Demog 110
Crude rate model of growth.
The Balancing Equation
Population today = population last year + changes in the interim
K (2006) = K(2005) + (births – deaths) + (in-migration – out-migration ) + adjustments + errors
Net migration = in-migration – out-migration
In this course we will concern ourselves with briths/deaths mostly and ignore migration,
adjustments, and errors.
K2006 = K2005 + Births (2005) – Deaths (2005)
= K(2005) + [births 2005/K2005 – deaths (2005)/K2005] * K2005
= K(2005) * [ 1 + b2005 – d2005]
= K (t) * [1 –b(t-1) – d(t-1)]
b = births/K(t) = crude birth rate
c = deaths/K(t) = crude death rate
b(t) – d(t) = r(t) = crude rate of natural increase
k(2006) = K(2000) * (1 + r2005) (1+r2004) (1+r2003) (1 + r2002) (1 + r2001) (1 + r2000)
if we assume unchanging r => r2000 = r2001 etc. => K2006 = K(2000) * r6
Critical assumption = unchanging r
K(t) = K(0) * (1 +r)t – an example of geometric growth
K(0) = initial year .
2000: 6.048 Billion
births – deaths = 75 million
r2000 = 75/6048 = .0124 = 1.24% or 12.4 per thousand
K (2001) = K(2001) * (1,0124)2
= K(2000) * (1.0124)6 = 6.512
r can be used for population projections.
K(t) = K(0) (1-r)t = geometric growth rate
K(t) = (K0) ert
Advisory comment:
When do we use geometric?
When do we use exponential?
They are close enough that usually we use the one that is simpler for that particular purpose.
Exponential growth model:
K(t) = K(0) ert
How long does it take for a population to double in size.
K(t) = 2K(0) = K(0)ert
2 = ert
log 2 = log (ert) = rt
log 2 = rt
log 2/r = t doubling
.6931/r = t ~ 70/1.2
.124 = 1.24% 12.4/1000
8000 BC 8million
1AD 300 million
Last time
• Lexis diagrams (lexis surfaces)
• Person years & ??
Today:
• Period years
• Cohort years
• Life expectancy
Lexis Diagram
#1 – China
# 2 – India
# 3 – US
#4 – Indonesia
Person year = unit of exposure
Age
TS Elliot – 1868 – 1965
WB Yates – 1865 – 1939
V Woolf – 1882 – 1941
J Joyce – 1882 – 1941
T Mann – 1875 – 1955
Period 1930 to 1940
10
9
10
10
10
Period 1940 to 1950
10
0
1
4
10
Total Person years
Avg
49
/5 =
22
/4 =
Crude birth rate and crude death rate = # events/ py’s (mid-period population).
So crude birth rate and crude death rate are period measures of mortality.
Cohort measures of mortality:
Count the # of Py’s lived in a cohort.
Divide the sum of the py’s by starting population
Average # of Py’s lived for each member of the cohort =
Cohort average age of death
Cohort expectation of life
We started counting at birth but we could have started ant any age. If we began at age 20, we
could capture the # of person years lived after age 20, divide the sum by the starting population.
We get the expectation of life at age 20.
Problem: You can’t calculate this until all members of a cohort die.
So, what do we do?
Due Tuesday, Sept 12 – 1.9, 2.1 – 5, Additional question chapter 2, table 2.1. Facts about the 10
most populations countries in the world. The 11th is Mexico. Complete the table for Mexico with
the same sources used for the table.
ex = label for the expectation of life.
ex
78
Stationary Population
r>0 Population increases
r< 0 Population decreases
r = 0 Stationary population
For a stationary population (closed migration), then b=d. In any population e0 = expectation of
life at age 0 # Py’s lived /# people born
B= # births/ total population
D = # deaths/ total population
The # of births * e0 = the total # of person years lived by those births.
Mean times are the reciprocal of rates
So if death rate in one population is d, then 1/d = average age of death = e0 = average
expectation of life at birth.
