5
Insurance Coverage Modifications
We will now turn our attention to common insurance practices and coverage modifications such as deductibles and
policy limits. Much of the “math” for these modifications was covered in Section 2.1 so things should seem at least
somewhat familiar as you read this section.
Recall from Section 2.1 the following definition.
Definition 5.1. A per-loss random variable Y L models the payout for losses X. A per-payment random variable
Y P models the payout for losses X conditional on the payout being positive, i.e. Y P = Y L |(Y L > 0).
Y L is called “per-loss” because it is defined even if the actual payment the insurance company makes is 0 (i.e. it
exists even if the payout is 0). In contrast, Y P is called “per-payment” because it is only defined if the insurer makes a
payment.
In the absence of modifications (deductibles, policy limits, etc.), Y L = X, since every loss results in a payout for
that loss. We’ll now see how these variables are defined for certain modifications.
5.1
Deductibles
Definition 5.2. Assume X to be a random variable denoting losses. A policy with an ordinary deductible of d pays
0 if X  d and pays X d if X > d.
Specifically,
1. A per-loss random variable Y L for a ordinary deductible d is:
(
0
X d
YL =
X d X >d
(5.1)
2. A per-payment random variable Y P for a ordinary deductible d is:
(
undefined X  d
P
Y =
X d
X >d
(5.2)
By definition of a deductible, losses below d are not covered, which explains why Y L = 0 for X  d, in which case
insurance companies do not pay a thing.
Now let’s look at some distributional properties of Y L and Y P based on our knowledge of X. We will use subscripts
to differentiate between the various distributions for Y L , Y P , and X.
For the per-loss random variable, note that Y L = 0 corresponds to X  d. For y > 0, Y L = y implies X = y + d.
Using this fact and the relationships between f (y), F(y), S(y), and h(y), we get:
8
(
<undefined y = 0
FX (d)
y=0
fY L (y) =
hY L (y) = fX (y + d)
:
fX (y + d) y > 0
y>0
SX (y + d)
FY L (y) = FX (y + d), y
0
SY L (y) = SX (y + d), y
0
Now for Y P , our calculations depend on the loss being large enough (larger than the deductible d). In considering
fY P (y), we basically take fY L (y) and divide it by the probability that the loss X exceeds d. All other derivations are
done in a similar fashion. Recalling that Y P is undefined if X  d, we set fY P (0) = 0:
8
8
y=0
<0
<undefined y = 0
fY P (y) = fX (y + d)
hY P (y) = fX (y + d)
:
:
y>0
y>0
SX (d)
SX (y + d)
FY P (y) =
FX (y+d) FX (d)
,
SX (d)
y
0
SY P (y) =
Notice that hY P (y) = hY L (y).
61
SX (y+d)
SX (d) ,
y
0
Example 5.3. Let X ⇠ exp(q ). Calculate fY P (y), SY P (y), FY P (y), and hY P (y). Also calculate fY L (y), SY L (y), FY L (y),
and hY L (y).
Answer
1
fY P (y) =
1
e (y+d)/q
fX (y + d)
(y+d)/q
= q
= q e d/q =
SX (d)
1 FX (d)
e
1
q e(y+d)/q
1
ed/q
=
ed/q
1
1
= y/q = e
(y+d)/q
q
qe
qe
y/q
Observe that fY P (y) is exactly the same as fX (y)! This reflects the memoryless property of the exponential distribution.
Given that the loss is already of size d, the probability of the total loss being y + d is the same as the probability of a
loss of size y. Basically, conditioning has no effect! We’ll elaborate on this after the example. Let’s see what happens
when we compute the other quantities.
SY P (y) =
d
SX (y + d) e (y+d)/q
=
= eq
d/q
SX (d)
e
y+d
q
y/q
=e
= SX (y)
Note that we could have arrived at the same conclusion by simply seeing that fY P (y) is exactly the same as fX (y). The
calculation confirms this, but is unnecessary.
FY P (y) =
FX (y + d) FX (d) 1
=
SX (d)
f P (y)
hY P (y) = Y
=
SY P (y)
e
y+d
q
e
fX (y+d)
SX (d)
SX (y+d)
SX (d)
(1
e
d
q
)
d/q
=1
e
fX (y + d)
=
=
SX (y + d) e
e
y/q
=
1
q
y+d
q
q
y+d
q
= FX (y)
Now, let’s do the same for the variable Y L .
fY L (y)
= fX (y + d) =
FY L (y) = FX (y + d) =
SY L (y)
hY L (y)
e (y+d)/q
q
1 e (y+d)/q
=1
FY L (y + d) = e (y+d)/q
⇠
⇠⇠
e (y+d)/q
⇠
f (y)
q
= SY LL (y) = (y+d)/q
⇠ = q1
⇠
e ⇠
⇠
Y
Remark. The memoryless property of the exponential distribution says that if X is exponentially distributed, then
(X d|X > d) is distributed the same way as X. As a concrete example, assume X represents time in hours you stand
in line at Macy’s on Black Friday before being served by a cashier, and you have already waited for 2 hours. Then the
amount of additional time you would have to wait before being served is distributed identically as X. The expected
wait time is the same as it was when you just got in line!
Now that we have the basic deductible down, let’s consider a franchise deductible.
Definition 5.4. Assume X to be a random variable denoting losses. A policy with a franchise deductible of d pays
0 if X  d and pays the full loss amount X if X > d.
