Chapter 3. Life tables.
Manual for SOA Exam MLC.
Chapter 3. Life tables.
Section 3.5. Interpolating life tables.
c
2008.
Miguel A. Arcones. All rights reserved.
Extract from:
”Arcones’ Manual for SOA Exam MLC. Fall 2009 Edition”,
available at http://www.actexmadriver.com/
1/114
c
2008.
Miguel A. Arcones. All rights reserved.
Manual for SOA Exam MLC.
Chapter 3. Life tables.
Section 3.5. Interpolating life tables.
Interpolating life tables.
Life tables only show the values of `x whenever x is a nonnegative
integer. In many computations, we need to know `x for each
x ≥ 0. We can estimate these values `x in several ways.
2/114
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2008.
Miguel A. Arcones. All rights reserved.
Manual for SOA Exam MLC.
Chapter 3. Life tables.
Section 3.5. Interpolating life tables.
Uniform distribution of deaths.
The simplest way is to assume a uniform distribution of deaths.
That is, assume that between integer–valued years x and x + 1 the
death rate is constant. This implies that the graph of `x+t ,
0 ≤ t ≤ 1 is linear. Hence,
`x+t = (1−t)`x +t`x+1 = `x +t(`x+1 −`x ) = `x −t ·dx , 0 ≤ t ≤ 1.
(1)
3/114
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2008.
Miguel A. Arcones. All rights reserved.
Manual for SOA Exam MLC.
Chapter 3. Life tables.
Section 3.5. Interpolating life tables.
Theorem 1
Under a linear form for the number of living, for each nonnegative
integer x and each 0 ≤ t ≤ 1:
(i) t px = 1 − tqx .
(ii) t qx = tqx , 0 ≤ t ≤ 1.
(iii) fT (x) (t) = qx .
qx
.
(iv) µx+t = 1−tq
x
4/114
c
2008.
Miguel A. Arcones. All rights reserved.
Manual for SOA Exam MLC.
Chapter 3. Life tables.
Section 3.5. Interpolating life tables.
Theorem 1
Under a linear form for the number of living, for each nonnegative
integer x and each 0 ≤ t ≤ 1:
(i) t px = 1 − tqx .
(ii) t qx = tqx , 0 ≤ t ≤ 1.
(iii) fT (x) (t) = qx .
qx
.
(iv) µx+t = 1−tq
x
Proof: (i) By (1),
t
px =
`x+t
`x − t · dx
dx
=
=1−t
= 1 − tqx ,
`x
`x
`x
0 ≤ t ≤ 1.
5/114
c
2008.
Miguel A. Arcones. All rights reserved.
Manual for SOA Exam MLC.
Chapter 3. Life tables.
Section 3.5. Interpolating life tables.
Theorem 1
Under a linear form for the number of living, for each nonnegative
integer x and each 0 ≤ t ≤ 1:
(i) t px = 1 − tqx .
(ii) t qx = tqx , 0 ≤ t ≤ 1.
(iii) fT (x) (t) = qx .
qx
.
(iv) µx+t = 1−tq
x
Proof: (ii)
t
qx = 1 − t px = tqx ,
0 ≤ t ≤ 1.
6/114
c
2008.
Miguel A. Arcones. All rights reserved.
Manual for SOA Exam MLC.
Chapter 3. Life tables.
Section 3.5. Interpolating life tables.
Theorem 1
Under a linear form for the number of living, for each nonnegative
integer x and each 0 ≤ t ≤ 1:
(i) t px = 1 − tqx .
(ii) t qx = tqx , 0 ≤ t ≤ 1.
(iii) fT (x) (t) = qx .
qx
.
(iv) µx+t = 1−tq
x
Proof: (iii)
fT (x) (t) = −
d
t px = qx .
dt
7/114
c
2008.
Miguel A. Arcones. All rights reserved.
Manual for SOA Exam MLC.
Chapter 3. Life tables.
Section 3.5. Interpolating life tables.
Theorem 1
Under a linear form for the number of living, for each nonnegative
integer x and each 0 ≤ t ≤ 1:
(i) t px = 1 − tqx .
(ii) t qx = tqx , 0 ≤ t ≤ 1.
(iii) fT (x) (t) = qx .
qx
.
(iv) µx+t = 1−tq
x
Proof: (iv)
µx+t = −
d
d
qx
log t px = − log(1 − tqx ) =
.
dt
dt
1 − tqx
8/114
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2008.
Miguel A. Arcones. All rights reserved.
Manual for SOA Exam MLC.
Chapter 3. Life tables.
Section 3.5. Interpolating life tables.
Example 1
Using the life table in Section 4.1 and assuming a uniform
distribution of deaths, find:
(i) 0.5 p35
(ii) 1.5 p35 .
9/114
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2008.
Miguel A. Arcones. All rights reserved.
Manual for SOA Exam MLC.
Chapter 3. Life tables.
Section 3.5. Interpolating life tables.
Example 1
Using the life table in Section 4.1 and assuming a uniform
distribution of deaths, find:
(i) 0.5 p35
(ii) 1.5 p35 .
Solution: (i) We have that p35 =
0.5
`36
`35
=
97126
97250
= 0.9987249357 and
p35 = 1 − 0.5q35 = 1 − 0.5(1 − 0.9987249357) = 0.9993624678.
(ii) We have that p36 =
0.5 p36
`37
`36
=
96993
97126
= 0.9986306447 and
= 1 − 0.5q36 = 1 − 0.5(1 − 0.9986306447) = 0.9993153224.
Hence,
1.5 p35
= p35 ·0.5 p36 = (0.9987249357)(0.9993153224) = 0.9980411311.
10/114
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2008.
Miguel A. Arcones. All rights reserved.
Manual for SOA Exam MLC.
Chapter 3. Life tables.
Section 3.5. Interpolating life tables.
Theorem 2
Given t ≥ 0, let k be the nonnegative integer such that
k ≤ t < k + 1. Under uniform interpolation,
(i) s(t) = ``k0 − (t − k) ``x0 .
(ii) fX (t) = d`0k = k |q0 .
(iii) fT (x) (t) =
dx+k
`x
= k |qx .
11/114
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2008.
Miguel A. Arcones. All rights reserved.
Manual for SOA Exam MLC.
Chapter 3. Life tables.
Section 3.5. Interpolating life tables.
Theorem 2
Given t ≥ 0, let k be the nonnegative integer such that
k ≤ t < k + 1. Under uniform interpolation,
(i) s(t) = ``k0 − (t − k) ``x0 .
(ii) fX (t) = d`0k = k |q0 .
(iii) fT (x) (t) =
dx+k
`x
= k |qx .
Proof: (i) By (1), `t = `k+t−k = `k − (t − k) · dk . Hence,
s(t) = ``0t = ``k0 − (t − k) d`0k .
12/114
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2008.
Miguel A. Arcones. All rights reserved.
Manual for SOA Exam MLC.
Chapter 3. Life tables.
Section 3.5. Interpolating life tables.
Theorem 2
Given t ≥ 0, let k be the nonnegative integer such that
k ≤ t < k + 1. Under uniform interpolation,
(i) s(t) = ``k0 − (t − k) ``x0 .
(ii) fX (t) = d`0k = k |q0 .
(iii) fT (x) (t) =
dx+k
`x
= k |qx .
Proof: (i) By (1), `t = `k+t−k = `k − (t − k) · dk . Hence,
s(t) = ``0t = ``k0 − (t − k) d`0k .
d
(ii) fX (t) = − dt
s(t) = d`0k = k |q0 .
13/114
c
2008.
Miguel A. Arcones. All rights reserved.
Manual for SOA Exam MLC.
Chapter 3. Life tables.
Section 3.5. Interpolating life tables.
Theorem 2
Given t ≥ 0, let k be the nonnegative integer such that
k ≤ t < k + 1. Under uniform interpolation,
(i) s(t) = ``k0 − (t − k) ``x0 .
(ii) fX (t) = d`0k = k |q0 .
(iii) fT (x) (t) =
dx+k
`x
= k |qx .
Proof: (i) By (1), `t = `k+t−k = `k − (t − k) · dk . Hence,
s(t) = ``0t = ``k0 − (t − k) d`0k .
d
(ii) fX (t) = − dt
s(t) = d`0k = k |q0 .
(iii) fT (x) (t) =
fX (x+t)
s(x)
=
dx+k
`x
= k |qx .
14/114
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2008.
Miguel A. Arcones. All rights reserved.
Manual for SOA Exam MLC.
Chapter 3. Life tables.
Section 3.5. Interpolating life tables.
Example 2
Consider the life table
x
`x
80
250
81
217
82
161
83
107
84
62
85
28
86
0
15/114
c
2008.
Miguel A. Arcones. All rights reserved.
Manual for SOA Exam MLC.
Chapter 3. Life tables.
Section 3.5. Interpolating life tables.
Example 2
Consider the life table
x
`x
80
250
81
217
82
161
83
107
84
62
85
28
86
0
(i) Calculate dx , x = 81, 82, . . . , 86.
16/114
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2008.
Miguel A. Arcones. All rights reserved.
Manual for SOA Exam MLC.
Chapter 3. Life tables.
Section 3.5. Interpolating life tables.
Example 2
Consider the life table
x
`x
80
250
81
217
82
161
83
107
84
62
85
28
86
0
(i) Calculate dx , x = 81, 82, . . . , 86.
Solution: (i) Using that dx = `x − `x+1 , we get that
x
`x
dx
80
250
33
81
217
56
82
161
54
83
107
45
84
62
34
85
28
28
86
0
0
17/114
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2008.
Miguel A. Arcones. All rights reserved.
Manual for SOA Exam MLC.
Chapter 3. Life tables.
Section 3.5. Interpolating life tables.
Example 2
Consider the life table
x
`x
80
250
81
217
82
161
83
107
84
62
85
28
86
0
(ii) Using linear interpolation, calculate `80+t , 0 ≤ t ≤ 6.
18/114
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2008.
Miguel A. Arcones. All rights reserved.
Manual for SOA Exam MLC.
Chapter 3. Life tables.
Section 3.5. Interpolating life tables.
Example 2
Consider the life table
x
`x
80
250
81
217
82
161
83
107
84
62
85
28
86
0
(ii) Using linear interpolation, calculate `80+t , 0 ≤ t ≤ 6.
Solution: (ii) Using that `x+t = `x − t · dx , 0 ≤ t ≤ 1,