Also in a stationary population b=d, so 1/b = e0 too.
Key relationship: In a stationary population b*e0=1
This is strictly true only for a stationary population but it’s a good approximation for a low
growing, or low declining population. Our population is growing very slowly so this is a good
approx. for the US population.
You can make estimates based on this relationship.
Year
1860
1970
Period
Age
65
=~ 20
Non Cohort Mortality - Chart 3
- function ?
Age - vertical.
Period - horizontal.
Cohort mortality
l20 = # of people age 20
l21 = l20 * probability of surviving to age 21
l22 = l20 (P20) (P21)
P20*P21 = 2P20 ??
In general Px = Probability forom age x to x+n
Observation # 1
Probabilities of surviving multiply
Typical pattern of lx
lx by age gragh = survivorship curve of a cohort. starts at the starting pop and goes to 0, starts
dying slow and then dies faster as the cohort ages, of course.
life table
x
lx
0
10
20
30
40
nPx
* Try to do this in R :)
ndx
nqx
nLx
Tx
ex
lx+n / lx = nPx
lx -lxn + # death s between age x and x+n = n d x
ndx/ lx = probablity of dying between x + x+n = nqx
Observation #2
npx + nqx = 11
npx = 1 - nqx
Mortality probabilty = nqx
mortality rate nmx
graph = close up of the survivorship curve bwtween l20 and l25, In the cohort we are counting
how many person years were lived between age 20 and age 25. And here we have a graph that
shows what percentage of the people who lived between age 20 and age 25.
We can count the cohort person years by counting area below the survivorship curve between l20
and l25.
Integral of age 20 to 25 of the lxdx (curve)
NLx is the number of person years between x and x+n.
Rule of thumb observation:
nLx ~ ~ n.lx or nlx+n
Sarah's office hours from 3 to 4 on Friday and 10 to 11 on Friday every other week thereafter.
... copy notes.
ex = expectation of remaining life at age x.
e0 = Tx/lx = Average # of years of remaining life.
Traditonally, LT was built up around nqx's. However, we often need to convert between different
intervals, n, and working with nqx' is a PITA
l25 = l20 * P20*P21*P22*P23.P24 = l205P20
If survival probs were the same,
5P20 = (P20) to the 5th power,
...
Neeeds notes.
hazard rate = m(x), h(x)
is related to the mortality rate, nmx
is also related to the growth rate in oru crude rate model. The hazard rate tells us how fast the lx
is changing.
In the crude rate model we took the log of the diff in pop divided by the time interval. It will turn
out that the hazard rate will also be the difference??
= - d/dx log lx = m(x) = h(x)
- (log l20 - log l21/ widht of the interval) = 1
In general for small n, the log of ln minus the log of lx+n divided by n is a good approximation
=>
log (lx+n/lx)/-n
The hazard rate curve is commonly u shape.
survivorship usuall goes down fast, becomes flattish and the goes down fast.
Mariana Horta
17978026
1.9 At mid-year 2006, Pakistan, Bangladesh, Russia, and
Nigeria had similar total populations, 166, 147,142, and
135 millions respectively. Suppose current growth rates in
these four countries, namely 0.024, 0.019, -0.006, and
0.024 continue for the next fourteen years to 2020. Use the
exponential model to project future populations, and find
how these countries would rank in population size at the
end of the period:
K(t)= K(0)* e^rt
log(K(t)) = log(K(0)) + (r*t)
K(t)= K(0)^rt
Pakistan
166m
r= 0.024
Bangladesh
Nigeria
147m
135m
r=0.019
Russia
142m
-0.006
0.024
t= 14
t= 14
t=14
K(t)=166m^(.024*14)
K(t)=147m^(.019*14)
.006*14)
K(t)=135m^(.024*14)
t= 14
K(t)=142m^(-
OK. I can do this.
2.1 Of the world’s ten most populous countries, which has
the highest rate of immigration today? Which has the lowest
Infant mortality rate? What fraction of the world’s
population is comprised by those 10 most populous
countries?