We’re now going to look at Y L and Y P for a franchise deductible.
1. A per-loss random variable Y L for a franchise deductible d is:
(
0 X d
L
Y =
X X >d
2. A per-payment random variable Y P for a franchise deductible d is:
(
undefined X  d
P
Y =
X
X >d
62
The properties of Y L for a franchise deductible can be derived in the same way as we did for an ordinary deductible.
8
(
>
<FX (d) y = 0
0
0yd
0<yd
fY L (y) = 0
hY L (y) =
>
hX (y) y > d
:
fX (y) y > d
FY L (y) =
(
FX (d) 0  y  d
FX (y)
SY L (y) =
y>d
(
SX (d) 0  y  d
SX (y)
y>d
We’ll briefly discuss the derivation of fY L (y), as the other quantities can be derived simply based off of fY L (y). If
payment Y L = 0, then X  d, which occurs with probability FX (d). Y L > d similarly corresponds to X > d. Note
that there will not be a payment of any amount between 0 and d, because the franchise deductible either pays all or
nothing, provided that “all” is above d.
FY L (y) can then be computed by integrating over fY L (y), and SY L (y) can be computed as the complement of FY L (y).
Similarly for Y P , we have:
8
(
0yd
<0
0
0yd
fY P (y) = fX (y)
hY P (y) =
:
hX (y) y > d
y>d
SX (d)
8
<0
FY P (y) = FX (y) FX (d)
:
SX (d)
0yd
y>d
8
<1
SY P (y) = SX (y)
:
SX (d)
0yd
y>d
Note that fY P (0) = 0 because we do not allow Y P to be 0. Also, with a franchise deductible, the minimum payout
occurs above d. Thus, we have to scale up the density above d by dividing the positive densities by SX (d).
We presented the formulas for completeness. If you have spare time before the exam and you finished learning
everything else, then learn these. If you can remember fY L and fY P , you should be able to derive the other functions.
The must-know formulas (which often get tested) are summarized in the following theorem:
Theorem 5.5. For an ordinary deductible d,
1. the expected cost per loss is: E(Y L ) = E(X)
2. the expected cost per payment is: E(Y P ) =
E(X ^ d)
E(X) E(X^d)
SX (d)
Similarly, for a franchise deductible d,
1. the expected cost per loss is: E(Y L ) = E(X)
2. the expected cost per payment is: E(Y P ) =
E(X ^ d) + d(SX (d))
E(X) E(X^d)
SX (d)
+d
Memorize the above four formulas. Let’s think about the reasoning behind them, in a “hand-wavy” manner, for an
intuitive understanding.
For an ordinary deductible, Y L = X X ^ d (see Definition 2.9, and write out what Y L would be for X  d and
X > d to see this). Taking the expectation of both sides, we get the formula for E(Y L ). For E(Y P ), we take the
expression for E(Y L ) and simply condition on X > d by dividing by S(d).
Now, considering E(Y L ) for a franchise deductible, we take the expected loss for an ordinary deductible and add
on the value of the deductible multiplied by the probability of a payout Pr(X > d) = SX (d). The difference between
a franchise deductible and an ordinary deductible is that when there is a payout, a franchise deductible always pays d
more than what an ordinary deductible pays. This also implies that E(Y P ) for a franchise deductible is d greater than
E(Y P ) for an ordinary deductible.
Example 5.6. Let X ⇠ exponential(q = 1000) and let d = 250. When d is an ordinary deductible, what is the expected
cost per loss and expected cost per payment?
63
Answer From the formula sheet, we know that E(X ^ d) = q (1
E(Y L ) = E(X)
E(X ^ d) = q
q (1
d
q
e
d/q ),
e
d
q
) = qe
so
250
1000
= 1000e
= 778.801
d
⇢
E(X ^ d) q⇢
e q
=
= q = 1000
d
⇢
F(d)
e q
⇢
Now what if this were a franchise deductible? Then we have
E(X)
E(Y ) =
1
P
E(Y L ) =
E(X)
q
=
q (1
E(X ^ d) + d(1
d
q
e
d
q
= (q + d)(e
) + de
)
250
1000
= (1250)(e
F(d))
d
q
)
973.501
⇡
E(Y P ) =
=
E(X)
q
E(X ^ d)
+d
S(d)
q (1
= q +d
e
e
d
q
)
d
q
+d
= 1250
Note that for any loss X, a franchise deductible always pays at least as much as an ordinary deductible.
Most exam problems pertaining to franchise deductibles involves expected value calculations. Be sure to know the
formulas in Theorem 5.5.
5.2
Coinsurance and Policy Limits
Definition 5.7. The loss elimination ratio (LER) is the percent decrease in the expected payment with a policy
modification. For losses X, and per-loss Y L ,
LER =
E(X) E(Y L )
E(X)
The numerator represents the decrease in expected payment when a deductible is included, and the denominator
represents the regular expected payment. For an ordinary deductible d, the LER is given by:
LER =
E(X)
[E(X) E(X ^ d)] E(X ^ d)
=
E(X)
E(X)
For a franchise deductible d, the LER is given by:
LER =
E(X)
[E(X)
E(X ^ d) + d(1
E(X)
Example 5.8. Let q = 1000 and d = 250. What is the LER?
