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We have already seen, e.g., the simple joint life status (x : n )along with the corresponding future life time r.v.’s T (x : n ) andK (x : n). We shall generalize the notion to the case when T (n
is not a degenerate r.v. We shall thus have T (x : y).
Definition 0.1 (Future lifetime of the joint life status)
The r.v. T (x : y) having an (absolutely) continuous c.d.f. isgiven by
T (x : y) := min(T (x), T (y)) ={
T (x), T (x) ≤ T (y)T (y), T (x) > T (y)
,
withtpx;y = P[T (x) ≥ t ,T (y) ≥ t], t ≥ 0,
and
tqx:y = tqx + tqy − P[T (x) < t ,T (y) < t], t ≥ 0.
Edward Furman Mathematics of life contingencies 1 2 / 30
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Note.
For independent T (x) and T (y), we certainly have that
tpx;y = tpx · tpy = 1 − tqx − tqy + tqx · tqy , t ≥ 0,
andtqx:y = tqx + tqy − tqx · tqy , t ≥ 0.
Question.
Can we calculate the p.d.f. of T (x : y)? Remember that
tpx;yµ(x : y + t) =ddt tqx;y = −
ddt tpx:y
ind= −
ddt tpx · tpx .
Proposition 0.1
Let T (x) and T (y) be independent future lifetimes of (x) and(y). Then
tpx;yµ(x : y + t) = tpx · tpy (µ(x + t) + µ(y + t)) , t > 0.
Edward Furman Mathematics of life contingencies 1 3 / 30
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Proof.
We have that
ddt
(tpx · tpy) = tp′x tpy + tpx · tp′
y
= −tpxµ(x + t)tpy − tpyµ(y + t)tpx
= −tpx · tpy(µ(x + t) + µ(y + t)),
which completes the proof.
Note.
It readily follows that
µ((x : y) + t) ind= µ(x + t) + µ(y + t), t > 0.
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Proposition 0.2
Let T (x) and T (y) be the future lifetimes of (x) and (y). Then
tpx:yµ((x : y)+t) =∫ ∞
tfT (x),T (y)(t , v)dv+
∫ ∞
tfT (x),T (y)(u, t)du.
Leibniz integral rule
Under certain continuity assumptions, we have that
ddt
∫ β(t)
α(t)ψ(t , v)dv =
ddtϕ(α(t), β(t), t)
=
∫ β(t)
α(t)
ddtψ(t , v)dv +
dβ(t)dt
ψ(t , β(t)) −dα(t)
dtψ(t , α(t))
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Proof of Proposition 0.2.
We are interested in the derivative
fT (x:y)(t) = −ddt
F T (x:y)(t) = −ddt
P[T (x) ≥ t ,T (y) ≥ t], t ≥ 0.
Let
ψ(t , v) =∫ ∞
tfT (x),T (y)(u, v)du,
we are then interested in the derivative of
F T (x),T (y)(t , t) =∫ ∞
t
∫ ∞
tfT (x),T (y)(u, v)dudv =
∫ ∞
tψ(t , v)dv .
So we can use Leibniz rule with β(t) = ∞ and α(t) = t , andhence,
dβ(t)dt
ψ(t , β(t)) = 0,dα(t)
dtψ(t , α(t)) =
∫ ∞
tfT (x),T (y)(u, t)du.
Also, using the Leibniz rule again or just due to the FTC,Edward Furman Mathematics of life contingencies 1 6 / 30
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Proof.
ddtψ(t , v) =
ddt
∫ ∞
tfT (x),T (y)(u, v)du = −fT (x),T (y)(t , v)
Hence
ddt
F T (x),T (y)(t , t)
= −
(∫ ∞
tfT (x),T (y)(t , v)dv +
∫ ∞
tfT (x),T (y)(u, t)du
)
,
that completes the proof.
Note.
In the case of independent future lifetimes, we have that∫ ∞
tfT (x),T (y)(t , v)dv = fT (x)(t)
∫ ∞
tfT (y)(v)dv = fT (x)(t)F T (y)(t)
and also thatEdward Furman Mathematics of life contingencies 1 7 / 30
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∫ ∞
tfT (x),T (y)(u, t)du = fT (y)(t)
∫ ∞
tfT (x)(u)du = fT (y)(t)F T (x)(t),
which gives
tpxµ(x + t)t py + tpyµ(y + t)t px = tpx tpy (µ(x + t) + µ(y + t)) ,
i.e., the p.d.f. of T (x : y) under independence.