From table 2.1:
Highest MIG = US
Lowest IMR = Japan
0.59 or just over half
2.2 Mexico is the country with the eleventh-largest
population in the world. Its CBR in 2003 was 2144/1000 and
there were 2,223,714 births. What figure for the mid-year
population would be consistent with those numbers?
(2,223,714/K) = (2144/1000)
K = (2,223,714 * 1000)/2144
K =
2.3 Poland has a nearly stationary population with e0=75.
There were about 300K births last year. Waht is the
approximate size of the population?
2.4 Study the data in Table 2.1 and determine whether be0
is typically greater than, equal to, or less than 1 in a
growing population.
b
.012
.025
.011
e0
72
62
66
be0
.864 <1
1.55>1
.71<1
r
.006
.017
-.006
2.5 From an almanach or other source find dates of birth
and death for the presidents of the US from Theodore
Roosevelt to George W. Bush. Draw a freehand Lexis diagram
with the lifelines of these presidents. Label the axes
clearly.
a) Draw and label a line representing the age 30
b) Draw and label a line representing the year 1945
c) Draw and label the area containing person-years lived by
people between the ages of 20 and 30 in the years between
1964 and 1968
d) Draw and label the area representing the lifetime
experience of the cohort aged 10 to 30 in 1917 (diagonal
rectangle).
Additional question: Chapter 2, table 2.1 lists facts about
the 10 most populous countries in the world. The 11th most
populous country is Mexico. Complete the table with data
for Mexico using the same sources the author used to build
table 2.1:
Overview of ps 2
Rules of logs
logs vs ln
ln = natural log = log e
log(a*b) = log(a) + log(b)
log(a/b) = log(a) - log(b)
e^a*e^b = e^(a+b)
(e^a)/(e^b) = e^a-b
log e^x = x
Problem from 2004 final
Germany
r= 1/t log (k(t)/k(0)
r = 1/30 log (79.380/72480)= 00303
Vietnam = .025
How large would Germany be if it had grown at Vietnam’s rate?
K(t)= K(0)*e^(rt)
K(t)= 72.480*e^(.025*30)
K(t)=
chapter 1 1 through 8 - do problem set 1!!!
Arrived late 10am
From last time: We defined al of the columns of the cohort life
table but we didn’t go through the exact calculations. Today we
do that.
Example 3.3.1
Children of Edward IV:
Edmund, 1330-1376, 46
Blanche, 1342-1342, 0
Isabel, 1332-1382, 50
Joan, 1335-1348, 13
William, 1336 - “died young” ?
Lionel, 1338-1368, 30
n
Agex lx
0
10
ndx
1
nqx
.10
npx
.9
nLx Tx
ex
x+ex nmx
90.5 33.8 33.8 33.8 .011
10
9
3
.333 .667 77.5 247.5
20
6
1
.167 .833 110.5
40
5
4
.800 .200 58.0 59.5 11.9 51.9 .070
60
1
1
1
10
27.5 37.5 .040
10
170.0
28.3 48.3 .009
20
20
0
1.5
1.5
1.5
61.5 .667
nd(x)= l(x)- l(x+n)
nq(x) = nd(x)/l(x) = probability of dying age x to x+n = (l(x)l(x+n))/l(x) = 1-(l(x+n)/l(x))
Not a bad
Approximation => nL(x) =n/2 (l(x) +l(x+n)) but when the sample is
small you can actually count the exact number of person years.
Typical shapes for life table functions
lx = monotonically decreasing
nqx = u-shaped
ndx = humped
ex = not monotonically decreasing
PS 3
Mariana Horta-Cappelli
17978026
nqx = ndx/lx probability of dying between age x and (x+n) => 10q20 = prob of dying between
ages 20 and 30.
nqx=1- (l(x+n)/lx)
1-nqx = lx+n/lx (probability of surviving)
lx = cohort survivors
h = hazard rate
h = -(log l1) – (loglo)/1
l(x+n) = lx (e^-(nhx)
nqx conversion = 1qy = 1 – (1-nqx)^1/n
l(x+n) = lx*e^-(n*h(x))
Survival probabilities multiply:
lx multiplies not qx. = multiply the 1-nqx values.