64
FX (d))]
=
E(X ^ d)
d(1
E(X)
FX (d))
Answer
LER
=
E(X)
= 1
e
= 1
e
[E(X) E(X ^ d)]
E(X)
d
q
250
1000
⇡ 0.221199
Theorem 5.9. Consider X, a random variable denoting losses for the current year. Given an ordinary deductible d
after inflation of r, the expected cost per loss variable for next year is:

✓
◆
d
E(Y L ) = (1 + r) E(X) E X ^
(5.3)
1+r
d
If SX ( 1+r
) > 0, then, the expected cost per payment for next year is:
P
E(Y ) =
⇥
(1 + r) E(X)
SX
d
E X ^ 1+r
d
1+r
⇤
(5.4)
Consider (5.3). Inflation increases all losses. If X represents the loss amounts for the current year, then (1 + r)X
models the loss amounts for the next year. The deductible of d is applied to (1 + r)X.
Note that if the formula above relates losses in the current year to expected costs in the next year. If we wanted
to consider the expected costs in t years, then we replace all occurrences of 1 + r with (1 + r)t . This assumes that the
annual rate of inflation is constant, an oversimplification that is permitted on the exam.
d
In (5.4), we condition on a payment being made. If SX ( 1+r
) = 0, then a payment is never made, and there is no
“cost per payment” variable.
Example 5.10. Suppose d = $242 and r = 10%. Also suppose losses X ⇠ exponential(q = 1000). What is the
expected cost per loss and the expected cost per payment next year?
Answer First 1 + r = 1.1 and
E(X ^
d
1+r
=
242
1.1
= 220. Then,
d
) = E(X ^ 220) = q (1
1+r
Thus,
(1 + r)[E(X)
d
Further, F( 1+r
) = F(220) = 1
e
220
1000
E(X ^
e
220
q
) = 1000(1
d
)] = 1.1[1000
1+r
e
220
1000
) ⇡ 197.481
197.481] = 882.771
⇡ 0.197481, which is less than 1. Thus the expected cost per payment is:
882.771
882.771
=
= $1, 100
d
1 F( 1+r ) 1 0.197481
Note that for an exponential distribution, the expected cost per payment is q (1 + r).
We’re now going to talk about a policy limit, which is closely tied to the limited loss variable (Definition 2.9).
Definition 5.11. A policy limit of u specifies u as the maximum payout, after the application of a deductible.
65
Policy limits are used by insurance companies to limit their upside risk. In that regard, it is the exact opposite of a
deductible. If u = 10, 000 and d = 5000, a loss of X = 25, 000 will pay out 10,000 since we apply the deductible first
(without the policy limit, the payout would be 20,000).
Let Y denote the payout for losses X after a policy limit of u is imposed. It is possible that X is a per-loss variable
that has already been adjusted for a deductible, or X could be simply the raw loss amount. Then,
(
FX (y), y < u
FY (y) =
1
y u
fY (y) =
(
fX (y),
1
y<u
FX (u) y = u
The most important result to remember here is the following.
Theorem 5.12. For a policy limit of u applied after inflation of r, the expected cost per loss is:
✓
◆
u
E(Y L ) = (1 + r)E X ^
1+r
Notice the similarities to a deductible with inflation (namely, you divide the limit by (1 + r) and then multiply the
entire quantity by (1 + r)).
Example 5.13. Let losses X ⇠ exponential(q = 1000) and let r = 10% with u = 2000. What is the expected cost per
loss? What is the proportional reduction to the case where there is no limit?
Answer The expected cost per loss is
⇣
⇣
u ⌘
(1 + r)E X ^
= 1.1E (X ^ 1818) = 1.1q 1
1.1
If there was no limit, then the cost per loss is
e
1818
q
⌘
⇣
= 1.1(1000) 1
e
1818
1000
⌘
⇡ 921.415
1.1E(X) = 1.1q = 1100
Thus, by imposing a policy limit, there is a reduction of
1100
921.415
⇡ 16.235%
1100
We now turn our attention to one final policy modification before combining all the modifications together.
Definition 5.14. A coinsurance of a is the proportion of a loss covered by the insurance company, after applying
inflation and the deductible. (The customer pays the rest). Thus, the insurance company is only responsible for aY of
each post-deductible/inflation-adjusted loss Y .
Definition 5.15. A maximum covered loss of u is the largest amount of your losses that will be covered.
Remark. The maximum covered loss is not to be confused with the policy limit, also denoted as u. The maximum
covered loss is applied before the deductible, unlike the policy limit. This means that if you have a maximum covered
loss of u and a deductible of d, the maximum payout on any loss will be u d. If instead u is the policy limit, the
maximum payout on any loss would be u.
In the remainder parts of this section, u will be used to denote maximum covered loss.
66
For losses X, subject to ordinary deductible d, maximum covered loss u, inflation r, and coinsurance a, the per-loss
variable is given by
8
d
> 0,
X < 1+r
<
L
d
u
Y =
a[(1 + r)X d] 1+r  X < 1+r
>
:
u
a(u d)
X 1+r
Using this, we get to the final formula which is really the only one you need to memorize; all preceding formulas are
simplifications of this one, as we’ll show.