250 − 33t
if 0 ≤ t ≤ 1,





217 − 56(t − 1) if 1 ≤ t ≤ 2,



161 − 54(t − 2) if 2 ≤ t ≤ 3,
`80+t =

107 − 45(t − 3) if 3 ≤ t ≤ 4,





62 − 34(t − 4)
if 4 ≤ t ≤ 5,



28 − 28(t − 5)
if 5 ≤ t ≤ 6.
19/114
c
2008.
Miguel A. Arcones. All rights reserved.
Manual for SOA Exam MLC.
Chapter 3. Life tables.
Section 3.5. Interpolating life tables.
Example 2
Consider the life table
x
`x
80
250
81
217
82
161
83
107
84
62
85
28
86
0
(iii) Using linear interpolation, calculate t p80 , 0 ≤ t ≤ 6.
20/114
c
2008.
Miguel A. Arcones. All rights reserved.
Manual for SOA Exam MLC.
Chapter 3. Life tables.
Section 3.5. Interpolating life tables.
Example 2
Consider the life table
x
`x
80
250
81
217
82
161
83
107
84
62
85
28
86
0
(iii) Using linear interpolation, calculate t p80 , 0 ≤ t ≤ 6.
Solution: (iii) Using that t px = `x+t
`x ,
t p80
=

250−33t


250


217−56(t−1)



250


 161−54(t−2)
250
107−45(t−3)


250


62−34(t−4)



250


 28−28(t−5)
250
if
if
if
if
if
if
0≤t
1≤t
2≤t
3≤t
4≤t
5≤t
≤ 1,
≤ 2,
≤ 3,
≤ 4,
≤ 5,
≤ 6.
21/114
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2008.
Miguel A. Arcones. All rights reserved.
Manual for SOA Exam MLC.
Chapter 3. Life tables.
Section 3.5. Interpolating life tables.
Example 2
Consider the life table
x
`x
80
250
81
217
82
161
83
107
84
62
85
28
86
0
(iv) Using linear interpolation, calculate the density function of the
future life T80 .
22/114
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2008.
Miguel A. Arcones. All rights reserved.
Manual for SOA Exam MLC.
Chapter 3. Life tables.
Section 3.5. Interpolating life tables.
Example 2
Consider the life table
x
`x
80
250
81
217
82
161
83
107
84
62
85
28
86
0
(iv) Using linear interpolation, calculate the density function of the
future life T80 .
t px )
Solution: (iv) Using that fT80 (t) = − d(dt
,

33

if 0 ≤ t ≤ 1,

250



56

if 1 ≤ t ≤ 2,

250


 54
if 2 ≤ t ≤ 3,
fT80 (t) = 250
45
 250 if 3 ≤ t ≤ 4,




34

if 4 ≤ t ≤ 5,

250


 28
if 5 ≤ t ≤ 6.
250
23/114
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2008.
Miguel A. Arcones. All rights reserved.
Manual for SOA Exam MLC.
Chapter 3. Life tables.
Section 3.5. Interpolating life tables.
Theorem 3
Under a linear form for the number of living,
(i) Lx = `x − d2x = `x+1 + d2x = `x +`2 x+1 .
P
(ii) Tx = `2x + ∞
k=x+1 `k .
(iii) mx = 1−qxqx .
◦
2
(iv) e x = ex + 21 .
24/114
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2008.
Miguel A. Arcones. All rights reserved.
Manual for SOA Exam MLC.
Chapter 3. Life tables.
Section 3.5. Interpolating life tables.
Theorem 3
Under a linear form for the number of living,
(i) Lx = `x − d2x = `x+1 + d2x = `x +`2 x+1 .
P
(ii) Tx = `2x + ∞
k=x+1 `k .
(iii) mx = 1−qxqx .
◦
2
(iv) e x = ex + 21 .
Proof:
(i)
Z 1
Z 1
dx
`x + `x+1
Lx =
`x+t dt =
(`x − t · dx ) dt = `x −
=
2
2
0
0
dx
=`x+1 + .
2
25/114
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2008.
Miguel A. Arcones. All rights reserved.
Manual for SOA Exam MLC.
Chapter 3. Life tables.
Section 3.5. Interpolating life tables.
Theorem 3
Under a linear form for the number of living,
(i) Lx = `x − d2x = `x+1 + d2x = `x +`2 x+1 .
P
(ii) Tx = `2x + ∞
k=x+1 `k .
(iii) mx = 1−qxqx .
◦
2
(iv) e x = ex + 21 .
Proof:
(i)
Z 1
Z 1
dx
`x + `x+1
Lx =
`x+t dt =
(`x − t · dx ) dt = `x −
=
2
2
0
0
dx
=`x+1 + .
2
P∞
P
P
`x +`x+1
(ii) Tx = k=x Lk = ∞
= `2x + ∞
k=x
k=x+1 `k .
2
26/114
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2008.
Miguel A. Arcones. All rights reserved.
Manual for SOA Exam MLC.
Chapter 3. Life tables.
Section 3.5. Interpolating life tables.
Theorem 3
Under a linear form for the number of living,
(i) Lx = `x − d2x = `x+1 + d2x = `x +`2 x+1 .
P
(ii) Tx = `2x + ∞
k=x+1 `k .
(iii) mx = 1−qxqx .
2
◦
(iv) e x = ex + 21 .
Proof:
(i)
Z 1
Z 1
dx
`x + `x+1
Lx =
`x+t dt =
(`x − t · dx ) dt = `x −
=
2
2
0
0
dx
=`x+1 + .
2
P∞
P
P
`x +`x+1
(ii) Tx = k=x Lk = ∞
= `2x + ∞
k=x
k=x+1 `k .
2
(iii) mx =
dx
Lx
=
dx
`x − d2x
=
qx
1− q2x
.
27/114
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2008.
Miguel A. Arcones. All rights reserved.
Manual for SOA Exam MLC.
Chapter 3. Life tables.
Section 3.5. Interpolating life tables.
Theorem 3
Under a linear form for the number of living,
(i) Lx = `x − d2x = `x+1 + d2x = `x +`2 x+1 .
P
(ii) Tx = `2x + ∞
k=x+1 `k .
(iii) mx = 1−qxqx .
2
◦
(iv) e x = ex + 21 .
Proof:
(i)
Z 1
Z 1
dx
`x + `x+1
Lx =
`x+t dt =
(`x − t · dx ) dt = `x −
=
2
2
0
0
dx
=`x+1 + .
2
P∞
P
P
`x +`x+1
(ii) Tx = k=x Lk = ∞
= `2x + ∞
k=x
k=x+1 `k .
2
(iii) mx =
◦
(iv) e x =
dx
Lx
=
Tx
`x
=
dx
= 1−qxqx .
`x − d2x
2
P∞
`k
1
k=x+1 `x
2 +
c
2008.
Miguel A. Arcones. All rights reserved.
= ex + 12 .
Manual for SOA Exam MLC.
28/114
Chapter 3. Life tables.
Section 3.5. Interpolating life tables.
Example 3
Consider the life table
x
`x
80
250
81
217
82
161
83
107
84
62
85
28
86
0
Assuming linear interpolation,
◦
R(i)∞ calculate the complete expected life at 80 using that e x =
0 tfT80 (t)dt.
29/114
c
2008.
Miguel A. Arcones. All rights reserved.
Manual for SOA Exam MLC.
Chapter 3. Life tables.
Section 3.5. Interpolating life tables.
Example 3
Consider the life table
x
`x
80
250
81
217
82
161
83
107
84
62
85
28
86
0
Assuming linear interpolation,
R(i)∞ calculate the complete expected
0 tfT80 (t)dt.
Solution: (i) By Example 2,