Question.
What happens with K (x) and K (y)?
Proposition 0.3
The p.n.f. of K (x : y) is
P[K (x : y) = k ] = kpx:y · qx+k :y+k = k |qx:y , k = 0,1, . . . .
Edward Furman Mathematics of life contingencies 1 8 / 30
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Proof.
We have already proven that for any life status (u), the p.m.f. ofK (u) is
P[K (u) = k ] = kpu · qu+k , k = 0,1, . . . .
Thus, for (u) = (x : y), we have that
P[K (x : y) = k ] = kpx:y · q(x:y)+k , k = 0,1, . . . .
Moreover, in the case of the joint life status
q(x:y)+k = qx+k :y+k ,
which completes the proof.
Note.
Under independence,
P[K (x : y) = k ] = kpx · kpy(1 − px+kpy+k ).
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In other words,
P[K (x : y) = k ] = kpx · kpy (qx+k + qy+k − qx+kqy+k).
Proposition 0.4
The probability that the joint life status will not survive k timeunits is
P[K (x : y) ≤ k ] = k+1qx:y , k = 0,1, . . . .
Note.
The joint life status can be generalized to take into accountmore than two lives. Then we shall have T (x1 : x2 : · · · : xn).
Note.
We have already seen the last survivor life status (x : y). In thecase of two lives we conveniently have that
T (x : y) + T (x : y) = T (x) + T (y).Edward Furman Mathematics of life contingencies 1 10 / 30
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Definition 0.2 (Future lifetime of the last survivor life status)
The r.v. T (x : y) having an (absolutely) continuous c.d.f. isgiven by
T (x : y) := max(T (x), T (y)) ={
T (x), T (x) > T (y)T (y), T (x) ≤ T (y)
,
withtqx:y = P[T (x) < t ,T (y) < t], t ≥ 0,
and
tpx:y = tpx + tpy − P[T (x) ≥ t ,T (y) ≥ t], t ≥ 0.
These can be further specialized under independence of T (x)and T (y).
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Proposition 0.5
Let (x : y) and x : y be the joint and last survivor life statuses,respectively. Then
tqx:y + tqx:y = tqx + tqy , t ≥ 0.
Proof.
Let A := {T (x) < t} and B := {T (y) < t}, t ≥ 0. Then
A ∩ B = {T (x : y) < t}
andA ∪ B = {T (x : y) < t}.
Then employ
P[{A ∪ B}] = P[{A}] + P[{B}]− P[{A ∩ B}]
to complete the proof.Edward Furman Mathematics of life contingencies 1 12 / 30
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Note.
We can use the proposition to obtain, say, tqx:y . Namely,
tqx:y = tqx + tqy − tqx:y
= tqx + tqy − (tqx + tqy − P[T (x) < t ,T (y) < t])
= P[T (x) < t ,T (y) < t], t ≥ 0.
Note.
As tqu = 1 − tpu for any t ≥ 0 and any life status (u), we havethat
tpx:y + tpx:y = tpx + tpy , t ≥ 0.
Hence
tpx:y = tpx + tpy − P[T (x) ≥ t ,T (y) ≥ t], t ≥ 0.
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Proposition 0.6
The force of mortality of the last survivor life status is
µ(x : y + t) = tpxµ(x + t) + tpyµ(y + t)− tpx:yµ((x : y) + t)
tpx:y,
where t > 0.
Proof.
Because of Proposition 0.5, it holds that
tpxµ(x + t)+ tpyµ(y + t) = tpx:yµ((x : y)+ t) + tpx:yµ(x : y + t),
for t > 0. Dividing by tpx:y 6= 0 throughout completes theproof.
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Corollary 0.1
Under independence we have that
µ((x : y) + t) = tpx · tqyµ(x + t) + tpy · tqxµ(y + t)
tpx · tqy + tpy · tqx + tpx · tpy,
where t > 0.
Proof.
The following completes the proof
µ(x : y + t)
=tpxµ(x + t) + tpyµ(y + t)− tpx tpy (µ(x + t) + µ(y + t))
tpx:y
=tpxµ(x + t) + tpyµ(y + t)− tpx tpyµ(x + t)− tpx tpyµ(y + t)
tpx + tpy − tpx · tpy
=tpx(1 − tpy )µ(x + t) + tpy(1 − tpx)µ(y + t)
tqx · tpy + tqy · tpx + tpx · tpy.