Theorem 5.16. Given a maximum covered loss u, inflation r, a deductible (ordinary) d and coinsurance a,
 ✓
◆
✓
◆
u
d
E(Y L ) = a(1 + r) E X ^
E X^
1+r
1+r
E(Y P ) =
E(Y L )
d
SX ( 1+r
)
(5.5)
(5.6)
If we rewrite (5.5) as E(Y L ) = (A) ⇥ (B)[(C) (D)], then the following simplifications can be made. If there is no
limit (u = •), simply modify (C) to be E(X). If there is no deductible (d = 0), simply remove all of (D). If there is
no coinsurance, set term a = 1 and if there is no inflation, set r in term (B) equal to 0. These modifications simplify
to yield the E(Y L ) expressions given earlier.
Remark. In the presence of an ordinary deductible d, coinsurance a, and maximum covered loss u, the effective policy
limit is a(u d), since that is the maximum amount that can be paid out for a given loss.
Example 5.17. Suppose losses are exponentially distributed with q = 1000. Let inflation r = 10%, maximum covered
loss u = 2000, d = 250, and a = 70%. What is the expected cost per loss and the expected cost per payment?
Answer First,
u
1+r
=
2000
1.1
= 1818.18 and
d
1+r
=
250
1.1
= 227.27. Thus,
u
d
) E(X ^
))
1+r
1+r
= (0.7)(1.1)[E(X ^ 1818.18) E(X ^ 227.27)]
E(Y L ) =
a(1 + r)(E(X ^
= (0.7)(1.1)[1000(1
= 0.77[837.679
e
1818.18
1000
)
1000(1
e
227.27
1000
)]
203.294]
= 488.48
The expected cost per payment is then
488.48
488.48
=
=
d
1
1 FX ( 1+r ) 1 FX ( 250
1.1 )
488.48
488.48
= 227.27 = 613.12
F(227.27) e 1000
While most problems will be a little more in depth (such as solving for the coinsurance or inflation, or setting
appropriate limits), the idea is the same. Write out the whole formula, plug in all the variables and solve. We’ll
include some difficult problems at the end of this section for practice.
5.3
Second Moments (semi-optional)
We now know how to compute the expected cost per loss given all the policy modifications. Computing higher
moments becomes much more complex and is not part of the syllabus. However, you should learn to compute the
second moment since then you are expected to compute the variance for Y L .
Theorem 5.18. For losses X, inflation r, maximum covered loss u, coinsurance a, and ordinary deductible d, the
second moment of the cost per loss (next year) variable is
⇥
⇤
u 2
E[(Y L )2 ] = a 2 (1 + r)2 E (X ^ 1+r
)
⇥
⇤
d 2
d
u
(5.7)
E (X ^ 1+r
)
2( 1+r
)E(X ^ 1+r
)
d
d
+ 2( 1+r
)E(X ^ 1+r
)
67
This formula has been tested once or twice, but it is certainly not common exam material. Memorize it if you have
time, but don’t worry about it if you don’t. However, keep in mind that because the formula is so complex, when it is
tested, it is usually a straight plug-and-chug problem. That could mean easy points if you remember it.
Example 5.19. For the setup of Example 5.17, compute the standard deviation of Y L ?
Answer In the previous example, we already computed E(Y L ) = 488.48. Now we need E[(Y L )2 ]. We proceed by
computing the various parts of the formula in (5.7).
a 2 = (0.7)2 = 0.49
(1 + r)2 = 1.12 = 1.21
u 2
2000 2
) ] = E[(X ^
) ] = E[(X ^ 1818.18)2 ]
1+r
1.1
If you look in the formula sheet, you get a big mess involving the gamma distribution. However, with a little bit of
work, we can compute the formula directly.
E[(X ^
In general,
Z v
E[(X ^ v)2 ] =
0
Z v
=
Z0 v
x2 f (x)dx +
Z •
v
x2 f (x)dx + v2
v2 f (x)dx
Z •
v
f (x)dx
x2 f (x)dx + v2 S(v)
| {z }
|
{z
}
(B)
=
0
(A)
So now we consider separately for an exponential(q ) distribution,R and work with
the terms in the above separately.
R
v
(B) becomes v2 e q . To calculate (A), we use integration by parts gdh = gh
hdg twice.
Z v
0
Let g = x2 and dh =
x
e q
q
dx ) h = e
x
q
x2 ( e
)
x
q
x2 f (x)dx =
Z v
0
x2
x
q
e
q
dx
(5.8)
and dg = 2xdx. Thus (5.8) becomes
Z v
v
0
0
e
x
q
x
v
2xdx = x2 ( e q ) + 2
|
{z
}0 |
Z v
0
(C)
To do (D), we do integration by parts one more time. Let g = x and dh = e
(D) becomes

2 (x( q e
x
q
))
Z v
v
0
0
x
q
qe
dx = 2( xq e
x
q
xe
{z
x
q
(D)
dx
}
dx ) dg = dx and h =
x
q
q 2e
x
q
)
qe
x
q
so that
v
0
Thus, (A) = (C) + (D), and our calculation of (A) becomes:
x2 ( e
x
q
) + 2( xq e
x
q
q 2e
x
q
)
v
0
(x2 + 2xq + 2q 2 )e
=
=
h
2
2
(v + 2vq + 2q )e
= [ (v2 + 2q v + 2q 2 )e
v
x
q
0
x
q
v
q
i
[ (02 + 0 + 2q 2 )e0 ]
] + 2q 2
Thus (A) + (B) combined equals
[ (v2 + 2q v + 2q 2 )e
v
q
] + 2q 2 + 2q 2 + v2 e
v
q
68
= [ (v◆2 + 2q v + 2q 2
v
v◆2 )e q ] + 2q 2
v
= [ (2q v + 2q 2 )e q ] + 2q 2
(5.9)
In our case, v = 1818.18,
E[(X ^ 1818.18)2 ] = [ (2 · 1000 · 1818.18 + 2 · 10002 )e
1818.18
1000
] + 2 · 10002 = 1, 085, 100.9
d 2
2
2
Now we need E[(X ^ 1+r
) ] = E[(X ^ 250
1.1 ) ] = E[(X ^ 227.27) ]. Plugging back into (5.9) computed a moment ago,
we get
227.27
[ (2 · 1000 · 227.27 + 2 · 10002 )e 1000 ] + 2 · 10002 = 44, 454.132
u
Now we need E(X ^ 1+r
) = E(X ^ 1818.18). Thankfully, this formula is available in the Exam Tables, namely
x
E(X ^ x) = q (1 e q ).