33

if

250



56

if

250


 54
if
fT80 (t) = 250
45
 250 if




34
 250
if



 28
if
250
c
2008.
Miguel A. Arcones. All rights reserved.
◦
life at 80 using that e x =
0≤t
1≤t
2≤t
3≤t
4≤t
5≤t
≤ 1,
≤ 2,
≤ 3,
≤ 4,
≤ 5,
≤ 6.
Manual for SOA Exam MLC.
30/114
Chapter 3. Life tables.
Section 3.5. Interpolating life tables.
Example 3
Consider the life table
x
`x
80
250
81
217
82
161
83
107
84
62
85
28
86
0
Assuming linear interpolation,
◦
R(i)∞ calculate the complete expected life at 80 using that e x =
0 tfT80 (t)dt.
Solution: (i) So,
◦
Z
∞
Z
1
33
t
+
250
Z
2
56
t
+
250
Z
3
54
t
+
250
ex =
tfT80 (t) =
0
0
1
2
Z 5
Z 6
34
28
+
t
+
t
250
250
4
4
1 33
3 56
5 54
7 45
9 34
11 28
=
+
+
+
+
+
.
2 250 2 250 2 250 2 250 2 250
2 250
Z
4
t
3
45
250
31/114
c
2008.
Miguel A. Arcones. All rights reserved.
Manual for SOA Exam MLC.
Chapter 3. Life tables.
Section 3.5. Interpolating life tables.
Example 3
Consider the life table
x
`x
80
250
81
217
82
161
83
107
84
62
85
28
86
0
Assuming linear interpolation,
◦
calculate the complete expected life at 80 using that e x =
R(ii)
∞
0 t px dt.
32/114
c
2008.
Miguel A. Arcones. All rights reserved.
Manual for SOA Exam MLC.
Chapter 3. Life tables.
Section 3.5. Interpolating life tables.
Example 3
Consider the life table
x
`x
80
250
81
217
82
161
83
107
84
62
85
28
86
0
Assuming linear interpolation,
calculate the complete expected
R(ii)
∞
0 t px dt.
Solution: (ii) By Example 2,

250−33t


250


217−56(t−1)



250


 161−54(t−2)
250
t p80 =
107−45(t−3)


250


62−34(t−4)