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Note.
The relation we had for c.d.f.’s and p.d.f.’s of T (x : y) andT (x : y) does not hold in the case of the force of mortality, i.e.,
µ((x : y) + t) + µ(x : y + t) 6= µ(x + t) + µ(y + t).
Proposition 0.7
The p.m.f. of K (x : y) is
P[K (x : y) = k ] = kpx · qx+k + kpy · qy+k − kpx:yq(x+k :y+k),
for k = 0, 1, . . ..
Proof.
Note that Proposition 0.5 does not assume any continuityrestrictions. Thus
kqx + kqy = kqx:y + kqx:y , k = 0, 1, . . . .
Edward Furman Mathematics of life contingencies 1 16 / 30
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Proof.
As the above is true for any k = 0, 1, . . ., we also have that
k+1qx + k+1qy = k+1qx:y + k+1qx:y .
Then, observing that, for any (u),
k+1qu − kqu = kpu · qu+k ,
we have that
P[K (x : y) = k ] = kpx · qx+k + kpy · qy+k − kpx:yq(x:y)+k ,
which along withq(x:y)+k = qx+k :y+k
completes the proof.
Edward Furman Mathematics of life contingencies 1 17 / 30
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Note.
Under independence and for k = 0, 1, . . ., we have that
P[K (x : y) = k ]
= kpxqx+k + kpyqy+k − kpx · kpy(qx+k + qy+k − qx+kqy+k)
= kpx(1 − kpy )qx+k + kpy (1 − kpx)qy+k + kpx · kpyqx+kqy+k
= kqy · kpxqx+k + kqx · kpyqy+k + kpx · kpyqx+kqy+k .
Question.
Can we find treat the life status (1x : y) in a similar manner?
Edward Furman Mathematics of life contingencies 1 18 / 30
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Definition 0.3 (Future lifetime of an ordered life status)
The r.v. T (1x : y) having an (absolutely) continuous c.d.f. is
given by
T (1x : y) :=
{
T (x), T (x) ≤ T (y)∞, T (x) > T (y)
,
with, for t ≥ 0,
tq1x:y
= P[T (x) < t ,T (x) ≤ T (y)] =∫ t
0
∫ ∞
ufT (x),T (y)(u, s)dsdu.
Proposition 0.8
The p.d.f. of the above r.v. is
fT (
1x :y)
(t) =∫ ∞
tfT (x),T (y)(t , s)ds, t ≥ 0.
Edward Furman Mathematics of life contingencies 1 19 / 30
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Proof.
Let
ψ(u) :=∫ ∞
ufT (x),T (y)(u, s)ds,
and take the derivative
ddt
∫ t
0ψ(u)du = ψ(t) = f
T (1x :y)
(t).
This completes the proof.
Corollary 0.2
Under independence,
tq1x:y
=
∫ t
0spy · spxµ(x + s)ds, t ≥ 0.
Edward Furman Mathematics of life contingencies 1 20 / 30
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Proof.
Indeed, for non-negative t , we have that
tq1x:y
=
∫ t
0
∫ ∞
ufT (x),T (y)(u, s)dsdu
=
∫ t
0
(∫ ∞
ufT (y)|T (x)(s|u)ds
)
fT (x)(u)du
=
∫ t
0P[(T (y) > u|T (x) = u)]upxµ(x + u)du
ind=
∫ t
0upy · upxµ(x + u)du,
as required.
Note.
The p.d.f. is easily obtained as:
fT (
1x :y)
(t) = tpy · tpxµ(x + t), t ≥ 0.
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Corollary 0.3
The c.d.f. of the r.v. T (1x : n ) is
tq1x:n
=
{
tqx , t ≤ nnqx , t > n
and tp1x:n
=
{
tpx , t ≤ nnpx , t > n
.
Proof.
Note that, say, for t > n,
tq1x:n
=
∫ n
0spn · spxµ(x + s)ds +
∫ t
nspn · spxµ(x + s)ds
=
∫ n
0spxµ(x + s)ds = nqx .
The other part follows in the same fashion, and this completesthe proof.
Edward Furman Mathematics of life contingencies 1 22 / 30
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Question.
Is the probability tqx:
2y
the same as tq1x:y
?