1818.18
E(X ^ 1818.18) = 1000(1 e 1000 ) = 837.68
d
We also need E(X ^ 1+r
) = E(X ^ 227.27). Using q (1
1000(1
e
e
x
q
227.27
1000
), we get
) = 203.29
We now have all of the parts to plug into (5.7). Thus,
✓
E[(Y L )2 ] = (0.49)(1.21) 1, 085, 100.9
✓
◆
◆
250
+ 2·
(203.29)
1.1
= 446, 031.4
44, 454.1
2·
✓
◆
250
(837.68)
1.1
From the previous example, we got that E(Y L ) = 488.48, so the variance is then equal to
446, 031.4
Hence, the standard deviation is
p
488.482 = 207, 418.7
207, 418 = 455.43. Whew!!
Although this problem was lengthy, none of the individual steps was too hard. On the test, you might be given
E[(X ^ u)2 ] and E[(X ^ d)2 ], from which you’ll be required to compute the rest. As I mentioned earlier, if you have
time, memorize the formula if you have time.
Also observe that if the formula in the Exam Tables is too complicated, it might be easier to compute it from first
principles. Check the Exam Tables first before making a decision.
5.4
Problems for Section 5
1. Suppose X ⇠ Pareto(a = 4, q = 5000). Compute fY P , SY P , FY P , hY P and fY L , SY L , FY L , hY L when an ordinary
deductible d = 750 is applied.
2. Repeat Problem 1 when d = 750 is a franchise deductible.
3. Suppose X ⇠ Pareto(a = 4, q = 5000). Compute E(Y L ) and E(Y P ) for both a regular and a franchise deductible
of d = 750.
4. Let
F(x) =
(
0
1
x<0
0.4e
0.00005x
x
0
Compute E(Y L ) and E(Y P ) for both a regular and franchise deductible of d = 1500.
5. Risk 1 has an exponential distribution with q = 2000. Risk 2 has an exponential distribution with q = 2500
with a deductible (ordinary) of d. What is the expected cost per loss for risk 1. What is the expected cost per
loss for risk 2?
What is the ratio of the E(Y L ) for risk 2 to the E(Y L ) for risk 1 as d ! •? What would happen to this ratio if
we were interested in per payment loss variables?
69
6. Consider the following table:
x
F(x) E(X ^ x)
20,000
0.5
10,000
25,000
0.6
18,000
30,000
0.7
21,000
35,000
0.8
24,000
•
1
27,000
There is a per-loss ordinary deductible of $20,000. The deductible is then raised so that 60% of the number
of losses exceed the new deductible as compared to the old deductible. What is the change in the expected cost
per payment (as a percentage).
7. Repeat Problem 6 for a franchise deductible.
8. Let losses be distributed by a Pareto(a = 4, q = 5000) distribution. Let d = 1000, where d is an ordinary
deductible. Compute the LER.
9. Repeat Problem 8 for a franchise deductible.
10. Suppose the coinsurance is 70%, the interest rate is 5%, the maximum covered loss is 30,000 and the deductible
is 5,000. Further suppose losses follow a Pareto(a = 2, q = 20, 000) distribution. What is the expected cost per
loss at the end of one year. What about the cost per payment?
11. Repeat Problem 10 but for the expected cost per loss and cost per payment at the end of 3 years.
Hint: You need to adjust the divisor when you see
u
1+r
and
d
1+r
as well as the (1 + r) term in front.
12. At the end of this year, total losses are expected to be $50,000,000. Individual losses are Pareto(a = 3, q =
10, 000). A reinsurer will pay the excess of each loss that is greater than 135% of the expected individual loss.
In return, the reinsurer is paid a premium P. How much should P be so that the reinsurer takes in enough to
cover expected losses? How much should P be if the reinsurer wants a 10% profit?
13. Reconsider Problem 12. At the end of this year, losses increase by 5% due to inflation. If the reinsurer wants to
keep the same profit margin, how much should the premium be increased by as a percentage of the old premium?
14. Losses are uniformly distributed between 0 and $10,000. For losses below $2,000, nothing is paid. For losses
between $2,000 and $8,000, the full loss is paid. For losses between $8,000 and $10,000, the policy limit of
$8,000 is paid.
