆE empirical
of
E all
). pairs
average
• Tstandard
he sampledeviation
covarianceestimators.
forC
allTˆ1000
of esCI (Confidence Interval) count to 1 and 0
estimators.
the
estimated
covariance
from
the
first
(last)
trials,
but
close
to
the
mean
of
all
covariance
ƒ
The
VaR
plug-in
estimator
is
biased
high
(average
is
4.50),
but
again
the
bias
is
muc
timators is 2.37,
is higher
(lower)
than
otherwise.a CTE estimate and an FSE estimate. Also
3. which
amples
calculate
For each
of the
m sub-s
A
number
of
more
realistic
examples
are
documented
in
the
paper
using
the
same
methodology
estimators.
than
the
sampling
error.
ƒ The VaR plug-in estimator is biased
high (average
is 4.50),
but
again
thebest
bias
issemuch
smaller
the estimated
covariance
from
first
(last)
4.
U
the
standard
of the
the CTE
esticheck
to
see
whether
our
answer
CTE
lies indeviation
the approximate
95%
confidence
N documented
A
number
ofthe
more
realistic
examples
are
in
paper
using
the s
as
described
above.
In
each
case
the
asymptotic
theory
worked
as
expected.
The
examples
were
A
number
of
more
realistic
examples
are
documented
the paper using
the
same count
methodolo
than the sampling error.
wefrom
setinthe
Interval)
2 u FSE . If it does
trials, but close to interval
the meanCTE
of allrcovariance
mates
StepCI3 (Confidence
to check the validity
of the to 1 a
The
sample
covariance
for
all
1000
pairs
of
estim
ƒ
ators
is
2.37,
which
is
higher
(lo
as
described
above.
In
each
case
the
asymptotic
theory
worked
as
expected.
T
chosen to test a wide range of
behaviours
andIn
practical
problems
facing insurers.
as
described
above.
each
case
the
theory
worked
examples
we
Apossible
number
of more
realistic
examples
areasymptotic
documented
in the
paper as
using
methodology
otherwise.
estimators.
asymptotic
formula.
Asexpected.
we the
will same
seeThe
shortly
a
the
estimated
covariance
from
first(lower)
(last)
trials,
but
close
to
the
mean ofwere
all c
chosen
tocase
test
athe
wide
range
of theory
possible
behaviours
and
practical
problems
facin
alldescribed
1000 pairs
of aestim
ƒ The sample covariance for as
ators
is
2.37,
which
is the
higher
than
chosen
toabove.
test
wide
range
of
possible
behaviours
and
practical
problems
insurers.
In each
asymptotic
worked
as expected.
The
examples
simple
adjustment
needs
to befacing
made
to
this
the estimated covariance from
thetofirst
(last)
but
close
the mean
ofand
all practical
covariance
4. trials,
dard to
deviation
of the
CTE
estimates
from Step 3 to check
the validity o
Use
the
stan
chosen
test
aestimators.
wide
range
of
possible
behaviours
insurers.
A number
of more
realistic
examples
are docunumber
before problems
comparingfacing
it to the
formula
asymptotic formula. As we will see shortly a simple adjustment needs to be made t
estimators.
mented in the papernumber
using the
samecomparing
method- it to the
estimates. estimates.
before
A number of more
realistic
examples are formula
documented in the paper using the same met
as described
above. In each case the
A More Practical Example – ology
Variance
Verification
as
described
above.
In
each
case
the
asymptotic
theory
worked
The exam
A number of more realistic examples
are
documented
in
the
paper
using
the
same
methodology
asymptotic theory
worked
asPractical
expected.
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The
table
below
shows
thessresults
of applying
The tab
toasanexpected.
inforce
portfolio
of U
leMore
below
shows
theExample
results
the
above
proce
A
– Variance
Verification
A
More
Practical
Example
–
Variance
Verification
to
test
a
wide
range
of
possible
behaviours
and
practical
problems
facing
insurer
as described
above.
each case
the
asymptotic
theory
worked
as
expected.
The
examples
were
Suppose you
haveIna model
thatexamples
takes
allchosen
night
to
run
n
=
1,000
scenarios.
It
would
be
impractical
ariable
annuities
with
GMDB,
GMAB
and
GMWB
features.