250


 28−28(t−5)
250
c
2008.
Miguel A. Arcones. All rights reserved.
◦
life at 80 using that e x =
if
if
if
if
if
if
0≤t
1≤t
2≤t
3≤t
4≤t
5≤t
≤ 1,
≤ 2,
≤ 3,
≤ 4,
≤ 5,
≤ 6.
33/114
Manual for SOA Exam MLC.
Chapter 3. Life tables.
Section 3.5. Interpolating life tables.
Example 3
Consider the life table
x
`x
80
250
81
217
82
161
83
107
84
62
85
28
86
0
Assuming linear interpolation,
◦
calculate the complete expected life at 80 using that e x =
R(ii)
∞
0 t px dt.
Solution: (ii) So,
∞
1
Z 2
250 − 33t
217 − 56(t − 1)
ex =
dt +
dt
t px dt =
250
250
0
0
1
Z 3
Z 4
107 − 45(t − 3)
161 − 54(t − 2)
dt +
dt
+
250
250
3
2
Z 6
Z 5
62 − 34(t − 4)
28 − 28(t − 5)
+
dt +
dt
250
250
4
5
=2.8
◦
Z
Z
c
2008.
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Manual for SOA Exam MLC.
34/114
Chapter 3. Life tables.
Section 3.5. Interpolating life tables.
Example 3
Consider the life table
x
`x
80
250
81
217
82
161
83
107
84
62
85
28
86
0
Assuming linear interpolation,
◦
(iii) calculate the complete expected life at 80 using that e x = ex + 21 .
35/114
c
2008.
Miguel A. Arcones. All rights reserved.
Manual for SOA Exam MLC.
Chapter 3. Life tables.
Section 3.5. Interpolating life tables.
Example 3
Consider the life table
x
`x
80
250
81
217
82
161
83
107
84
62
85
28
86
0
Assuming linear interpolation,
◦
(iii) calculate the complete expected life at 80 using that e x = ex + 21 .
Solution: (iii) We have that
e80 =
∞
X
`80+k
k=1
`80
=
217 + 161 + 107 + 62 + 28
= 2.3
250
◦
So, e 80 = 2.3 + 0.5 = 2.8.
36/114
c
2008.
Miguel A. Arcones. All rights reserved.
Manual for SOA Exam MLC.
Chapter 3. Life tables.
Section 3.5. Interpolating life tables.
Example 3
Consider the life table
x
`x
80
250
81
217
82
161
83
107
84
62
85
28
86
0
Assuming linear interpolation,
◦
(iv) calculate e 80:3| .
37/114
c
2008.
Miguel A. Arcones. All rights reserved.
Manual for SOA Exam MLC.
Chapter 3. Life tables.
Section 3.5. Interpolating life tables.
Example 3
Consider the life table
x
`x
80
250
81
217
82
161
83
107
84
62
85
28
86
0
Assuming linear interpolation,
◦
(iv) calculate e 80:3| .
Solution: (iv)
◦
e 80:3| =
L80 + L81 + L82
=
`80
250+217
2
+
217+161
2
+
161+107
2
250
= 2.226.
38/114
c
2008.
Miguel A. Arcones. All rights reserved.
Manual for SOA Exam MLC.
Chapter 3. Life tables.
Section 3.5. Interpolating life tables.
Theorem 4
For each x,
(i) P{K (x) = k} = dx+k
`x , k = 0, 1, 2 . . .
(ii) Sx has a uniform distribution on the interval (0, 1).
(iii) K (x) and Sx are independent r.v.’s.
39/114
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2008.
Miguel A. Arcones. All rights reserved.
Manual for SOA Exam MLC.
Chapter 3. Life tables.
Section 3.5. Interpolating life tables.
Proof: For each x, each k ≥ 0 and each 0 ≤ t < 1,
P{K (x) = k, Sx ≤ t} = P{k < T (x) ≤ k+t} =
`x+k − `x+k+t
tdx+k
=
.
`x
`x
Letting t → 1, we get that
P{K (x) = k} =
dx+k
.
lx
We also have that
P{Sx ≤ t} =
∞
X
P{K (x) = k, Sx ≤ t} =
k=0
∞
X
tdx+k
k=0
`x
= t.
Hence, for each k ≥ 0 and each 0 ≤ t ≤ 1,
P{K (x) = k, Sx ≤ t} = P{K (x) = k}P{Sx ≤ t}
which implies that K (x) and Sx are independent r.v.’s.
c
2008.
Miguel A. Arcones. All rights reserved.
Manual for SOA Exam MLC.
40/114
Chapter 3. Life tables.
Section 3.5. Interpolating life tables.
Corollary 1
Under the assumption of uniform distribution of deaths:
◦
(i) e x = ex + 21 .
1
(ii) Var(T (x)) = Var(K (x)) + 12
.
41/114
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2008.
Miguel A. Arcones. All rights reserved.
Manual for SOA Exam MLC.
Chapter 3. Life tables.
Section 3.5. Interpolating life tables.
Corollary 1
Under the assumption of uniform distribution of deaths:
◦
(i) e x = ex + 21 .
1
(ii) Var(T (x)) = Var(K (x)) + 12
.
Proof: (i) Since T (x) = K (x) + Sx ,
◦
e x = E [T (x)] = E [K (x)] + E [Sx ] = ex + 12 .
42/114
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2008.
Miguel A. Arcones. All rights reserved.
Manual for SOA Exam MLC.
Chapter 3. Life tables.
Section 3.5. Interpolating life tables.
Corollary 1
Under the assumption of uniform distribution of deaths:
◦
(i) e x = ex + 21 .
1
(ii) Var(T (x)) = Var(K (x)) + 12
.
Proof: (i) Since T (x) = K (x) + Sx ,
◦
e x = E [T (x)] = E [K (x)] + E [Sx ] = ex + 12 .
(ii) Since T (x) = K (x) + Sx and K (x) and Sx are independent,
Var(T (x)) = Var(K (x)) + Var(S(x)) = Var(K (x)) +
1
.
12
43/114
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2008.
Miguel A. Arcones. All rights reserved.
Manual for SOA Exam MLC.
Chapter 3. Life tables.
Section 3.5. Interpolating life tables.
Exponential interpolation.
Under exponential interpolation, we assume that `x+t = abt , for
0 ≤ t ≤ 1, where a and b depend on x. Since `x = ab0 and
`x+1 = ab1 , we get that a = `x , b = `x+1
`x = px , and
t
`x+t = ab =
`x pxt
= `x
`x+1
`x
t
= (`x )1−t (`x+1 )t .
(2)
We will see that force of mortality is constant between x and x + 1.
The form obtained using exponential interpolation is also called the
constant force of mortality form of the number of living.
44/114
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2008.
Miguel A. Arcones. All rights reserved.
Manual for SOA Exam MLC.
Chapter 3. Life tables.
Section 3.5. Interpolating life tables.
Theorem 5
Under a exponential form for the number of living, for each
nonnegative integer x and each 0 ≤ t < 1:
(i) t px = pxt .
(ii) t qx = 1 − (1 − qx )t .
(iii) fTx (t) = −pxt log px .
(iv) µx+t = − log px .
45/114
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2008.
Miguel A. Arcones. All rights reserved.
Manual for SOA Exam MLC.
Chapter 3. Life tables.
Section 3.5. Interpolating life tables.
Theorem 5
Under a exponential form for the number of living, for each
nonnegative integer x and each 0 ≤ t < 1:
(i) t px = pxt .
(ii) t qx = 1 − (1 − qx )t .
(iii) fTx (t) = −pxt log px .
(iv) µx+t = − log px .
Proof: (i) t px =
`x+t
`x
= pxt .
46/114
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2008.
Miguel A. Arcones. All rights reserved.
Manual for SOA Exam MLC.
Chapter 3. Life tables.
Section 3.5. Interpolating life tables.
Theorem 5
Under a exponential form for the number of living, for each
nonnegative integer x and each 0 ≤ t < 1:
(i) t px = pxt .
(ii) t qx = 1 − (1 − qx )t .
(iii) fTx (t) = −pxt log px .
(iv) µx+t = − log px .
t
Proof: (i) t px = `x+t
`x = px .
(ii) t qx = 1 − t px = 1 − pxt = 1 − (1 − qx )t .
47/114
c
2008.
Miguel A. Arcones. All rights reserved.
Manual for SOA Exam MLC.
Chapter 3. Life tables.
Section 3.5. Interpolating life tables.
Theorem 5
Under a exponential form for the number of living, for each
nonnegative integer x and each 0 ≤ t < 1:
(i) t px = pxt .
(ii) t qx = 1 − (1 − qx )t .
(iii) fTx (t) = −pxt log px .
(iv) µx+t = − log px .
t
Proof: (i) t px = `x+t
`x = px .
(ii) t qx = 1 − t px = 1 − pxt = 1 − (1 − qx )t .
d
(iii) fT (x) (t)(t) = − dt
log(t px ) = −pxt log px .
48/114
c
2008.
Miguel A. Arcones. All rights reserved.
Manual for SOA Exam MLC.
Chapter 3. Life tables.
Section 3.5. Interpolating life tables.
Theorem 5
Under a exponential form for the number of living, for each
nonnegative integer x and each 0 ≤ t < 1:
(i) t px = pxt .
(ii) t qx = 1 − (1 − qx )t .
(iii) fTx (t) = −pxt log px .
(iv) µx+t = − log px .
t
Proof: (i) t px = `x+t
`x = px .
(ii) t qx = 1 − t px = 1 − pxt = 1 − (1 − qx )t .
d
(iii) fT (x) (t)(t) = − dt
log(t px ) = −pxt log px .
d
(iv) µx+t = − dt log(t px ) = − log px .
49/114
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2008.
Miguel A. Arcones. All rights reserved.
Manual for SOA Exam MLC.
Chapter 3. Life tables.
Section 3.5. Interpolating life tables.
Example 4
Using the life table in Section 4.1 and exponential interpolation,
find:
(i) 0.75 p80
(ii) 2.25 p80 .
50/114
c
2008.
Miguel A. Arcones. All rights reserved.
Manual for SOA Exam MLC.
Chapter 3. Life tables.
Section 3.5. Interpolating life tables.
Example 4
Using the life table in Section 4.1 and exponential interpolation,
find:
(i) 0.75 p80
(ii) 2.25 p80 .
Solution: (i) We have that
0.75 p80
=
0.75
p80
=
`81
`80
0.75
=
50987
53925
0.75
= 0.958852885.
51/114
c
2008.
Miguel A. Arcones. All rights reserved.
Manual for SOA Exam MLC.
Chapter 3. Life tables.
Section 3.5. Interpolating life tables.
Example 4
Using the life table in Section 4.1 and exponential interpolation,
find:
(i) 0.75 p80
(ii) 2.25 p80 .
Solution: (i) We have that
0.75 p80
=
0.75
p80
=
`81
`80
0.75
=
50987
53925
0.75
= 0.958852885.
(ii) We have that
2.25 p80
=
= p80 p81 · 0.25 p82
50987 47940
53925 50987
44803
47940
`81 `82
=
`80 `81
`83
l82
0.25
0.25
= 0.8450154997.
52/114
c
2008.
Miguel A. Arcones. All rights reserved.
Manual for SOA Exam MLC.
Chapter 3. Life tables.
Section 3.5. Interpolating life tables.
Theorem 6
Given t ≥ 0, let k be the nonnegative integer such that
k ≤ t < k + 1. Under exponential interpolation:
(i) s(t) = ``k0 pkt−k .
(ii) fX (t) = ``k0 pkt (− log pk ).
t
(iii) fT (x) (t)(t) = k px · px+k
(− log px+k ), 0 ≤ t ≤ 1.
53/114
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2008.
Miguel A. Arcones. All rights reserved.
Manual for SOA Exam MLC.
Chapter 3. Life tables.
Section 3.5. Interpolating life tables.
Theorem 6
Given t ≥ 0, let k be the nonnegative integer such that
k ≤ t < k + 1. Under exponential interpolation:
(i) s(t) = ``k0 pkt−k .
(ii) fX (t) = ``k0 pkt (− log pk ).
t
(iii) fT (x) (t)(t) = k px · px+k
(− log px+k ), 0 ≤ t ≤ 1.
Proof: (i) By (2), for each integer x and each 0 ≤ t ≤ 1,
`x+t
`x
s(x + t) =
=
`0
`0
`x+1
`x
t
=
`x t
p .
`0 x
Hence, for t ≥ 0 and k ≤ t < k + 1,
s(t) = s(k + t − k) =
`k t−k
p .
`0 k
54/114
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2008.
Miguel A. Arcones. All rights reserved.
Manual for SOA Exam MLC.
Chapter 3. Life tables.
Section 3.5. Interpolating life tables.
Theorem 6
Given t ≥ 0, let k be the nonnegative integer such that
k ≤ t < k + 1. Under exponential interpolation:
(i) s(t) = ``k0 pkt−k .
(ii) fX (t) = ``k0 pkt (− log pk ).
t
(iii) fT (x) (t)(t) = k px · px+k
(− log px+k ), 0 ≤ t ≤ 1.
Proof: (ii)
fX (t) = −
d
`k
s(t) = pkt (− log pk ).
dt
`0
55/114
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2008.
Miguel A. Arcones. All rights reserved.
Manual for SOA Exam MLC.
Chapter 3. Life tables.
Section 3.5. Interpolating life tables.
Theorem 6
Given t ≥ 0, let k be the nonnegative integer such that
k ≤ t < k + 1. Under exponential interpolation:
(i) s(t) = ``k0 pkt−k .
(ii) fX (t) = ``k0 pkt (− log pk ).
t
(iii) fT (x) (t)(t) = k px · px+k
(− log px+k ), 0 ≤ t ≤ 1.
Proof: (iii)
fX (x + t)
fT (x) (t) =
=
s(x)
`x+k t
`0 px+k (− log px+k )
`x
`0
t
= k px · px+k
(− log px+k )
56/114
c
2008.
Miguel A. Arcones. All rights reserved.
Manual for SOA Exam MLC.
Chapter 3. Life tables.
Section 3.5. Interpolating life tables.
Example 5
Consider the life table
x
`x
80
250
81
217
82
161
83
107
84
62
85
28
86
0
57/114
c
2008.
Miguel A. Arcones. All rights reserved.
Manual for SOA Exam MLC.
Chapter 3. Life tables.
Section 3.5. Interpolating life tables.
Example 5
Consider the life table
x
`x
80
250
81
217
82
161
83
107
84
62
85
28
86
0
(i) Using exponential interpolation, calculate `80+t , 0 ≤ t ≤ 6.
58/114
c
2008.
Miguel A. Arcones. All rights reserved.
Manual for SOA Exam MLC.
Chapter 3. Life tables.
Section 3.5. Interpolating life tables.
Example 5
Consider the life table
x
`x
80
250
81
217
82
161
83
107
84
62
85
28
86
0
(i) Using exponential interpolation, calculate `80+t , 0 ≤ t ≤ 6.
t
Solution: (i) Using that `x+t = `x `x+1
, 0 ≤ t ≤ 1,
`x
`80+t

t

250 217

250 


161 t−1

217 217



161 107 t−2
161 =
62 t−3

107 107




28 t−4

62

62



0
if 0 ≤ t ≤ 1,
if 1 ≤ t ≤ 2,
if 2 ≤ t ≤ 3,
if 3 ≤ t ≤ 4,
if 4 ≤ t ≤ 5,
if 5 < t ≤ 6.
59/114
c
2008.
Miguel A. Arcones. All rights reserved.
Manual for SOA Exam MLC.
Chapter 3. Life tables.
Section 3.5. Interpolating life tables.
Example 5
Consider the life table
x
`x
80
250
81
217
82
161
83
107
84
62
85
28
86
0
(ii) Using exponential interpolation, calculate t p80 , 0 ≤ t ≤ 6.
60/114
c
2008.
Miguel A. Arcones. All rights reserved.
Manual for SOA Exam MLC.
Chapter 3. Life tables.
Section 3.5. Interpolating life tables.
Example 5
Consider the life table
x
`x
80
250
81
217
82
161
83
107
84
62
85
28
86
0
(ii) Using exponential interpolation, calculate t p80 , 0 ≤ t ≤ 6.
Solution: (ii) Using that t px = `x+t
`x ,
t p80
=