What is the expected cost per loss and expected cost per payment?
15. Losses are uniform on [0, 100] for cars, [300, 500] for houses and [800, 1000] for luxury yachts.
50% of losses come from cars, 40% from houses, and 10% from yachts. What is the E[(Y ^ 450)2 ]?
16. Losses are exponentially distributed with q = 10, 000. From 0 to 3,000, the insured pays everything. Losses
from 3,000.01 to 10,000 are paid by the insurer with a coinsurance of 65%. Losses above that are paid by the
insured until the insured pays a total of $9,000. After that, the insurer covers everything with a coinsurance
factor of 95%. What is the expected cost per loss? What about the expected cost per payment?
17. Losses are Pareto(a = 2, q = 10, 000) distributed. Policy 1 has a 2,000 ordinary deductible. Policy 2 has no
deductible but a coinsurance factor of a.
What should a be so that the expected cost per loss is the same for policies 1 and 2?
70
5.5
Solutions for Section 5
1. From the formula sheet, we see that a Pareto(a, q ) variable has the following attributes:
aq a
(x + q )a+1
✓
◆a
q
F(x) = 1
x+q
f (x) =
From the second equation, we can deduce that
S(x) = 1
F(x) =
✓
q
x+q
◆a
Hence, we can substitute the above information directly into the formulas presented earlier in this section.
fX (y + d)
, y>0
SX (d)
fY P (y) =
aq a
(y+d+q )a+1
a
q
d+q
=
=
a(d + q )a
(y + d + q )a+1
=
4(750 + 5000)4
(y + 750 + 5000)4+1
=
4(5750)4
, y>0
(y + 5750)5
SX (y + d)
, y>0
SX (d)
⇣
⌘a
SY P (y) =
q
y+d+q
=
a
q
d+q
✓
◆a
d +q
y+d +q
✓
◆4
5750
, y>0
y + 5750
=
=
FY P (y) =
1
=
1
SY P (y), y > 0
✓
◆4
5750
, y>0
y + 5750
fY P (y)
, y>0
SY P (y)
hY P (y) =
4(5750)4
(y+5750)5
=
⇣
5750
y+5750
⌘4
4
, y>0
y + 5750
=
71
fY L (y) =
fX (y + d), y > 0
aq a
(y + d + q )a+1
=
4(50004 )
, y>0
(y + 5750)5
=
SY L (y) =
SX (y + d), y > 0
✓
◆a
q
=
y+d +q
✓
◆4
5000
=
, y>0
y + 5750
FY L (y) =
1
=
1
SY L (y), y > 0
✓
◆4
5000
, y>0
y + 5750
hY L (y) =
=
hY P (y)
4
, y>0
y + 5750
2. We can use the corresponding set of equations for a franchise deductible in the same way.
(
FX (d) y = 0
fY L (y) =
fX (y) y > d
(
a
q
1
y=0
d+q
=
aq a
y>d
(y+q )a+1
8
5000 4
< 1
y=0
5750
=
4
4(5000)
:
y > 750
5
(y+5000)
FY L (y) =
(
FX (d) 0  y  d
FX (y) y > d
8
a
q
< 1
0yd
d+q ⌘
⇣
a
=
q
: 1
y>d
y+q
8
5000 4
< 1
0  y  750
5750
⇣
⌘4
=
5000
: 1
y > 750
y+5000
SY L (y) = 1 FY L (y)
8
< 5000 4
⇣5750 ⌘4
=
5000
:
y+5000
72
0  y  750
y > 750
hY L (y) =
=
=
(
>
>
:
(
=
=
=
=
0yd
hX (y) y > d
8
>
0  y  750
>
< 0
fY P (y) =
SY P (y) =
0
4(5000)4
(y+5000)5
⇣
5000
y+5000
0
⌘4
4
y+5000
(
y > 750
0  y  750
y > 750
undefined y = 0
fX (y)
SX (d)
y>d
8
>
< undefined y = 0
aq a
(y+q )a+1
>
y>d
:
a
q
( d+q
)
(
undefined y = 0
a(d+q )a
(y+q )a+1
(
(
y>d
undefined y = 0
4(5750)4
(y+5000)5
1
SX (y)
SX (d)
8
>
< 1⇣
⌘a
q
y+q
a
q
d+q
y > 750
0yd
y>d
0yd
>
y>d
:
( )
8
< 1
0yd
⇣
⌘a
=
d+q
: y+q
y>d
8
< 1
0yd
⇣
⌘4
=
5750
:
y > 750
y+5000
FY P (y) = 1 SY P (y)
8
< 0
⇣
⌘4
=
5750
: 1
y+5000
hY P (y) =
=
=
(
(
(
0
0  y  750
y > 750
0yd
hX (y) y > d
0
a
y+q
0
4
y+5000
73
0yd
y>d
0  y  750
y > 750
3. This question simply applies Theorem 5.5. Note that for a Pareto(a, q ) distribution, the exam formula sheet
yields:
✓
◆a
q
F(x) = 1
x+q
q
E(X) =
a 1
"
✓
◆a 1 #
q
q
E(X ^ d) =
1
a 1
d +q
For an ordinary deductible, we use:
E(Y L ) =
E(X)
E(X ^ d)
"
q
q
1
a 1 a 1
✓
◆a
q
q
a 1 d +q
=
=
=
1
✓
q
d +q
◆a
1
#
q4
1)(d + q )3
(a
50004
3(5750)3
1095.86
=
=
E(X) E(X ^ d)
1 F(d)
E(Y L )
1 F(d)
1095.86
E(Y P ) =
=
=
a
q
d+q
1095.86
=
5000 4
5750
= 1916.67
For a franchise deductible, we use:
E(Y L ) =
E(X)
E(X ^ d) + d(1 F(d))
◆
✓
5000 4
= 1095.86 + 750
5750
= 1524.67
E(X) E(X ^ d)
+d
1 F(d)
= 1916.67 + 750
E(Y P ) =
2666.67
=
4. Note f (x) = F 0 (x) = 0.00002e
0.00005x .