Definition 0.4 (Future lifetime of an ordered life status)
The r.v. T (x :2y) having an (absolutely) continuous c.d.f. is
given by
T (x :2y) :=
{
T (y), T (x) < T (y)∞, T (x) ≥ T (y)
,
with
tqx:
2y= P[T (x) ≤ T (y) ≤ t] =
∫ t
0
∫ t
ufT (x),T (y)(u, s)dsdu, t ≥ 0.
Edward Furman Mathematics of life contingencies 1 23 / 30
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Proposition 0.9
The p.d.f. of T (x :2y) is
fT (x:
2y)(t) =
∫ t
0fT (x),T (y)(u, t)du, t ≥ 0.
Proof.
Let
ψ(u, t) :=∫ t
ufT (x),T (y)(u, s)ds.
Then we need to find the derivative
ddt tq
x:2y=
ddt
∫ t
0ψ(u, t)du = ψ(t , t) +
∫ t
0
∂
∂tψ(u, t)du,
which results in
ddt tq
x:2y=
∫ t
0fT (x),T (y)(u, t)du.
and thus completes the proof.Edward Furman Mathematics of life contingencies 1 24 / 30
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Corollary 0.4
Under independence, we have that
nqx:
2y=
∫ n
0spy · spxµ(x + s)ds −n py
∫ n
0spxµ(x + s)ds.,
for n ≥ 0. Also,
npx:
2yµ((x :
2y) + n) = npyµ(y + n) · nqx ,
for n ≥ 0.
Proof.
For the density, the result follows immediately. Further,
Edward Furman Mathematics of life contingencies 1 25 / 30
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Proof.
Under independence,
nqx:
2y
=
∫ n
0
∫ n
sfT (x),T (y)(s, t)dtds
=
∫ n
0
∫ n
sfT (y)|T (x)(t |s)fT (x)(s)dtds
=
∫ n
0P[s < T (y) ≤ n|T (x) = s]fT (x)(s)ds
=
∫ n
0P[s < T (y) ≤ n|T (x) = s]spxµ(x + s)ds
ind=
∫ n
0P[s < T (y) ≤ n]spxµ(x + s)ds
=
∫ n
0(spy −n py )spxµ(x + s)ds,
as required.
Edward Furman Mathematics of life contingencies 1 26 / 30
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Note.
It readily follows that
T (1x : y) ≤ T (x :
2y).
Hence
P[T (1x : y) < t] =t q1
x:y≥ tq
x:2y= P[T (x :
2y) < t], t ≥ 0,
and alsotp1
x:y≤ tp
x:2y, t ≥ 0.
Edward Furman Mathematics of life contingencies 1 27 / 30
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Proposition 0.10
The full expectancy of life is
◦ex:y =
∫ ∞
0tpx:ydt
for the joint life status, and
◦ex:y =
∫ ∞
0tpx:ydt
for the last survivor life status. Also, it holds that
◦ex:y +
◦ex:y =
◦ex +
◦ey .
Proof.
Recalling the formula for◦eu along with the fact that
Tx + Ty = Tx:y + Tx:y completes the proof.
Edward Furman Mathematics of life contingencies 1 28 / 30
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Proposition 0.11
We have that
Var[T (x : y)] = 2∫ ∞
0t tpx:ydt − (
◦ex:y )
2,
and
Var[T (x : y)] = 2∫ ∞
0t tpx:ydt − (
◦ex:y )
2.
Also
Cov[T (x : y),T (x : y)] = Cov[T (x),T (y)]+(◦ex−
◦ex:y )(
◦ey−
◦ex:y ).
Proof.
The formulas for the variances are straight forward. As to thecovariance:
Edward Furman Mathematics of life contingencies 1 29 / 30
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Proof.
Cov[T (x : y),T (x : y)]
= E[T (x : y)T (x : y)]− E[T (x : y)]E[T (x : y)]
= E[T (x)T (y)]− E[T (x : y)]E[T (x : y)]
= E[T (x)T (y)]
− E[T (x : y)] (E[T (x)] + E[T (y)]− E[T (x : y)]) .
Further, adding and subtracting E[T (x)]E[T (y)], we obtain
Cov[T (x : y),T (x : y)]
= Cov[T (x),T (y)]
+ (E[T (x)]− E[T (x : y)]) (E[T (y)]− E[T (x : y)]) ,
which completes the proof.
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Corollary 0.5
If T (x) and T (y) are uncorrelated, then
Cov[T (x : y),T (x : y)] = (◦ex −
◦ex:y )(
◦ey −
◦ex:y ).
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