The
book
is
slightly
out
of
v
were chosen to test a wide range of
the above process to an inforce portfolio of U.S.
A
More
Practical
Example
–
Variance
Verification
to
repeat
the
run
process
hundreds
of
time
in
order
to
test
the
validity
of
the
variance
formulas
as
money.
5000
real
world
(P
measure)
scenarios
were
generated
and
for
each
scenario
the
pr
chosen to test a wide range of possible
behaviours Sand
practical
problems
facing
insurers.
uppose
youthat
have
a model
that
all=night
to
runGMDB,
n = 1,000
scenarios.
It wou
possible
and
practical
problems
variable
with
GMAB
andimpractic
Supposebehaviours
you
have
a model
takes
all night
totakes
runpresent
nannuities
1,000
scenarios.
Itfees
would
be
described above.
value
of
guarantee
benefits
(claims)
less
the
value
of
related
was
captured.
repeat the
run process
hundreds
time
inThe
order
to istest
thevariance
validity
of the var
facing
insurers.
GMWBof
features.
book
slightly
out of theformulas
to repeat
runatomodel
process
hundreds
ofnight
time
in
test the
validity
of
the
Suppose
you the
have
that
takes all
to order
run n to
= 1,000
scenarios.
It would
be impractical
described
above.
money.
5000
world
(P the
measure)
described
above.
A more practical process forto
confirming
asymptotic
variance
formula,
we
call
The
claims have
been normalized
so the
mean
CTE(0)
is 1,000.
Thevariance
feesscenarios
were
scaled soastha
repeat
thethe
run
process
hundreds
of
time inwhich
order
to
test
thereal
validity
of
formulas
Variance Verification, is as follows:
A More
Practical
Example—
TE(75)
of
the
net
(PV
claims
–
PV
fees)
is
zero.
The
example
itself
is
for
CTE(90).
C
described
above.
A More Practical Example – Variance Verification
Aprocess
more practical
process the
for asymptotic
confirming variance
the asymptotic
variance
A more practical
for confirming
formula,
which formula,
we call whic
Variance
Verification
A More Practical Example – Variance
Verification
Variance
is as2:follows:
Variancevariance
Verification
Variance
Verification,
is Verification,
as
follows: Table
A more
practical
process
for
confirming
asymptotic
which we It
call
Syou
uppose
you
have
a model thatthe
takes
all night to run n5formula,
= 1,000 scenarios.
would be im
Suppose
have
a model
that
Variance
Verification,
is
as
follows:
to repeat
then run
process
hundreds
of time
in
order
to
test
the
validity
of
the
takesall
allnight
night
Suppose you have a model that takes
to run
run
= 1,000
scenarios.
It would
be
impractical
N = 5,000 Samples variance for
D 90%
described
above.
CTE
=
FSE
2214
to repeat the run process hundredsscenarios.
of time in
orderbe
toimpractical
test the validity of the variance
formulas as
N
N = 159
It would
described above.
m = 100 Random Sub Samples of Size n = 1,000
to repeat the run process hundreds
5
A
more
practical
process
for
confirming
the
asymptotic
variance
formula,
which
we
call
of time in order to test the validity of
Variance
Verification,
as follows:
FSE
CI Count
A more practical process for confirming
the
asymptotic
variance isformula,
which we call CTE
the variance
formulas
as described
Mean
2,111
346
94%
Variance Verification, is as follows:
above.
First
Last
Min
Max
Std Dev'n
Adjusted Std Dev'n
w Page 40
2,004
2,242
1,558
3,120
316
354
5
355
353
262
376
30
1
1
0
0
2%
=Std Dev'n/ (1-n/N)^1/2
The first thing we note is that if the FSE based on the sample of size 5,000 is correct then w
expect an FSE of about 159 5 | 355 when dealing with a sample size of 1,000. The FSE
estimates are clearly consistent with this, but the standard deviation of 316 (of the 100 CTE
estimates) is not. One possible explanation for this discrepancy is sampling error, but more
The table below shows the results of applying the above process to an inforce portfolio of U.S.
variable annuities with GMDB, GMAB and GMWB features. The book is slightly out of the
money. 5000 real world (P measure) scenarios were generated and for each scenario the present
value of guarantee benefits (claims) less the present value of related fees was captured.