We proceed as follows for an ordinary deductible.
E(Y L ) =
=
Z •
Zd •
d) f (x)dx
(x
1500
0.00002(x
= 7421.95
74
1500)e
0.00005x
dx
E(X) E(X ^ d)
1 F(d)
7421.95
=
0.4e 0.00005(1500)
= 20, 000
E(Y P ) =
Now for a franchise deductible:
E(Y L ) =
=
=
E(X)
E(X ^ d) + d(1
F(d))
0.00005(1500)
7421.95 + 1500 · 0.4e
7978.60
E(X) E(X ^ d)
+d
1 F(d)
= 20, 000 + 1500
E(Y P ) =
= 21, 500
5. From the formula sheet, we get the following for an exponential(q ) distribution:
F(x) = 1
e
x/q
E(X) = q
E(X ^ d) = q (1
Hence,
E(Y1L ) =
E(Y2L ) =
E(X)
e
E(X ^ d)
= 2000
2000(1
= 2000e
d/2000
2500e
d/q
d/2500
)
e
d/2000
)
by a similar calculation
The ratio then becomes
E(Y1L )
2000e
= lim
d!• E(Y2L )
d!• 2500e
lim
d/2000
d/2500
Now similarly,
E(Y1P ) =
= lim
d!•
4
e
5
d/2000+d/2500
=
4
lim e
5 d!•
d/10,000
=
4
5
E(Y1L )
2000e d/2000
=
= 2000
1 F(d)
e d/2000
E(Y2P ) = 2500
The ratio between these two expectations is independent of d. Hence, taking the limit as d ! • is exactly equal
4
to the ratio 2000
2500 = 5 .
6. At the old deductible of 20, 000, 50% of losses resulted in payments since
1
F(20, 000) = 0.5
At the new deductible, we know that 60% · 50% = 30% of losses will result in payments. This implies that at
the new deductible d, we have
1 F(d) = 0.3 ) d = 30, 000
75
Before we proceed to calculations, it is important to note that E(X ^ •) = E(X). Intuitively, this makes sense
because E(X ^ d) is essentially the expected payout on a loss with a maximum payout d. If d is infinite, then
that is equivalent to a payout on a loss with no cap, as would be represented by E(X).
We proceed to calculate expected cost per payment for both deductibles.
E(Y1P ) =
E(X)
1
E(X ^ 20, 000) 27, 000 10, 000
=
= 34, 000
F(20, 000)
1 0.5
E(Y2P ) =
E(X)
1
E(X ^ 30, 000) 27, 000 21, 000
=
= 20, 000
F(30, 000)
1 0.7
The change in expected cost per payment is thus
20, 000 34, 000
= 41.2%
34, 000
7. For a franchise deductible, we simply modify our calculation of the expected costs per payment.
E(Y1P ) =
E(X)
1
E(X ^ 20, 000)
27, 000 10, 000
+ 20, 000 =
+ 20, 000 = 54, 000
F(20, 000)
1 0.5
E(Y2P ) =
E(X)
1
E(X ^ 30, 000)
27, 000 21000
+ 30, 000 =
+ 30, 000 = 50, 000
F(30, 000)
1 0.7
The change in expected cost per payment is thus
50, 000 54, 000
= 7.4%
54, 000
8. Recall the formula for the LER:
LER =
E(X)
[E(X) E(X ^ d)] E(X ^ d)
=
E(X)
E(X)
Using what we know about the Pareto distribution, the LER is simply
h
i
a 1
q
q
✓
◆a 1
✓
◆3
a 1 1
d+q
q
5000
LER =
=1
=1
= 0.42
q
d +q
1000 + 5000
a 1
9. We can modify the LER as follows to accommodate for a franchise deductible.
LER =
E(X)
[E(X)
E(X ^ d) + d(1
E(X)
F(d))]
Again, using our knowledge of the Pareto distribution, we get
h
i
a 1
q
q
1
a 1
d+q
LER =
q
=
1
✓
✓
=
1
=
0.132
q
d +q
◆a
a 1
1
5000
1000 + 5000
76
d
◆3
=
d
E(X ^ d)
d(1
E(X)
a
q
d+q
a
q
d+q
q
a 1
5000 4
6000
5000
3
1000
F(d))
10. This question simply applies Theorem 5.16. Refer back to the theorem should you require a reminder of what
each of the variables are.