Assume x<y<(x+n)
lx = 1-q
(1-1qy)^n=1-nqx => 1qy = 1-(1-nqx)^(1/n)
1981 cohort 1q0 = 0.38528 and 4q1= 0.015878
1-5q0 = (1-.038528)*(1-.015878)
5q0= .0537942524
1985 cohort 5q0= .020320 and 5q5= .002795
10q0 =?
1-10q0 = (1-.020320)*(1-.002795)
10q0 = .0230582056
1780 cohort 5q40= .062756,
1q40 =?
Assuming the probability of dying stays the same throughout the period 40 to 45 years of
age:
1q40 = 1-((1-5q40)^(1/5))
1q40 = .0128786759
3.3
lo = 1, l5=.91301, l10 = .90394 1q5 =?
5q5=l10/l5 = .0099341738
1q5 = 1-((1-5q5)^(1/5)
1q5 = .001994777
b) l40= .91264, l41=.91046, 10q30=.01485
1q40 = 1-(l41/l40) = .0023886746
1q39=1 –((1-10q39)^(1/10)
1q39 = .0014950179
2q39 = 1 – ((1-1q39)*(1-1q40) = .0038801214
2q39=?
Survivors: .8888, .8640, .8248, .7651, .4554
Radix= unity => l0= 1
T80=4.530 years
x
lx
n
0.8888 5
79.54826733
0.009284699
0.072381183
65
0.7651
17.884917
NA
4.53
nqx
ndx
nax
nLx
nmx
Tx
ex
x+ex
0.02790279
0.0248 2.5
4.382
0.005659516
26.2625
55
0.864
5
0.04537037
0.0392 2.5
21.8805 25.32465278
80.32465278
60
0.8248
0.0597 2.5
3.97475 0.015019813
17.6585 21.40943259
15
0.404783688
0.3097 7.5
9.15375 0.033833128
82.884917
80
0.4554 Infinity 1
1
9.947299078
89.94729908
50
29.54826733
4.222
5
81.40943259
13.68375
NA
NA
x = age
lx = survivorship , l0 = radix (initial size of the cohort) = 1
n = interval
nqx = probability of dying = 1- (l(x+n)/lx)
ndx = deaths between ages x and x+n = lx – l(x+n)=
nax= average #of years lived in the interval from x to x+n by those dying within the interval = If
we can’t calculate it directly from data, we make nax= n/2 assuming that those who die in the
interval die on average half-way through it.
nLx= person-years lived = (n)*(lx+n) + (nax)*(ndx) or ( n/2)*(lx+lx+n) or nLx= Integr(?)
nmx= ratio of deaths to person years lived or age specific crude death rate = ndx/nLx
Tx= Person-years of life remaining for cohort members who reach age x. =
Tx= nLx+nLx+n+nLx+2n…
ex= expectation of further life beyond age x= life expectancy = Tx/lx
x+ex = average age of death for cohort members who all survive to age x.
T85=.125, T80=.525, T75=1.15, and T70=2, l90=0
5L75 =?, l75=? And e75=?
nLx = (n/2)*(lx+lx+n)
Tx = Sum nLX from bottom up
T75=1.15=5L90+5L85+5L80+5L75
1.15=0 + 5L85 + 5L80 + 5L75
T85= .125 = 0 + 5L85 => 5L85 = .125
T80= .525 = 0 + .125 + 5L80 => 5L80 = .525 - .125 = .4
T75 = 1.15 = 0 + .125 + .4 + 5L75 => 5L75 = .625
5L75 = .625
5L75 = .625 = 2.5*(l75+l80)
l90= 0 => 5L85 =.125 = 2.5*(l85+0) => l85 = .05
5L80 = .4 = 2.5*(l80+.05) => l80 = .11
5L75 = .625 = 2.5 * (l75 + .11) => l75 = .14
l75 = .14
ex = Tx/lx
e75 = T75/l75 = 1.15/.14 = 8.214285714
e75 = 8.21
3.6
The expectation of life at birth would change very significantly if we have imputed William of
Hatfield with a death age of one month.