August 2008 w Risk Management
The claims have been normalized so the mean CTE(0) is 1,000. The fees were scaled so that
CTE(75) of the net (PV claims – PV fees) is zero. The example itself is for CTE(90).
Table 2: Variance Verification
D
90%
N = 5,000 Samples
were generated and for each scenario
the presOur
variance verification test is therefore to
CTE N = 2214
FSE N = 159
ent value of guarantee benefits
compare
empirical
errorofestimate
354
from
universe
N=5,000
sowith
they are not independent. Because the various CTE
m = 100(claims)
Randomless
Subthe
Samples
of
Size the
nthe
= same
1,000
are using
some346.
of the
data they are positively correlated and so the set of
present value of related fees was captured.
theestimates
mean formula
estimate
Wesame
conclude
more
tightly
clustered
than if they were truly independent. This intuitive result
CTE
FSEasymptotic
CI is
Count
thatestimates
the
theory
appears
to be working
Mean
2,111
346easy 94%
very
to understand as n o N .
The claims have been normalized
so the mean
for this model.1That is, the asymptotic variance
First
2,004
from355
the same universe of N=5,000 so they are not independent. Because the various CTE
Lastwere scaled so 2,242
353
1
2
CTE(0) is 1,000. The fees
that
of the
Estimator
agrees
with “experiment,”
It 262
isCTE
possible
to use
theof
meth
our paper
topositively
show that,correlated
in the large
sample
in data
estimates
are using
some
theods
same
they are
and
so thelimit,
set ofthe
Min
1,558
0
CTE(75) of the net (PV claims
–
PV
fees)
is
zero.
after
adjusting
for
the
non-independence
of
the
orrelation
of
two
sub-sample
estimates
is
just
c
U
n
/
N
.
A
better
estimate
for
theresult
sampli
estimates
is
more
tightly
clustered
than
if
they
were
truly
independent.
This
intuitive
is
Max
3,120
376
0
from
the same universe
N=5,000
sounderstand
they are not
Because the various CTE
The example itself is for CTE(90).
sub-samples.
Std
Dev'n
316 of
30 to
2%
very
easy
asindependent.
noN .
316
316
estimates are using some of error
the same
they
are positively
and so the
of but3
whendata
using
a sample
size of correlated
1,000 is therefore
notset
316,
| 35
n
1000
Dev'n/ (1-n/N)^1/2
Adjusted
Std Dev'n
354 =Std
2 This intuitive result is
estimates
is more tightly
clustered
than
if
they
were
truly
independent.
1 the
It
is possible
use theismeth
limit,
in our paper
The first thing we note is that if the FSE based
The
CI Counttoresult
alsoods
consistent
with to show that, in the large1sample
N for the 5000
very easy to understand as
n o N . of two sub-sample estimates is just U n / N . A better estimate
orrelation
c
thething
sample
sizeis5,000
correct
theOur
idea
that
theverification
standard
errors areto compare the empirical error estimate sampling
Theonfirst
weofnote
that ifisthe
FSE then
basedwe
onexthe sample
of size
5,000
isformula
correct then
weis therefore
variance
test
354 with
316
316
2count
| 355
pectan
anFSE
FSEofofabout
about 159 It5 is
when
dealing
working.
The
actual
of
94
very
expect
whento
dealing
with
a sample
size
of
1,000.
FSE
mean
estimate
346
. ofis
We
conclude
that
the
asymptotic
possible
use the
meth
our
toThe
show
that,
in
theclose
sample
limit,
ods
informula
paper
error
when
using
a sample
size
1,000
islarge
therefore
not
316,the
but3 theory appears to be workin
| 354
n with 1000
estimates
are clearly
with
the standard
ofvalue
316That
the
100
CTE
thisestimates
model.
is,U(i.e.,
the nasymptotic
variance
of theforCTE
agrees
“experiment
of
two
sub-sample
is(of
just
correlation
N . confidence
A better
estimate
the Estimator
sampling1 with a sample
sizeconsistent
of 1,000.
Thethis,
FSEbut
estimates
todeviation
expected
of
95
a/95%
1
estimates) is not. One possible explanation for this discrepancy
sampling error,
moreuniverse of N=5,000
from
thenon-independence
same
so they
are316
not independent. Because
CTE
afterisadjusting
for
thebut
of the
sub-samples.