 ✓
◆
✓
◆
u
d
L
E(Y ) = a(1 + r) E X ^
E X^
1+r
1+r
and
E(Y P ) =
E(Y L )
d
1 FX ( 1+r
)
From the formula sheet we compute:
"
u
30, 000
20, 000
E(X ^
) = E(X ^
) = E(X ^ 28, 571) =
1
1+r
1.05
1
"
d
5000
20, 000
E(X ^
) = E(X ^
) = E(X ^ 4762) =
1
1+r
1.05
1
Hence
E(Y L ) = 0.7(1.05)[11, 765
and
E(Y P ) =
1
✓
✓
20, 000
48, 571
20, 000
24, 762
◆1 #
◆1 #
= 11, 765
= 3846
3846] = 5820
5820
5820
=⇣
⌘2 = 8921
FX (4762)
20,000
24,762
11. By the hint, we simply replace (1 + r) with (1 + r)3 in each equation. Thus, we need to calculate:
"
✓
◆ #
u
30, 000
20, 000
20, 000 1
E(X ^
) = E(X ^
) = E(X ^ 25, 915) =
1
= 11, 288
(1 + r)3
1.053
1
45, 915
"
d
5000
20, 000
E(X ^
) = E(X ^
) = E(X ^ 4319) =
1
1+r
1.053
1
Now, substituting, we get:
E(Y L ) = 0.7(1.053 )[11, 288
and
E(Y P ) =
1
12. The expected loss for an individual is
E(X) =
✓
20, 000
24, 319
◆1 #
= 3552
3552] = 6269
6269
6269
=⇣
⌘2 = 9269
5000
FX ( 1.053 )
20,000
24,319
q
a
1
=
10, 000
= 5000
2
Thus, the deductible is d = 1.35(5000) = 6750. Per loss, the reinsurer’s expected cost is calculated as
"
"
✓
◆a 1 #
✓
◆ #
q
q
10, 000 2
E(X) E(X ^ 6750) = 5000
1
= 5000 1 1 +
= 1782
a 1
6750 + q
16, 750
Since the total expected loss is 50,000,000, and an individual expected loss is 5000, dividing the two yields a
total of 10,000 expected claims. Thus, the total premiums should be set to 10, 000(1782) = 17, 820, 000 in order
to cover expected losses. If the reinsurer wants a 10% profit, the premiums should be 10, 000(1.10)(1782) =
19, 602, 000.
77
13. Similar to the above, the new expected cost per loss is:
1.05(E(X)
"
✓
10, 000
1+
16, 429
E(X ^ 6750/1.05)) = 1.05(5000) 1
◆2 #
= 1945
Total premiums would then be set to 10, 000(1.10)(1945) = 21, 395, 000. As a percentage of the old premium,
19,602,000
this is 21,395,000
= 9.15% higher.
19,602,000
14. Given the losses come from a uniform[0, 10, 000] distribution, we know that
1
10, 000
x
F(x) =
10, 000
f (x) =
Thus,
L
E(Y ) =
Z 8000
2000
x
1
dx + 8000(1
10, 000
E(Y P ) =
15. The density functions over the intervals are
E(Y L )
4600
=
= 5750
1 F(2000)
0.8
0.5 0.4 0.1
100 , 200 , 200 .
2
E((X ^ d) ) =
Thus, we compute
Z 450
E((Y ^ 450)2 ) =
Z d
0
Recall that
2
x f (x)dx +
Z •
d
d 2 f (x)dx
Z •
4502 f (y)dy
✓Z 100
◆ ✓Z 500
◆
Z 1000
0.5 2
0.4 2
0.4
0.1
2
2
=
y dy +
y dy +
450 dy +
450 dy
100
0
300 200
450 200
800 200
= (1666.67 + 42, 750) + (20, 250 + 20, 250)
0
y2 f (y)dy +
F(8000)) = 3000 + 8000(0.2) = 4600
450
Z 450
84, 916.67
=
16. We first compute where the breakpoint above which the insurer covers everything with a coinsurance factor of
95% lies. To do so, realize that this coverage begins after the insured has paid a total of $9,000 from the previous
few brackets. This means the breakpoint can be computed as
(3000
0) + (1
0.65)(10, 000
3000) + (x
10, 000) = 9000 ) x = 13, 550
To summarize, let Y L be the amount covered by the insurer. Then,
8
0
X < 3000
>
>
>
<0.65(X 3000)
3000  X < 10, 000
YL =
>
0.65(10, 000 3000) = 4550 10, 000  X < 13, 550
>
>
:
4550 + 0.95(X 13, 550)
X 13, 550
Thus,
E(Y L ) =
=
Z 10,000
3000
e
0.65(x
x/10,000
3000)
e
x/10,000
10, 000
(0.65x + 4550)
= 4874.6
10,000
3000
dx +
Z •
10,000
+ 4550(e
4550
e
x/10,000
10, 000
10,000/10,000
dx +
Z •
13,550
) + 0.95(10, 000)(e
Carrying this over to our calculation of expected cost per payment, we get
E(Y P ) =
4874.6
E(Y L )
=
= 6580.0
1 F(3000) e 3000/10,000
78
0.95(x
13, 550)
13,550/10,000
)
e
x/10,000
10, 000
dx
17. For Policy 1, we have
L
E(Y ) = E(X)
10, 000
E(X ^ 2000) =
1
For Policy 2, we have
E(Y L ) = aE(X) = a
Setting the two equations equal, we get that a = 0.833.
79
"
10, 000
1
1
✓
10, 000
12, 000
10, 000
= 10, 000a
1
◆1 #
= 8333.33