Mathematics of life contingencies 1 - Multiple life life ...

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

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.


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.


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))


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

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

∫ ∞

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, . . . .


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 ,


Note.

Under independence,

P[K (x : y) = k ] = kpx · kpy(1 − px+kpy+k ).


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

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).


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

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.


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.


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.


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, . . . .


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.


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?


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.


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.


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.


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.


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.


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

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,


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.


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.


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.


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:


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)]) ,



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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