N the various
5000
316
are clearly consistent with this, but the standard
interval).
3
estimates
are
using
some
of
the
same
data
they
are
positively
correlated
and
so
the set of
testing shows this is not the case.
error when using a sample
size
of
1,000
is
therefore
not
316,
but
.
|
354
Our variance verification
test is therefore to compare
empirical
error estimate 354 with the
deviation of 316 (of the 100 CTE estimates) is
estimates is more tightly clustered than
were
truly independent. This intuitive result is
n if theythe
1000
ulaappears
standardtoerrors
are worfo
Theformula
CI Count
result
is
also
consistent
with
thethe
idea
that
the form
1
1
mean
estimate
346
.
We
conclude
that
asymptotic
theory
be working
very
easy
to
understand
as
.
n
o
N
e universeThe
of not.
N=5,000
so they
are not independent.
Because
the various
CTEis a biased
standard
deviation
of the m=10
0 sub-samples
estimate
of the sampling
error Table
and 2 that N
One
possible
explanation
for
this
discrepFinally,
it might
appear from
5000
T
he
actual
count
of
94
is
very
close
to
expected
value
of
95
(i.e.,
a
95%
confidence
interval
this
model.
That
is,
the
asymptotic
variance
of
the
CTE
Estimator
agrees
with
“experiment”,
using some
of not
the same data
are positively
correlated
and
so the set
of
it is
tothey
understand
why.
The
various
sub-samples
(each
ofevidence
size n=1,000)
were all
ancyhard
isthansampling
but
more
testing
shows
there
istherefore
of
small
sample
It ismaterial
possible
todrawn
use
the meth
Our
variance
verification
test
is
compare
the
empirical
error
354 that,
withinthe
the large sample limit, the
ods
in ourestimate
paper 2 to show
more tightly clustered
if they wereerror,
truly independent.
This
intuitive
result is
after
adjusting
fortothe
non-independence
of
the
sub-samples.
of two sub-sample
estimates
is just
correlation
/ N . A better
for the sampling
understand as this
noN
mean formula estimate 346
. We
conclude
that
theory
appears
toUbe nworking
for estimate
is. not the case.
bias
since
mean
ofthe
theasymptotic
100Table
sub-sample
Finally,
itthe
might
appear
from
2 that there
is evidence
of material
small sample
bias si
316
316
6
universe of N=5,000 so they are not independent.
Because
the
various
CTE
3
this
model.
That
is, the asymptotic
of
agrees
with
“experiment”,
2
error
when
usingEstimator
a sample
size
of is
1,000
is that
therefore
316,
butstandard
| workin
354 .
thCI
e mean
ofresult
the 100
sub-sample
estimates
2,111,
with
an
apparent
precision
of
estimates
isvariance
2,111,
with
an CTE
apparent
precision
ula
errors
are
to usesome
the meth
The
Count
is the
also
consistent
with
the
idea
the not
form
that, incorrelated
the large and
sample
limit,
odssame
in our
paper
sing
of the
data
they to
areshow
positively
so the
setthe
of
n
1000
adjusting
theis non-independence
ofofthe
sub-samples.
1
1
ore
tightly
clustered
than if is
they
This intuitive
result
two
sub-sample
estimates
justwere
U truly
n / N independent.
. A after
estimate
for thefor
sampling
T
actual
count
is very
close less
to
expected
95 (i.e.,
a 95%from
confidence
interval).
deviation
ofbetter
the m=100
sub-samofhe316
which
isismuch
than
/ 100
| 32
,94
which
much
less
than thevalue
valueof2,214
obtained
of size
N the sample
5000
nderstand as nThe
.
o N standard
316
316
sing a sample size of 1,000 is therefore not 316, but3
Our
variance
verification
test
is
therefore
to
compare
the
empirical
error
estimate
354
with
| 354 .
5,000.
However,
this
analysis
misleading,
againerrors
due to
the
non-independence of thethesubples is a biased estimate The
of large
the
sampling
erroris also
theconsistent
value 2,214
from
theisform
sample
of
nCount
1000
ula
standard
are
working.
withobtained
the
idea
thatTable
the
o use the methods in our paper 2 to show that, in the
the
mean
formula
estimate
346
. We
conclude
that the asymptotic
theory
appears
to be working
for
1CI
sample
1limit,
result
Finally,
it
might
appear
from
2
that
there
is
evidence
of
material
small
sample
bias since
samples.