In the lifetable where he is omitted the expectation of life at birth was 33.8 years and the
expectation of life at birth when he was included with a death age of one month is 8.2.
Here is the revised lifetable:
x
lx
n
nqx
ndx
nax
nLx
Tx
10
0.1818182
0.181818182
0.291666667
8.23484848
10
0.818181818
10
7.03636364
7.03636364
8.6
18.6
0.1666667
0.090909091
10.5
10.0454545
40
0.454545455
20
0.8
0.363636364
11.6
51.6
60
0.090909091
Infinity
0.13636364
0.13636364
1.5
61.5
ex
x+ex
0
1
8.23484848
8.23484848
8.234848
0.3333333
0.272727273
5.8
20
0.545454545
20
10.0454545
18.41667 38.4166667
9.5
5.27272727
5.27272727
1
0.090909091
1.5
nqx:
5q9
5q10 =>
---------------------Application of LT’s
Insurance and annuities
- Life insurance, any kind of risk insurance. Any kind.
In this case we’ll talk about life insurance.
1. Term insurance
2. Whole insurance (vested insurance/ funded insurance)
Applications of the risks of mortality that we see on the life
table.
Annuities - Consol (pension system that was pop. in the 19th
century)
Annuity is just insurance on it’s head.
You pay a lump sum and then at a particular moment (when you
retire?) it starts sending you a check each month until you
retire.
Assume they are “actuarily fair”
Term insurance, the simple case (w/o profit)
You buy 1 year term insurance policy x years old.
1qx = probability of dying the next year
If
P= cost of insurance policy, premium
B= benefit
Then p/1qx = Benefit. => Premium = Benefit *1qx
Moral Hazard = you may know more about your true probability of
dying than does the insurance company.
lx
Plx = Blx*1qx
Suppose that as an insurance company you charge a dollar more per
person for profit:
Plx = Blx *1qx +lx + (profit)
Whole Funded Insurance
P= premium
lx
Age x you buy an insurance policy
P.nLx = total amount of money collected in first n years
P.nLxm =
How much does the insurance ocmpany collect in total?
??? P. Sum??Lx = P*Tx
They live to Tx years, how much is the benefit?
The Benefit = (P*Tx)/lx = P*ex
Annuities:
Insurance: Smallish stream of payment, one big pay out.
Annuities: You buy an annuity for a lump sum and get back a
stream of payments.
Suppose you’re the annuity company.
You’re selling $1/yr annuities.
How much do you charge?
- To someone age x, you pay ex dollars.
Without interest, i=0, annuities and insurance are mirrored.
Term with interest <> 0
P= premium
B= Benefit
P=(B*1qx)/ (1+i)^1/2, Because on average we get money for half
the period. => n=1
Annuity w/ i<>0
(P/lx)*[((nLx)/(1+i)^n/2) + ((nLx+n)/(1+i)^2.5n) +
((nLx+3n)/(1+1)^3.5n)...] = Cost of the annuity
where P= periodic payment or benefit.
Observation about annuities. If the interest rate, i, is big,
then the nL10n (old age mortality) matters less.
PS3 = 3 2-6
Sarahz@demog.berkeley.edu
Standardization (by Age)
One of the key concepts that occurs in demography that does not
occur in other fields.
lx, nqx, nmx = age specific mortality rate or propability.
ASMR or ASDR
Crude death rate = # of Deaths in a year/ average population in
that year (year or t interval)
Age Standardized Crude Death Rate = A population’s death rates
standardized by the age structure of a standard population. E.g.
Use US age structure to standardize a state’s crude death rate or
the world’s age structure to standardize a specific country’s
crude death rate.