250


250



217


250


 161
250
107


250



62


250


0
217 t
250 161 t−1
217 107 t−2
161 62 t−3
107
28 t−4
62
if 0 ≤ t ≤ 1,
if 1 ≤ t ≤ 2,
if 2 ≤ t ≤ 3,
if 3 ≤ t ≤ 4,
if 4 ≤ t ≤ 5,
if 5 < t ≤ 6.
61/114
c
2008.
Miguel A. Arcones. All rights reserved.
Manual for SOA Exam MLC.
Chapter 3. Life tables.
Section 3.5. Interpolating life tables.
Example 5
Consider the life table
x
`x
80
250
81
217
82
161
83
107
84
62
85
28
86
0
(iii) Using exponential interpolation, calculate µ(80 + t), 0 ≤ t ≤ 6.
62/114
c
2008.
Miguel A. Arcones. All rights reserved.
Manual for SOA Exam MLC.
Chapter 3. Life tables.
Section 3.5. Interpolating life tables.
Example 5
Consider the life table
x
`x
80
250
81
217
82
161
83
107
84
62
85
28
86
0
(iii) Using exponential interpolation, calculate µ(80 + t), 0 ≤ t ≤ 6.
Solution: (iii) Using that µ(80 + t) = − d(log(dtt p80 )) ,

217

− log 250
if 0 ≤ t ≤ 1,




161

− log 217
if 1 ≤ t ≤ 2,



− log 107 if 2 ≤ t ≤ 3,
161 µ(80 + t) =
62
− log 107
if 3 ≤ t ≤ 4,




28
− log 62
if 4 ≤ t ≤ 5,



∞
if 5 < t ≤ 6.
63/114
c
2008.
Miguel A. Arcones. All rights reserved.
Manual for SOA Exam MLC.
Chapter 3. Life tables.
Section 3.5. Interpolating life tables.
Example 5
Consider the life table
x
`x
80
250
81
217
82
161
83
107
84
62
85
28
86
0
(iv) Using exponential interpolation, calculate the density function
of the future life T80 .
64/114
c
2008.
Miguel A. Arcones. All rights reserved.
Manual for SOA Exam MLC.
Chapter 3. Life tables.
Section 3.5. Interpolating life tables.
Example 5
Consider the life table
x
`x
80
250
81
217
82
161
83
107
84
62
(iv) Using exponential interpolation, calculate
of the future life T80 .
Solution: (iv) Using that fT80 (t) = − d(tdtp80 ) ,

217
250 217 t


250 250 − log 250



217 161 t−1

− log 161

250 217 217 

 161 107 t−2 − log 107
250 161 161 fT80 (t) = 107
62 t−3
62

−
log

250 107
107



62 28 t−4

− log 28
 250 62
62


0
85
28
86
0
the density function
if 0 ≤ t ≤ 1,
if 1 ≤ t ≤ 2,
if 2 ≤ t ≤ 3,
if 3 ≤ t ≤ 4,
if 4 ≤ t ≤ 5,
if 5 < t ≤ 6.
65/114
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2008.
Miguel A. Arcones. All rights reserved.
Manual for SOA Exam MLC.
Chapter 3. Life tables.
Section 3.5. Interpolating life tables.
Theorem 7
Under an exponential form for `x+t ,
dx
.
(i) Lx = − log
P∞px
dk
(ii) Tx = k=x − log
pk .
(iii) mx = − log px .
P
◦
dk
(iv) e x = ∞
k=x −`x log pk .
66/114
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2008.
Miguel A. Arcones. All rights reserved.
Manual for SOA Exam MLC.
Chapter 3. Life tables.
Section 3.5. Interpolating life tables.
Theorem 7
Under an exponential form for `x+t ,
dx
.
(i) Lx = − log
P∞px
dk
(ii) Tx = k=x − log
pk .
(iii) mx = − log px .
P
◦
dk
(iv) e x = ∞
k=x −`x log pk .
Proof: (i)
Z
Lx =
1
Z
`x+t dt =
0
1
`x pxt
0
`x pxt 1 `x (px − 1)
dt =
=
log px 0
log px
`x qx
dx
=
=
.
− log px
− log px
67/114
c
2008.
Miguel A. Arcones. All rights reserved.
Manual for SOA Exam MLC.
Chapter 3. Life tables.
Section 3.5. Interpolating life tables.
Theorem 7
Under an exponential form for `x+t ,
dx
.
(i) Lx = − log
P∞px
dk
(ii) Tx = k=x − log
pk .
(iii) mx = − log px .
P
◦
dk
(iv) e x = ∞
k=x −`x log pk .
Proof: (i)
Z
Lx =
1
Z
`x+t dt =
0
1
`x pxt
0
`x qx
dx
=
=
.
− log px
− log px
P
P∞
(ii) Tx = ∞
k=x Lx =
k=x
`x pxt 1 `x (px − 1)
dt =
=
log px 0
log px
dk
− log pk .
68/114
c
2008.
Miguel A. Arcones. All rights reserved.
Manual for SOA Exam MLC.
Chapter 3. Life tables.
Section 3.5. Interpolating life tables.
Theorem 7
Under an exponential form for `x+t ,
dx
.
(i) Lx = − log
P∞px
dk
(ii) Tx = k=x − log
pk .
(iii) mx = − log px .
P
◦
dk
(iv) e x = ∞
k=x −`x log pk .
Proof: (i)
Z
Lx =
1
Z
`x+t dt =
0
1
`x pxt
0
`x qx
dx
=
=
.
− log px
− log px
P
P∞
(ii) Tx = ∞
k=x Lx =
k=x
(iii) mx = Ldxx = − log px .
`x pxt 1 `x (px − 1)
dt =
=
log px 0
log px
dk
− log pk .
69/114
c
2008.
Miguel A. Arcones. All rights reserved.
Manual for SOA Exam MLC.
Chapter 3. Life tables.
Section 3.5. Interpolating life tables.
Theorem 7
Under an exponential form for `x+t ,
dx
.
(i) Lx = − log
P∞px
dk
(ii) Tx = k=x − log
pk .
(iii) mx = − log px .
P
◦
dk
(iv) e x = ∞
k=x −`x log pk .
Proof: (i)
Z
Lx =
1
Z
`x+t dt =
0
1
`x pxt
0
`x pxt 1 `x (px − 1)
dt =
=
log px 0
log px
`x qx
dx
=
=
.
− log px
− log px
P
P∞
dk
(ii) Tx = ∞
k=x Lx =
k=x − log pk .
(iii) mx = Ldxx = − log px .
P
◦
dk
(iv) e x = T`xx = ∞
k=x −`x log pk .
70/114
c
2008.
Miguel A. Arcones. All rights reserved.
Manual for SOA Exam MLC.
Chapter 3. Life tables.
Section 3.5. Interpolating life tables.
Example 6
Consider the life table
x
`x
80
250
81
217
82
161
83
107
84
62
85
28
86
0
Assuming exponential interpolation,
71/114
c
2008.
Miguel A. Arcones. All rights reserved.
Manual for SOA Exam MLC.
Chapter 3. Life tables.
Section 3.5. Interpolating life tables.
Example 6
Consider the life table
x
`x
80
250
81
217
82
161
83
107
84
62
85
28
86
0
Assuming exponential interpolation,
◦
R(i)∞ calculate the complete expected life at 80 using that e x =
0 t px dt.
72/114
c
2008.
Miguel A. Arcones. All rights reserved.
Manual for SOA Exam MLC.
Chapter 3. Life tables.
Section 3.5. Interpolating life tables.
Example 6
Consider the life table
x
`x
80
250
81
217
82
161
83
107
84
62
85
28
86
0
Assuming exponential interpolation,
◦
R(i)∞ calculate the complete expected life at 80 using that e x =
0 t px dt.
Solution: (i) By Example 5,