N for
5000
wo sub-sample
estimates
is just
U nto/ Nunderstand
. A better
the
sampling
and
it is not
hard
why.
The
varisize
5,000.
However,
analysis
misleadthisthis
model.
That
is,isthe
asymptotic
variance
of the CTE
Estimator agrees with “experiment”,
The estimate
actual
count
of 94 is th
very
close
to
expected
value
of
95
(i.e.,
a
95%
confidence
interval).
e mean of the 100after
sub-sample
is 2,111,ofwith
an apparent precision of
verification test is therefore to compare the empirical
error
354 with the
316 estimate
316
adjusting forestimates
the non-independence
the sub-samples.
3
ng a sample size ofsub-samples
1,000 is therefore not 316,
|all
354 .
ofbut
size
n=1,000)
were
a estimate 346ous
. We conclude that the(each
asymptotic
theory
appears
to 1000
be
working
for ing, again due to the non-independence of the
n
316
/
100
|
32
,
which
is
much
less
than
value
2,214
obtained
theissample
When
sam
ples
are
positively
correlated
the
variance
of
the
samplefrom
mean
larger of size
1 with
1“experiment”,
That is, the asymptotic
of thesame
CTE Estimator
agrees
Finally,
might
Table 2 that2 there
material
small
bias
The is
CI evidence
Count resultof
is also
consistent
with sample
the idea that
the since
formula standard errors are working.
drawn variance
from the
universe
ofitN=5,000
so from
sub-samples.
N
5000appear
gerification
for the non-independence
of
the
sub-samples.
5,000.
However,
this
analysis
is
misleading,
again
due
to
the
non-independence
of interval).
the precisio
subVARestimates
( x ) V [ Uis (1actual
Uwith
)count
/ m]an
would
be.ofA95better
estimate
for
test is therefore to compare the empirical
estimate
354 100
with the
T2,111,
he
ofthan
94
is it
very
close
tootherwise
expected
(i.e., a 95%
confidence
thBecause
e error
mean
of the
sub-sample
apparent
precision
of value
are notthat
independent.
various
estimate 346 .they
We conclude
the asymptotic theory
appearsthe
to be
working for samples.
ula
tatresult
is asymptotic
also consistent
with the
ideaCTE
thatEstimator
the form
1000
1000
is, the
variance
of the
agrees
witherrors
“experiment”,
316
/standard
| 32aredata
, working.
which When
is muchsamples
less than are
the
value
2,214appear
obtained
from
the sample
of size material small sample bias since
Finally,
it might
from
CTE estimates
are
using
some
of100
the
same
the 2,111 number
ispositively
then
which is not materially differ
346 correlated
(Table
1 2 that)there
/ 100is evidence
| 158 of
unt the
of 94
is very close to expected
value of 95 (i.e., a 95% confidence interval).
for
non-independence
of the sub-samples.
thagain
e meandue
of theto100
sub-sample
estimates
is 2,111,
with suban apparent precision of
5,000.
However,
this
analysis
is
misleading,
the
non-independence
o
f
the
5000
5000
they are positively correlated and so the set of
the variance
sample correlated
mean is larger
When
samplesofarethe
positively
the
variance
of
the
sample
mean is larger
316 / 100 | 32 , which is much less than the value 2,214 obtained from the sample of size
ula standard
working.
result
is also
consistent
idea
that the form
ght appear
from
Table 2with
that the
there
is evidence
ofsamples.
material
small errors
sampleare
bias
since
2
2,214
value.otherwise
Thus, while
ther
from
theVsampling
error
in] the
e is some
evidence
of small sam
estimates
is
more
tightly
clustered
than
if
they
than
would
nt
of
94
is
very
close
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expected
value
of
95
(i.e.,
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95%
confidence
interval).
VAR
(
x
)
[
U
(
1
U
)
/
m
it
would
be.
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better
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for the
5,000. However, this analysis is misleading, again due to the non-independence
of theprecision
subhe 100 sub-sample estimates is 2,111, with an apparent precision of
bias
in
that
2,111
<
2,214,
there
is
not
enough
data
here
to
quantify
it.
