2. Consider a lifetime random variable W with p.d.f. f(w) and survival function S(w). We shall use the integration by parts equation w f(w) dwE(W) = 0.

2. Consider a lifetime random variable W with p.d.f. f(w) and survival function S(w).

We shall use the integration by parts equation

w f(w) dwE(W) =0

(a) Use integration by parts to find a formula for E(W) which is an alternative to the basic definition.

The basic definition is

u v / = u v – u / v

with u = w and v / = f(w). u / = and v =1 – S(w)

w f(w) dw = – w S(w) – (1)[– S(w)]dw =

– w S(w) + S(w)dw

w f(w) dw =E(W) =0

– w S(w) + S(w)dw =

0

S(w)dw0

S(w)dw0

E(W) =

We shall use the integration by parts equation

w2 f(w) dwE(W 2) =0

(b) Use integration by parts to find a formula for E(W 2) which is an alternative to the basic definition.

The basic definition is

u v / = u v – u / v

with u = w2 and v / = f(w). u / = and v =2w – S(w)

w2 f(w) dw = – w2 S(w) – 2w[– S(w)]dw =

– w2 S(w) + 2w S(w)dw

w2 f(w) dw =E(W 2) =0

– w2 S(w) + 2w S(w)dw =

0

2w S(w)dw0

2 w S(w)dw0

E(W 2) =

Next, we shall use the integration by parts equation

u v / = u v – u / v

with u = w and v / = S(w). u / = and v =1

2 w S(w)dw

E(W 2) =0

– S(y)dyw

w S(w) dw = – w S(y)dy – (1) dw =w

– S(y)dyw

– w + dwS(y)dyw

S(y)dyw

2 w S(w)dw =0

E(W 2) =

– w + dw =S(y)dyw

S(y)dyw

w = 0

2

dwS(y)dyw

0

E(W 2) = 2 dwS(y)dyw

0

2

Let X be a lifetime random variable with survival function S(x). Let Kx denote the time interval of failure for (x), and let K(x) denote the curtate duration of failure, that is, Kx = K(x) + 1.

k = 0

ex = E[K(x)] = k P[K(x) = k] = k p q =k = 0

k x x + k

k = 0

k p (1 – p ) =k x x + k

k = 0

p – p p =k x x + kk x

k = 0

p – p =k x k + 1 x

k = 0

p – p =k x k + 1 x

k = 0

k

k k k

k = 0

p – p =k x k + 1 x

k = 0

k (k + 1 – 1)

k = 0

p – p +k x k + 1 x

k = 0

k (k + 1) p =k + 1 x

k = 0

p =k + 1 x

k = 0

E[K(x)2] = k = 0

k2 P[K(x) = k] = k2 p q =k = 0

k x x + k

k = 0

k2 p (1 – p ) =k x x + k

k = 0

p – p p =k x x + kk x

k2

k = 1

k xp

k = 0

p – p =k x k + 1 x

k = 0

p – p =k x k + 1 x

k = 0

k2 k2 k2

k = 0

p – p =k x k + 1 x

k = 0

k2 (k + 1)2 – (2k + 1)

k = 0

p – p +k x k + 1 x

k = 0

k2 (k + 1)2 p =k + 1 x

k = 0

(2k + 1)

k = 0

p =k + 1 x

(2k + 1) k = 1

pk x

(2k – 1)

Var[K(x)] = k = 1

p – ex2

k x(2k – 1)

ex : n =

k = 0

n k P[K(x) = k] = k p q + n p =k = 0

n – 1

k x x + k n x

k = 0

k p (1 – p ) +k x x + k

k = 0

p – p p +k x x + kk x

k

n – 1

n – 1

n p =

n p =

k = 0

p – p +k x k + 1 x

k n p =n – 1

n x

n x

n x

k = 0

p – p +k x k + 1 x

k = 0

k k n p =

k = 0

p – p +k x k + 1 x

k = 0

k (k + 1 – 1)

n – 1 n – 1

n p =

k = 0

p – p +k x k + 1 x

k = 0

k (k + 1) p +k + 1 x

k = 0

p =k + 1 x

k = 0

k = 1 k x

p

n – 1 n – 1

n – 1 n – 1 n – 1

n xn p =

n – 1 n

n x

n x

2. - continued

i = 0

m

ti Δlx + t + lx + t where Δlx + t = i m i

lx + t – lx + ti – 1 i

limm

i = 0

m

ti Δlx + t + lx + t = i m

The expected number of years lived from age x to age x + 1 is

i = 0

m

ti Δti + lx + t = m

Δlx + t—— Δti

i

m lim

0

1

t dt + lx + 1 = d— lx + tdt

0

1

t dt + lx + 1 = d— l0s(x + t)dt

0

1

t dt + lx + 1 = s/(x + t)

l0s(x + t) ——— s(x + t)

0

1

t lx + t (x + t) dt + lx + 1

Section 3.6 (Uniform Assumption Only)1.

(a)

Use Table 3.3.1 with the uniform assumption to find each of the probabilities.

P(13.4 < X 29.8)

= s(13.4) – s(29.8) =

[(1 – 0.4)s(13) + (0.4)s(14)] –[(1 – 0.8)s(29) + (0.8)s(30)] =

[(1 – 0.4)(0.98285) + (0.4)(0.98248)] – [(1 – 0.8)(0.96604) + (0.8)(0.96477)] =

0.982702 – 0.965024 = 0.017678

(b) P[13.4 < T(10) 29.8]

s(23.4) – s(39.8)= ——————— =

s(10)

[(1 – 0.4)(0.97370) + (0.4)(0.97240)] – [(1 – 0.8)(0.95129) + (0.8)(0.94926)]

0.98347

= 0.023909

Section 3.81.

(a)

(b)

Example 3.8.1 makes use of formulas from Section 3.4 which we do not cover. Instead of using these formulas to get the answers, use probability identities by completing the following:

2 p

[30] = p p =

[30]

[30] + 1

(1 – 0.000222)(1 – 0.000330) = 0.999448

5 p

[30] = p p p p p =

[30]

[30] + 1 [30] + 2 [30] + 3 [30] + 4

(1–0.000222)(1–0.000330)(1–0.000422) (1–0.000459)(1–0.000500) =

0.998068

(c)

(d)

1| q

[31] =

3 q

[31] + 1 =

p q =[31]

[31] + 1

(1 – 0.000234)(0.000352) = 0.000344

1 – p =3 [31] + 1

1 – p p p =

[31] + 1

[31] + 2 [31] + 3

1 – (1 – 0.000352)(1 – 0.000459)(1 – 0.000500) = 0.0001310