250 217 t

if 0 ≤ t ≤ 1,

250 250 


217 161 t−1

if 1 ≤ t ≤ 2,

250 217


 161 107 t−2 if 2 ≤ t ≤ 3,
250 161 t p80 =
107 62 t−3

if 3 ≤ t ≤ 4,

250 107


t−4

62
28

if 4 ≤ t ≤ 5,
 250 62


0
if 5 < t ≤ 6.
73/114
c
2008.
Miguel A. Arcones. All rights reserved.
Manual for SOA Exam MLC.
Chapter 3. Life tables.
Section 3.5. Interpolating life tables.
Example 6
Consider the life table
x
`x
80
250
81
217
82
161
83
107
84
62
85
28
86
0
Assuming exponential interpolation,
◦
R(i)∞ calculate the complete expected life at 80 using that e x =
0 t px dt.
Solution: (i) So,
∞
Z
◦
ex =
Z
t px dt
0
Z
3
+
2
Z
+
4
5
161
250
1
=
0
t−2
Z 2
217 t
217 161 t−1
dt +
dt
250
217
1 250
Z 4
107 62 t−3
dt +
dt
107
3 250
250
250
107
161
t−4
62 28
dt = 2.712484924.
250 62
74/114
c
2008.
Miguel A. Arcones. All rights reserved.
Manual for SOA Exam MLC.
Chapter 3. Life tables.
Section 3.5. Interpolating life tables.
Example 6
Consider the life table
x
`x
80
250
81
217
82
161
83
107
84
62
85
28
86
0
Assuming exponential interpolation,
◦
(ii) calculate the complete expected life at 80 using that e x =
P
∞
dk
k=x −`x log pk .
75/114
c
2008.
Miguel A. Arcones. All rights reserved.
Manual for SOA Exam MLC.
Chapter 3. Life tables.
Section 3.5. Interpolating life tables.
Example 6
Consider the life table
x
`x
80
250
81
217
82
161
83
107
84
62
85
28
86
0
Assuming exponential interpolation,
◦
(ii) calculate the complete expected life at 80 using that e x =
P
∞
dk
k=x −`x log pk .
Solution: (ii) Using that dx = `x − `x+1 and px =
x
`x
dx
px
80
250
33
81
217
56
82
161
54
83
107
45
84
62
34
217
250
161
217
107
161
62
107
28
62
85
28
28
0
`x+1
`x ,
we get that
86
0
0
0
76/114
c
2008.
Miguel A. Arcones. All rights reserved.
Manual for SOA Exam MLC.
Chapter 3. Life tables.
Section 3.5. Interpolating life tables.
Example 6
Consider the life table
x
`x
80
250
81
217
82
161
83
107
84
62
85
28
86
0
Assuming exponential interpolation,
◦
(ii) calculate the complete expected life at 80 using that e x =
P
∞
dk
k=x −`x log pk .
Solution: (ii) Hence,
◦
e 80 =
∞
X
k=80
=
33
250 − log
dk
−`x log pk
56
54
+
161
250 − log 217
250 − log 107
161
45
34
+
= 2.712484924.
+
62
250 − log 107
250 − log 28
62
217
250
+
77/114
c
2008.
Miguel A. Arcones. All rights reserved.
Manual for SOA Exam MLC.
Chapter 3. Life tables.
Section 3.5. Interpolating life tables.
Example 6
Consider the life table
x
`x
80
250
81
217
82
161
83
107
84
62
85
28
86
0
Assuming exponential interpolation,
◦
(iii) calculate e 80:3| .
78/114
c
2008.
Miguel A. Arcones. All rights reserved.
Manual for SOA Exam MLC.
Chapter 3. Life tables.
Section 3.5. Interpolating life tables.
Example 6
Consider the life table
x
`x
80
250
81
217
82
161
83
107
84
62
85
28
86
0
Assuming exponential interpolation,
◦
(iii) calculate e 80:3| .
Solution: (iii)
◦
e 80:3| =
82
X
k=80
33
=
250 − log
dk
−`80 log pk
217
250
+
56
250 − log
161
217
+
54
250 − log
107
161
=2.211545729.
79/114
c
2008.
Miguel A. Arcones. All rights reserved.
Manual for SOA Exam MLC.
Chapter 3. Life tables.
Section 3.5. Interpolating life tables.
Harmonic interpolation.
1
, for
Under harmonic interpolation, we assume that `x+t = a+tb
1
0 ≤ t ≤ 1, where a and b depend on x. This implies that `x+t is a
linear function. Hence,
1
1
1
= (1 − t) + t
`x+t
`x
`x+1
and
`x+t =
1
(1 −
t) `1x
1
+ t `x+1
.
(3)
1
A function of the form a+bx
is called a hyperbolic function.
Harmonic interpolation of the number of living is also called the
hyperbolic form of the number of living. If the number of living
follows harmonic interpolation, we say that it satisfies the
Balducci assumption.
80/114
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2008.
Miguel A. Arcones. All rights reserved.
Manual for SOA Exam MLC.
Chapter 3. Life tables.
Section 3.5. Interpolating life tables.
Theorem 8
Under the Balducci assumption for `x+t ,
px
1−qx
(i) t px = t+(1−t)p
= 1−(1−t)q
, 0 ≤ t ≤ 1.
x
x
tqx
1−(1−t)qx , 0 ≤ t ≤ 1.
qx
1−px
= 1−(1−t)q
.
µx+t = t+(1−t)p
x
x
px (1−px )
qx (1−qx )
fT (x) (t) = (t+(1−t)p
2 = (1−(1−t)q )2 ,
x)
x
(ii) t qx =
(iii)
(iv)
0 ≤ t ≤ 1.
81/114
c
2008.
Miguel A. Arcones. All rights reserved.
Manual for SOA Exam MLC.
Chapter 3. Life tables.
Section 3.5. Interpolating life tables.
Theorem 8
Under the Balducci assumption for `x+t ,
px
1−qx
(i) t px = t+(1−t)p
= 1−(1−t)q
, 0 ≤ t ≤ 1.
x
x
tqx
1−(1−t)qx , 0 ≤ t ≤ 1.
qx
1−px
= 1−(1−t)q
.
µx+t = t+(1−t)p
x
x
px (1−px )
qx (1−qx )
fT (x) (t) = (t+(1−t)p
2 = (1−(1−t)q )2 ,
x)
x
(ii) t qx =
(iii)
(iv)
0 ≤ t ≤ 1.
Proof: (i) We have that
t px
=
1
px
`x+t
1
=
=
.
=
1
`
x
`x
t + (1 − t)px
(1 − t) + t px
(1 − t) + t `x+1
and
px
1 − qx
1 − qx
=
=
.
t + (1 − t)px
t + (1 − t)(1 − qx )
1 − (1 − t)qx
82/114
c
2008.
Miguel A. Arcones. All rights reserved.
Manual for SOA Exam MLC.
Chapter 3. Life tables.
Section 3.5. Interpolating life tables.
Theorem 8
Under the Balducci assumption for `x+t ,
px
1−qx
(i) t px = t+(1−t)p
= 1−(1−t)q
, 0 ≤ t ≤ 1.
x
x
tqx
1−(1−t)qx , 0 ≤ t ≤ 1.
qx
1−px
= 1−(1−t)q
.
µx+t = t+(1−t)p
x
x
px (1−px )
qx (1−qx )
fT (x) (t) = (t+(1−t)p
2 = (1−(1−t)q )2 ,
x)
x
(ii) t qx =
(iii)
(iv)
0 ≤ t ≤ 1.
Proof: (ii)
t
qx = 1 − t px = 1 −
tqx
1 − qx
=
.
1 − (1 − t)qx
1 − (1 − t)qx
83/114
c
2008.
Miguel A. Arcones. All rights reserved.
Manual for SOA Exam MLC.
Chapter 3. Life tables.
Section 3.5. Interpolating life tables.
Theorem 8
Under the Balducci assumption for `x+t ,
px
1−qx
(i) t px = t+(1−t)p
= 1−(1−t)q
, 0 ≤ t ≤ 1.
x
x
tqx
1−(1−t)qx , 0 ≤ t ≤ 1.
qx
1−px
= 1−(1−t)q
.
µx+t = t+(1−t)p
x
x
px (1−px )
qx (1−qx )
fT (x) (t) = (t+(1−t)p
2 = (1−(1−t)q )2 ,
x)
x
(ii) t qx =
(iii)
(iv)
0 ≤ t ≤ 1.
Proof: (iii) We have that
d
px
d
log t px = − log
dt
dt
t + (1 − t)px
d
1 − px
= log(t + (1 − t)px ) =
,
dt
t + (1 − t)px
d
d
1 − qx
µx+t = − log t px = − log
dt
dt
1 − (1 − t)qx
d
qx
= log(1 − (1 − t)qx ) =
.
dt
1 − (1 − t)qx
µx+t = −
84/114
c
2008.
Miguel A. Arcones. All rights reserved.
Manual for SOA Exam MLC.
Chapter 3. Life tables.
Section 3.5. Interpolating life tables.
Theorem 8
Under the Balducci assumption for `x+t ,
px
1−qx
(i) t px = t+(1−t)p
= 1−(1−t)q
, 0 ≤ t ≤ 1.
x
x
tqx
1−(1−t)qx , 0 ≤ t ≤ 1.
qx
1−px
= 1−(1−t)q
.
(iii) µx+t = t+(1−t)p
x
x
px (1−px )
qx (1−qx )
(iv) fT (x) (t) = (t+(1−t)p
2 = (1−(1−t)q )2 , 0 ≤ t ≤ 1.
x)
x
px (1−px )
qx (1−qx )
Proof: (iv) fT (x) (t) = t px µx+t = (t+(1−t)px )2 = (1−(1−t)q
2.
x)
(ii) t qx =
85/114
c
2008.
Miguel A. Arcones. All rights reserved.
Manual for SOA Exam MLC.
Chapter 3. Life tables.
Section 3.5. Interpolating life tables.
Theorem 9
Under the Balducci assumption,
1−t qx+t
= (1 − t)qx .
86/114
c
2008.
Miguel A. Arcones. All rights reserved.
Manual for SOA Exam MLC.
Chapter 3. Life tables.
Section 3.5. Interpolating life tables.
Theorem 9
Under the Balducci assumption,
1−t qx+t
= (1 − t)qx .
Proof: We have that
s(x + t) − s(x + 1)
px
=1−
s(x + t)
t px
t + (1 − t)px
= (1 − t)qx .
=1 − px
px
1−t qx+t
=
87/114
c
2008.
Miguel A. Arcones. All rights reserved.
Manual for SOA Exam MLC.
Chapter 3. Life tables.
Section 3.5. Interpolating life tables.
Example 7
Using the life table in Section 4.1 and harmonic interpolation, find:
(i) 0.75 p80
(ii) 2.25 p80 .
88/114
c
2008.
Miguel A. Arcones. All rights reserved.
Manual for SOA Exam MLC.
Chapter 3. Life tables.
Section 3.5. Interpolating life tables.
Example 7
Using the life table in Section 4.1 and harmonic interpolation, find:
(i) 0.75 p80
(ii) 2.25 p80 .
Solution: (i) We have that
0.75 p80
=
`81
`80
81
0.75 + (1 − 0.75) ``80
=
50987
53925
0.75 + (0.25) 50987
53925
= 0.9585734295.
(ii) We have that
`
2.25 p80
= p80 p81 · 0.25 p82
83
`81 `82
`82
=
`80 `81 0.25 + (1 − 0.25) ``83
82
=
44803
47940
50987 47940
= 0.8737185905.
53925 50987 0.25 + (0.75) 44803
47940
89/114
c
2008.
Miguel A. Arcones. All rights reserved.
Manual for SOA Exam MLC.
Chapter 3. Life tables.
Section 3.5. Interpolating life tables.
Theorem 10
Given t ≥ 0, let k be the nonnegative integer such that
k ≤ t < k + 1. Under the Balducci assumption
pk
(i) s(t) = ``k0 1−(1−t+k)(1−p
.
k)
pk (1−pk )
`k
`0 (1−(1−t+k)(1−pk ))2 .
px+k (1−px+k )
= k px · (1−(1−t+k)(1−p
2,
x+k ))
(ii) fX (t) =
(iii) fTx (t)
0 ≤ t ≤ 1.
90/114
c
2008.
Miguel A. Arcones. All rights reserved.
Manual for SOA Exam MLC.
Chapter 3. Life tables.
Section 3.5. Interpolating life tables.
Theorem 10
Given t ≥ 0, let k be the nonnegative integer such that
k ≤ t < k + 1. Under the Balducci assumption
pk
(i) s(t) = ``k0 1−(1−t+k)(1−p
.
k)
pk (1−pk )
`k
`0 (1−(1−t+k)(1−pk ))2 .
px+k (1−px+k )
= k px · (1−(1−t+k)(1−p
2,
x+k ))
(ii) fX (t) =
(iii) fTx (t)
0 ≤ t ≤ 1.
Proof: (i) For each integer x and each 0 ≤ t ≤ 1,
s(x + t) =
`x+t
`x
1
`x
px
=
=
`
x
`0
`0 (1 − t) + t `
`0 (1 − t)px + t
x+1
`x
px
=
.
`0 1 − (1 − t)(1 − px )
Hence, for t ≥ 0 and k ≤ t < k + 1,
s(t) = s(k + t − k) =
`k
pk
.
`0 1 − (1 − t + k)(1 − pk )
91/114
c
2008.
Miguel A. Arcones. All rights reserved.
Manual for SOA Exam MLC.
Chapter 3. Life tables.
Section 3.5. Interpolating life tables.
Theorem 10
Given t ≥ 0, let k be the nonnegative integer such that
k ≤ t < k + 1. Under the Balducci assumption
pk
(i) s(t) = ``k0 1−(1−t+k)(1−p
.
k)
pk (1−pk )
`k
`0 (1−(1−t+k)(1−pk ))2 .
px+k (1−px+k )
= k px · (1−(1−t+k)(1−p
2,
x+k ))
(ii) fX (t) =
(iii) fTx (t)
0 ≤ t ≤ 1.
Proof: (ii)
fX (t) = −
d
`k
pk (1 − pk )
s(t) =
.
dt
`0 (1 − (1 − t + k)(1 − pk ))2
92/114
c
2008.
Miguel A. Arcones. All rights reserved.
Manual for SOA Exam MLC.
Chapter 3. Life tables.
Section 3.5. Interpolating life tables.
Theorem 10
Given t ≥ 0, let k be the nonnegative integer such that
k ≤ t < k + 1. Under the Balducci assumption
pk
(i) s(t) = ``k0 1−(1−t+k)(1−p
.
k)
pk (1−pk )
`k
`0 (1−(1−t+k)(1−pk ))2 .
px+k (1−px+k )
= k px · (1−(1−t+k)(1−p
2,
x+k ))
(ii) fX (t) =
(iii) fTx (t)
0 ≤ t ≤ 1.
Proof: (iii)
fX (x + t)
fX (x + k + t − k)
fTx (t) =
=
=
s(x)
s(x)
=k px ·
`x+k
px+k (1−px+k )
`0 (1−(1−t+k)(1−px+k ))2
`x
`0
px+k (1 − px+k )
.
(1 − (1 − t + k)(1 − px+k ))2
93/114
c
2008.
Miguel A. Arcones. All rights reserved.
Manual for SOA Exam MLC.
Chapter 3. Life tables.
Section 3.5. Interpolating life tables.
Example 8
Using harmonic interpolation for the life table
x
`x
80
250
81
217
82
161
83
107
84
62
85
28
86
0
94/114
c
2008.
Miguel A. Arcones. All rights reserved.
Manual for SOA Exam MLC.
Chapter 3. Life tables.
Section 3.5. Interpolating life tables.
Example 8
Using harmonic interpolation for the life table
x
`x
80
250
81
217
82
161
83
107
84
62
85
28
86
0
(i) Calculate `80+t , 0 ≤ t ≤ 6.
95/114
c
2008.
Miguel A. Arcones. All rights reserved.
Manual for SOA Exam MLC.
Chapter 3. Life tables.
Section 3.5. Interpolating life tables.
Example 8
Using harmonic interpolation for the life table
x
`x
80
250
81
217
82
161
83
107
(i) Calculate `80+t , 0 ≤ t ≤ 6.
Solution: (i) Using that `x+t = `x
`80+t =
`x+1
`x