The
bias
is lost in th
samples. estimate for the
, which
iswere
much
than
the value
2,214 of
obtained
from
thesample
sample
of size
truly
independent.
This
intuitive
result
is
otherwise
be.
A
better
ht32appear
from
Tableless
2 that
there
is evidence
material
small
bias
since
1000
1000is larger
When
sam
are positively correlated
the variance
of the
sample
mean
error.
This
usually
thecase.
analysis
misleading,
againwith
due to
non-independence
of the sub-CTE thesampling
0ever,
sothis
they
areis not
independent.
Because
theples
100
sub-sample
2,111,
anthe
apparent
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2,111 number
thenis
which
is not materially different
346
correlated) /the
100
| 158of the
2
veryestimates
easy tois understand
asVAR
n →precision
N.
precision
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theisWhen
2,111
is (1then
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plesbe.
are
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sampleofmean is larger
5000
5000
(
x
)
V
[
U
(
1
U
)
/
m
]
than
it
otherwise
A
better
estimate for
the precision
2
,
which
is
much
less
than
the
value
2,214
obtained
from
the
sample
of
size
me data they are positively correlated and so the set of
2
VAR( x ) V [ U (1 U ) / m] than it would otherwise be. A better estimate for the precision of
er, this analysis is misleading, again due to the non-independence of the subn exercise
fewetimes
a practicing
kn
going
through
verificatio
the 2,214
value. Thus,
whilea ther
from
the
sampling
errorainvariance
is some
evidence actuary
of smallwill
samp
if theycorrelated
were truly
independent.
This
intuitive result
isAfter
esthan
are positively
the variance
of the sample mean
is larger
1000
1000
1000herematerially
1000working
2
dependent.
Because
the
various
CTE
the
is then
which
different
346
(1asymptotic
< the
) / 100
| 158
whether
formulas
described
are
forwhich
his/her
situation.
possible
to use be.
theAmethods
innumber
ourthepaper
in that the
2,111
2,214,
there
isis not
enough
data
quantify
it. The
bias
is lost
in the
2,111
number
then
is notparticular
materially
different
346 is not
(1 here
)to
/ 100
| 158
[.U (1 U ) / It
m] isthan
it would otherwise
better2,111
estimate
for
precision
of bias
5000
5000
5000
5000
are positively
the variance
the sample
mean is larger
sitively
correlated
and soof the
set of
sampling
error. This
is usually
theerror
case.
tocorrelated
show
1000 that,
1000in the large sample limit, the
the 2,214
value. Thus,
while ther
from thewhile
sampling
in some
e is some evidence of small sample
U (1isthen
U) / m
] than it would
mber
whichestimate
isthe
not materially
different
346
(1 otherwise
) / 100 be.
| 158
A better
for the precision
of in the 2,214 value. Thus,
ther
from
sampling
error
e
is
evidence
of small
sample
uly
independent.
This
intuitive
result
is limit, isthe which is not materially
2 5000
5000
bias different
in that 2,111from
< 2,214,
there
is not enough data here to quantify it. The bias is lost in the
correlation
of two
sub-sample
estimates
the
samo
to
show
that,
in
the
large
sample
ur
paper
1000
1000
bias
innot
that
2,111
<sample
2,214,After
there
is notthrough
enough
data
here
quantify
The
is lost
in the
while
therewhich
pling
in the 2,214value.
is some
evidence
of small
ber
is error
then 346
is
materially
different
(1 Thus,
) / 100
| 158
sampling
error. to
This
is usually
case. bias
Conclusions
nit.the
exercise
a few
times
a practicing actuary will know
going
variance
verificatio
5000
mates
is just
Unot enough
n /5000
N data
. AA
better
estimate
for
the
just
better
estimate
the
sampling error
in the 2,214a value.
Thus,
while there
111 < 2,214,
there
is
here
to quantify
it. The for
bias
is lost
in sampling
the is usually
sampling
error.
This
the
case.
the 2,214
value. Thus, while there is some evidence of small sample whether the asymptotic formulas described here are working for his/her particular situation.
ingThis
errorisinusually
or.
the case.