Exemple:
3 desert Islands:
age groups
A
pop
deaths ASMR
B
pop
deaths ASMR
C
pop deaths
ASMR
0-4
1500 120
.08
500 40
.08
.10
5-39
4000 40
.01
5000 50
.01
4000 20
.005
40+
500 40
.08
500 40
.08
1500 60
.04
------------------------------------------------------------------------Total
6000 200
6000 130
6000
130
500
50
d(A)= 200/6000= .0333
d(B)= 130/6000= .0217
d(C)= 130/6000= .0217
1. How many deaths would there be in c (standard) if had B’s
rates? = indirect standardization
C’s age structure
B’s rates Answer = B standardized crude death
rate
500
.08
=40
4000
.01
=40
1500
.08
=120
--------------------------------------6000
200
200/6000 = Age-standardized crude death rate in B (standardized
on C’s age structure).
2. How many deaths in B (standard) if it had C’s age structure? =
Indirect Standardization.
When the standard is the age structure, that’s what we call
direct. When the standard are the rates, we call it indirect.
Fertility
Many of the things we’ve done so far has an analog in fertility.
Crude Birth Rate = b = number of births/total population
General Fertility Rate = GFR
number of births/# of women of reproductive age
Age Specific Fertility Rate = ASFR = nfx
nfx = function (fertility and age) => more classic shape to the
curve than for men because the women’s reproductive years (15-44
or 15-49).
Total Fertility Rate = TFR
= SUM 1fx = n* SUM nfx
AKA Completed family size = average family size.
PS4 Due 9/26
3.7-11
4.1-4
6.3
Mariana Horta
17978026
Problem Set 4
3.7
nqx < (2nqx+2n)./2?
Yes. It’s unlikely but it’s possible. If there the mortality rate goes up abruptely in the later
(s2nqx_2n) period, then nqx will be < 2nqx+2n)./2.
Example:
nqx = 1q20 = .0025
Then there’s a war, genocide, famine, epidemy:
2nqx+2n = 2q22 = .098
.0025<.098/2 =.0025<.049
3.8
nLx < (2nLx+2n)/2 ?
nLx = (n/2)*(lx+lx+n)
(n/2)*(lx+lx+n) <[(2n/2)*(lx+lx+2n)]/2
No. This is impossible because for all values of lx, lx+n will be equal to (if no one dies) or
smaller than lx. The cohort can only get smaller, it doesn’t get bigger, the lx curve is always
monotonically decreasing
3.9 If 1q0=.0500; 2q1 = .0100,; 3q2 =.0043; 5q3= .0098 and 2q3=.0040, then what is = 5q0
age
0
1
3
n
1
2
nqx
.05
.01
3q2
2
5
3
.0043
5q3
3
5
.0098
2q3
3
2
.0040
1q0
2q1
5q0 = prob of dying between age 0 and age 5.
1-5q0 = (1-1q0) * (1-2q1) * (1-2q3)
5q0 = 1 – [(1-.05)*(1-.01)*(1-.004)]
5q0 = .063262
3.10 5d0=.2, 10d5=.1, 15d10=.05, l0=1, What’s l15?
lx
n
ndx
l0=1 5
5d0=.2
l5= .8 10
10d5=.1
l15=.7
l15=.7
3.11
x
n
lx
nLx [nLx/(1+i)^(n/2)]
528734.877027567000
70
249596.442989313000
80
73706.663038074700
90
8777.167164899950
100
103
0
0
860976.861753586000
Interest = 8%
Profit = 0 => P = Price = total Benefit
Benefit = 20K/year
10
10
10
3
60
10
86306 776885
69071 538860
38701 233810
8061 40910
121 181.5 161.711533731354
nLx = (n/2)*(lx+lx+n)
P = B/lx * { [nLx/(1+i)^(n/2)] + [nLx+n/(1+i)^(n+(n/2))]+[nLx+2n/(1+i)^(2n+(n/2))]…}
P = (20,000/86306) * { [nLx/(1+i)^(n/2)] +
[nLx+n/(1+i)^(n+(n/2))]+[nLx+2n/(1+i)^(2n+(n/2))]…}=
P = .2317335991 * 860976.861753586000 = $199,517.27
= ~ 200K for a 20K a year annuity. Those who died before age 70 lost money. Those who died
after age 75 made a good profit.