1

1
1

(1−t) 250
+t 217




 (1−(t−1)) 11 +(t−1) 1


217
161


1

1
1
(1−(t−2)) 161 +(t−2) 107
1


1
1

(1−(t−3)) 107
+(t−3) 62



1



 (1−(t−4)) 621 +(t−4) 281


0
84
62
t
85
28
86
0
, 0 ≤ t ≤ 1,
if 0 ≤ t ≤ 1,
if 1 ≤ t ≤ 2,
if 2 ≤ t ≤ 3,
if 3 ≤ t ≤ 4,
if 4 ≤ t ≤ 5,
if 5 < t ≤ 6.
96/114
c
2008.
Miguel A. Arcones. All rights reserved.
Manual for SOA Exam MLC.
Chapter 3. Life tables.
Section 3.5. Interpolating life tables.
Example 8
Using harmonic interpolation for the life table
x
`x
80
250
81
217
82
161
83
107
84
62
85
28
86
0
(ii) Calculate t p80 , 0 ≤ t ≤ 6.
97/114
c
2008.
Miguel A. Arcones. All rights reserved.
Manual for SOA Exam MLC.
Chapter 3. Life tables.
Section 3.5. Interpolating life tables.
Example 8
Using harmonic interpolation for the life table
x
`x
80
250
81
217
82
161
(ii) Calculate t p80 , 0 ≤ t ≤ 6.
Solution: (ii) Using that t px =
t p80
=
83
107
85
28
86
0
`x+t
`x ,

1

250

(1−t) 250
+t 250

217


1


250

(1−(t−1)) 217 +(t−1) 250

161


1

250
250
(1−(t−2)) 161 +(t−2) 107
1



(1−(t−3)) 250
+(t−3) 250

107
62


1


250
250

 (1−(t−4)) 62 +(t−4) 28


0
84
62
if 0 ≤ t ≤ 1,
if 1 ≤ t ≤ 2,
if 2 ≤ t ≤ 3,
if 3 ≤ t ≤ 4,
if 4 ≤ t ≤ 5,
if 5 < t ≤ 6.
98/114
c
2008.
Miguel A. Arcones. All rights reserved.
Manual for SOA Exam MLC.
Chapter 3. Life tables.
Section 3.5. Interpolating life tables.
Example 8
Using harmonic interpolation for the life table
x
`x
80
250
81
217
82
161
83
107
84
62
85
28
86
0
(iii) Calculate the density function of T80 .
99/114
c
2008.
Miguel A. Arcones. All rights reserved.
Manual for SOA Exam MLC.
Chapter 3. Life tables.
Section 3.5. Interpolating life tables.
Example 8
Using harmonic interpolation for the life table
x
`x
80
250
81
217
82
161
83
107
84
62
(iii) Calculate the density function of T80 .
t px )
Solution: (iii) Using that fT80 (t) = − d(dt
,

250
− 250

217
250

2

250

(1−t) 250 +t 250
(

217 )

250
250

− 217

161


250 2
250


((1−(t−1)) 217 +(t−1) 161
)


250
250

− 161

107
250
250 2
+(t−2) 107
)
fT80 (t) = ((1−(t−2))
161
250

− 250

62
107

2


+(t−3) 250
((1−(t−3)) 250

107
62 )

250

− 250

28
62


2
250

(1−(t−4)) 62 +(t−4) 250

(
28 )