After
going
throughbias
a variance
verification exercise a few times a practicing actuary will know
pling
error
when
a3 sample
size
ofis1,000
is
is some evidence
small
sample
in have
that
11 < 2,214, there
is not
enough
data using
here to quantify
it. The
bias
lost
316
316in the
experience,
so far, we
yet to see
practical
situation
wheresituation.
the theory
authors’ of
whether the asymptotic
formulas
described
hereaare
working for
his/her particular
therefore
316, but
|variance
354In. the
w1,000
in
the
sample
limit,
the
. that,
This is is
usually
thelarge
case. not
3
n
exercise
a
few
times
a
practicing
actuary that
will practitioners
know
After
going
through
a
verificatio
n exercise
a few times a practicing
actuary 1000
will know 2,111
hrough a variance
verificatio
therefore
not
316, but
< 2,214,
enough data
to we believe
herethere
failsisinnot
a material
way.here
Hence,
can use the asym
outlined
n
/ough
N .a variance
Aformulas
better
estimate
symptotic
described
here arefor
for
his/her
particular
1asymptotic
situation.
n exercise
aworking
fewthe
times
a1
practicing
actuary
will know formulas described here are working for his/her particular situation.
verificatio
wsampling
hether
the
quantify
it. The bias is lost in the sampling error.
C
onclusions
Nparticular situation.
5000
ymptotic formulas described here are working for his/her
Conclusions
316
316
3
2 is usually the case.
This
erefore
compare the empirical error
ot 316,tobut
| estimate
354 . 354 with the This result is not in the original paper.
3
experience,
so far,
wetohave
to see a practical
situation
where the
the theory
In the authors’
sodistributed,
far,
we have
seeyet
a practical
situation
where
theory
In for
the
n
1000 appears to be working
conclude that the1 asymptotic
If authors’
we have experience,
m identically
butyet
correlated,
samples
from
a distribution
with
finite
Csituation
onclusions
theory
here
fails
in aon
material
way.
Hence,
we believe
that practitioners
canuse
use the
asymptotic
outlined
continued
page 42
utlined
here
fails
in
a
material
way.
Hence,
we
believe
that
practitioners
can
the
asympto
o
a practical
where the theory
s’ experience, so far, we have yet to see 1
2
cexperience,
variance
of
the
CTE
Estimator
agrees
with
“experiment”,
N
5000
(x x)
so far,way.
we have
yet we
to see
a practical
situation where
thethe
theory
fails in a material
Hence,
believe
that practitioners
can use
asymptotic
mean and variance
V 2 then E[¦ i
] V 2 (1 U ) where U is the correlation betwee
ails
in a material
Hence, we believe that practitioners can use the asymptotic
2
dence
of theway.
sub-samples.
the
empirical
error
estimate 354
the experience, so far, we have yet
This
result
ispractical
not in the
original
paper.
to
see
a
situation
where the theory
In with
the authors’
1
m
i
2
Ifthe
webelieve
have m identically
distributed, but
correlated,
from a distribution with finite
This result
is
not in3 we
original
paper.
s not in the original
paper. appears to be o
utlined here
way.
Hence,
that
practitioners
can
use thesamples
asymptotic
2 theory
ymptotic
for fails in a material
not in the original
samples.
Thispaper.
result is not in the originalworking
paper.
3
m
identically
distributed,
but
correlated,
samples
from
adistribution
distribution
with finite
finite
( xi x ) 2 samples
2
2
If
we
have
m
identically
distributed,
but
correlated,
from
with
finite
3
ula
standard
errors
are
working.
ent
with
the
idea
that
the
form
identically
distributed,
but
correlated,
samples
from
a
with
then E[¦
] V (1 U ) wherea Udistribution
is the correlation
between
If we agrees
have m identically
distributed, but correlated, samples from a distribution with finitemean
meanand
andvariance
variance V then
2 with “experiment”,
TE Estimator
m 1
2
2
2 i
x( xixx) ) of
where
is
the
correlation
between
samples.
U
E[ [¦(value
then
Vexpected
] V95
V
(1(i.e.,
UU) )where
where
is
the
correlation
between
eiance
to
a
95%
confidence
interval).
(
)
x
x
U
E
ance
then
V
]
(
1
is
the
correlation
between
2
¦i mm11
mples.
This result is not in the original
paper.
mean and
variance samples.
V 2 then E[¦ i
] V 2 (1 U ) where U is the correlation between
3
1
m
i
If we have m identically distributed, but correlated, samples from a distribution with finite
7
e 2 that there is evidence of material small sample bias 7since
7 samples. ( x x ) 2
ula
standard
errors
are
working.
at
the
form
2
2
timates is 2,111, with an apparent
precision
of
and variance
V then E[¦ i
] V (1 U ) where U is the correlation between
of 95 (i.e., a 95% confidence mean
interval).
m 1
i
ess than the value 2,214 obtained from the sample of size
samples.
misleading, again due to the non-independence
of the sub2
i
2
2
i
dence of material small sample bias since
ith an apparent precision of
7
Page 41 w
Risk Management
w August 2008
Chairman’s
Variance ofCorner
the CTE Estimator
CTE Estimator …
w continued
from page 41
After going through a variance verification exercise a few times a practicing actuary will know
whether the asymptotic formulas described here
are working for his/her particular situation.
If simply increasing the run size is impractical
then there are other tools that can be used to improve the precision of the CTE estimator without
significantly increasing the computational cost.
That will be the subject of our next article. F
Conclusions
In the authors’ experience, so far, we have yet
to see a practical situation where the theory
outlined here fails in a material way. Hence, we
believe that practitioners can use the asymptotic formulas presented here to understand the
sampling error in a given CTE estimate. A practitioner can go through a variance verification
exercise if they want to prove to themselves, or
others, that the asymptotic formulas are working
in their particular situation.
formulas presented here to understand the sampling error in a given CTE estimate. A
practitioner can go through a variance verification exercise if they want to prove to themselves,
Finally, we note that if a practitioner were
or others, that the asymptotic formulas are working in their particular situation.
using this tool then, on the basis of the first run
of 1,000
scenarios,
theythat
would
a CTEwere using this tool then, on the basis of the first run of
Finally,
we note
if a report
practitioner
estimate
of 2,004+355.
Is a relative
sampling
scenarios, they
would report
a CTE estimate of 2,004 r 355 . Is a relative sampling error
1,000
formulas presented here to understand the sampling error in a given CTE estimate. A
errorpractitioner
ofofroughly
acceptroughly
acceptable?
The answer
to that
question
depends on the
355
/
2
,
004
|
18
%
can go through a variance verification exercise
if they want
to prove
to themselves,
or
others,
that
the
asymptotic
formulas
are
working
in
their
particular
situation.
circumstances.
able? The answer to that question depends on
the circumstances.
firstoption
run of is to increase the
Finally,
we is
note
practitionererror
werethen
usingwhat
this tool
on the basis of the
arethen,
the alternatives?
One
If 18%
notthat
anifaacceptable
1,000 scenarios, they would report a CTE estimate of 2,004 r 355 . Is a relative sampling error
number of scenarios. If we increase the number of scenarios by a factor of K then the sampling
If 18%
is not an
acceptable
then whatThe
are answer to that question depends on the
of roughly
355
/ 2,004 | 18error
% acceptable?
error scales by a factor of 1 / K . Cutting the sampling error by a factor of 4 would require us t
circumstances.
the alternatives?
One option is to increase the
run 16,000 scenarios.
number
of scenarios.
If we increase
thewhat
number
error then
are the alternatives? One option is to increase the
If 18%
is not an acceptable
of scenarios
by
a
factor
of
K
then
the
sampling
K then
number
of
scenarios.
If
we
increase
the
number
of scenarios
a factor
the sampling
thenbythere
areofother
tools
that can be used to
If simply increasing the run size is impractical
afactor
factor
of 1 / ofKthe
. Cutting
Cutting
the
error bysignificantly
a factor of 4 would
requirethe
us to
errorerror
scales
bybyathe
of
thesampling
imscales
prove
precision
CTE estimator
without
increasing
computational
run
16,000
scenarios.
cost.
That
will
be
the
subject
of
our
next
article.
sampling error by a factor of 4 would require us
to run
16,000increasing
scenarios.the run size is impractical then there are other tools that can be used to
If simply
improve the precision of the CTE estimator without significantly increasing the computational
cost. That will be the subject of our next article.
w Page 42