0
c
2008.
Miguel A. Arcones. All rights reserved.
85
28
86
0
if 0 ≤ t ≤ 1,
if 1 ≤ t ≤ 2,
if 2 ≤ t ≤ 3,
if 3 ≤ t ≤ 4,
if 4 ≤ t ≤ 5,
if 5 < t ≤ 6.
Manual for SOA Exam MLC.
100/114
Chapter 3. Life tables.
Section 3.5. Interpolating life tables.
Theorem 11
Under the Balducci assumption for `x+t ,
(i) Lx = −`x+1qxlog px .
P
−`k+1 log pk
(ii) Tx = ∞
.
k=x
qk
2
x
(iii) mx = −pxqlog
px .
P
◦
∞ −`k+1 log pk
.
(iv) e x = k=x
`x qk
101/114
c
2008.
Miguel A. Arcones. All rights reserved.
Manual for SOA Exam MLC.
Chapter 3. Life tables.
Section 3.5. Interpolating life tables.
Theorem 11
Under the Balducci assumption for `x+t ,
(i) Lx = −`x+1qxlog px .
P
−`k+1 log pk
(ii) Tx = ∞
.
k=x
qk
2
x
(iii) mx = −pxqlog
px .
P
◦
∞ −`k+1 log pk
.
(iv) e x = k=x
`x qk
Proof: (i)
Z
Lx =
1
Z
`x+t dt =
0
0
1
1
(1 −
t) `1x
1
+ t `x+1
dt
1
1
1
log (1 − t) `1x + t `x+1
− log `1x
log `x+1
=
=
1
1
1
1
0
`x+1 − `x
`x+1 − `x
=
`x
`x `x+1 log `x+1
`x − `x+1
=
−`x+1 log px
.
qx
102/114
c
2008.
Miguel A. Arcones. All rights reserved.
Manual for SOA Exam MLC.
Chapter 3. Life tables.
Section 3.5. Interpolating life tables.
Theorem 11
Under the Balducci assumption for `x+t ,
(i) Lx = −`x+1qxlog px .
P
−`k+1 log pk
(ii) Tx = ∞
.
k=x
qk
2
x
(iii) mx = −pxqlog
px .
P
◦
∞ −`k+1 log pk
.
(iv) e x = k=x
`x qk
P
P∞
Proof: (ii) Tx = k=x Lx = ∞
k=x
−`k+1 log pk
.
qk
103/114
c
2008.
Miguel A. Arcones. All rights reserved.
Manual for SOA Exam MLC.
Chapter 3. Life tables.
Section 3.5. Interpolating life tables.
Theorem 11
Under the Balducci assumption for `x+t ,
(i) Lx = −`x+1qxlog px .
P
−`k+1 log pk
(ii) Tx = ∞
.
k=x
qk
2
x
(iii) mx = −pxqlog
px .
P
◦
∞ −`k+1 log pk
.
(iv) e x = k=x
`x qk
Proof: (iii) mx =
dx
Lx
=
dx qx
−`x+1 log px
=
`x dx qx
−`x+1 `x log px
=
qx2
−px log px .
104/114
c
2008.
Miguel A. Arcones. All rights reserved.
Manual for SOA Exam MLC.
Chapter 3. Life tables.
Section 3.5. Interpolating life tables.
Theorem 11
Under the Balducci assumption for `x+t ,
(i) Lx = −`x+1qxlog px .
P
−`k+1 log pk
(ii) Tx = ∞
.
k=x
qk
2
x
(iii) mx = −pxqlog
px .
P
◦
∞ −`k+1 log pk
.
(iv) e x = k=x
`x qk
P
◦
Proof: (iv) e x = T`xx = ∞
k=x
−`k+1 log pk
.
`x qk
105/114
c
2008.
Miguel A. Arcones. All rights reserved.
Manual for SOA Exam MLC.
Chapter 3. Life tables.
Section 3.5. Interpolating life tables.
Example 9
Use harmonic interpolation for the life table
x
`x
80
250
81
217
82
161
83
107
84
62
85
28
86
0
106/114
c
2008.
Miguel A. Arcones. All rights reserved.
Manual for SOA Exam MLC.
Chapter 3. Life tables.
Section 3.5. Interpolating life tables.
Example 9
Use harmonic interpolation for the life table
x
`x
◦
80
250
81
217
82
161
◦
(i) Calculate e 80 using that e x =
83
107
R∞
0
84
62
85
28
86
0
t px dt.
107/114
c
2008.
Miguel A. Arcones. All rights reserved.
Manual for SOA Exam MLC.
Chapter 3. Life tables.
Section 3.5. Interpolating life tables.
Example 9
Use harmonic interpolation for the life table
x
`x
80
250
81
217
82
161
83
107
84
62
R∞
◦
◦
(i) Calculate e 80 using that e x = 0 t px dt.
Solution: (i) By Example 5,

1

if
250

+t 250
(1−t) 250

217


1


if

(1−(t−1)) 250
+(t−1) 250

217
161


1

if
(1−(t−2)) 250
+(t−2) 250
161
107
p
=
t 80
1

if


+(t−3) 250
 (1−(t−3)) 250
107
62


1

if

 (1−(t−4)) 250
+(t−4) 250

62
28


0
if
85
28
86
0
0 ≤ t ≤ 1,
1 ≤ t ≤ 2,
2 ≤ t ≤ 3,
3 ≤ t ≤ 4,
4 ≤ t ≤ 5,
5 < t ≤ 6.
108/114
c
2008.
Miguel A. Arcones. All rights reserved.
Manual for SOA Exam MLC.
Chapter 3. Life tables.
Section 3.5. Interpolating life tables.
Example 9
Use harmonic interpolation for the life table
x
`x
80
250
81
217
◦
82
161
◦
(i) Calculate e 80 using that e x =
Z
◦
e 80 =
0
Z
4
+
3
+
4
+
t 250
217
250
2)) 161
84
62
85
28
2
dt +
1
+ (t − 2) 250
107
86
0
t px dt.
0
1
(1 − (t −
2
Z
(1 −
t) 250
250
R∞
Z
1
3
+
Z
1
83
107
1
(1 − (t −
1)) 250
217
+ (t − 1) 250
161
dt
dt
1
dt
250
(1 − (t − 3)) 107
+ (t − 3) 250
62
5
1
(1 − (t −
4)) 250
62
+ (t − 4) 250
28
c
2008.
Miguel A. Arcones. All rights reserved.
dt = 2.681292266.
Manual for SOA Exam MLC.
109/114
Chapter 3. Life tables.
Section 3.5. Interpolating life tables.
Example 9
Use harmonic interpolation for the life table
x
`x
◦
80
250
81
217
82
161
◦
(ii) Calculate e 80 using that e x =
83
107
P∞
k=x
84
62
85
28
86
0
−`k+1 log pk
.
`x qk
110/114
c
2008.
Miguel A. Arcones. All rights reserved.
Manual for SOA Exam MLC.
Chapter 3. Life tables.
Section 3.5. Interpolating life tables.
Example 9
Use harmonic interpolation for the life table
x
`x
80
250
81
217
◦
82
161
◦
(ii) Calculate e 80 using that e x =
83
107
P∞
k=x
84
62
85
28
−`k+1 log pk
.
`x qk
Solution: (ii) Using that dx = `x − `x+1 , px =
`x −`x+1
,
`x
86
0
`x+1
`x
and qx =
we get that
x
`x
dx
px
qx
80
250
33
81
217
56
82
161
54
83
107
45
84
62
34
217
250
33
250
161
217
56
217
107
161
54
161
62
107
45
107
28
62
34
62
85
28
28
0
0
86
0
0
0
0
111/114
c
2008.
Miguel A. Arcones. All rights reserved.
Manual for SOA Exam MLC.
Chapter 3. Life tables.
Section 3.5. Interpolating life tables.
Example 9
Use harmonic interpolation for the life table
x
`x
◦
80
250
81
217
82
161
◦
(ii) Calculate e 80 using that e x =
Solution: (ii)Hence,
83
107
P∞
k=x
84
62
85
28
86
0
−`k+1 log pk
.
`x qk
∞
X
−`k+1 log pk
e 80 =
`80 qk
k=80
107
−(217) log 217
−(161) log 161
−(107) log 161
250
217
+
+
=
33
56
54
(250) 250
(250) 217
(250) 161
62
−(62) log 107
−(28) log 28
62
+
= 2.681292266.
+
45
34
(250) 107
(250) 62
◦
112/114
c
2008.
Miguel A. Arcones. All rights reserved.
Manual for SOA Exam MLC.
Chapter 3. Life tables.
Section 3.5. Interpolating life tables.
Example 9
Use harmonic interpolation for the life table
x
`x
80
250
81
217
82
161
83
107
84
62
85
28
86
0
◦
(iii) Calculate e 80:3| .
113/114
c
2008.
Miguel A. Arcones. All rights reserved.
Manual for SOA Exam MLC.
Chapter 3. Life tables.
Section 3.5. Interpolating life tables.
Example 9
Use harmonic interpolation for the life table
x
`x
80
250
81
217
82
161
83
107
84
62
85
28
86
0
◦
(iii) Calculate e 80:3| .
Solution: (iii)
82
161
X
−(217) log 217
−(161) log 217
−`k+1 log pk
250
+
e 80:3| =
=
33
56
`80 qk
(250)
(250)
250
217
k=80
107
−(107) log 161
+
54
(250) 161
◦
=2.197149575.
114/114
c
2008.
Miguel A. Arcones. All rights reserved.
Manual for SOA Exam MLC.