4.1 No. When we compare two countries the country with the higher expectation of life at birth
may also have the higher crude death rate. Since life expectancy is higher, the population is also
older and an older population will have a higher crude death rate than a young population. High
life expectancy correlates with low age specific death rate at young ages but not with a low crude
death rate and low life expectancy correlates with a higher age specific death rate at younger
ages where the crude death rate may not be particularly high because the population is young.
High crude death rates and high life expectancy countries are those coutries who are in the 3rd
phase of the democratic transition where they experience population aging and low population
growth. These are countries like the US and most Wetern European countries.
4.2
TFR? GRR? How close is the GRR to the NRR?
Cohort of 1000 women born in 1934 - NRR = 1.502
5fx*5Lx
5
5
4712 0
20
5
923
4662
481.7843
40
5
4
4503
0
0
0.2396 1133
922.6098
0
4726
15
4681
35
0.0046 21
4561
39.6264
0.6673 3091
TFR = Sum (nfx)*(n) = 5*Sum nfx =
x
0
n
0
5
0
5
0.0815 383
1121.5676
30
5
5
0.031 145
20.9806
50
5
50975
Table 4.3
Babies 5Lx
0
5
0
0
382.887
25
5
0.1979
0.1039 482
4637
4604 142.724
45
5
0.0088
0
0
4421 0
5fx
4770
10
4698
3.3365
GRR = Sum(nfx)*(n)*(ffab) = TFR*(ffab)
ffab = #babies* proportion female = .4886
1.6302139
0.4886
The GRR is quite similar to the NRR being a bit larger because the GRR does not take into the
consideration the mortality of the mothers, meaning it does not count the potential mothers who
died before childbearing age.
2.25 and 1.25
Swedish Women born in 1800
Radix = 1
Find NRR, TFR and GRR. How close is the GRR to the NRR
Cohort of 1000 women born in 1934 - NRR = 1.502
Table 4.3
x
n
5fx
5Lx 5fx*5Lx
15
5
0.012200
3134 38.234800
20
5
0.103800
3036
315.136800
25
5
0.221100
2930 647.823000
30
5
0.240800
2808 676.166400
35
5
0.213100
2663
567.485300
40
5
0.113600
2509 285.022400
45
5
0.018200
2351 42.788200
0.922800
19431
2572.656900
TFR = Sum (nfx)*(n) = 5*Sum nfx =
4.614
GRR =
Sum(nfx)*(n)*(ffab) = TFR*(ffab) 2.2544004
ffab = #babies* proportion
female = .4886
l0= 1000 => NNR=.[4886*SUM(nfx*nLx)]/1000 =
1.257000161
Now the GRR and the NRR are not very close to each other indicating that a much smaller
proportion of this 1800 female cohort survived to childbearing age than that of the 1934 cohort
given in table 4.3 (exercise 4.2).
number of girls born = 34
1 dies at infancy
33 survive through childbearing.
Sum (data)/34
39/34 = 1.14 = NRR
6.3 Use as your standard the age structure for the Oakland MSA in 1990 and calculate a
standardized crude death rate for Togo based on this standard.
Age specific Death rates for women in Togo in 1961:
Age
Togo ASDR
0-5
.063014 = deaths/ pop this age => deaths =140K*.063014 =
.011403 = deaths/ pop this age
60+
.071386 = deaths/ pop this age
Oakland MSA 2million people
Age lx
%
0-5
140K .07
5-60 1570K .785
60+ 290K .145
2000K
Togo’s standardized crude death rate based on Oakland’s population age structure would be:
(SUM TogoASDR*Oakland’s pop)/Oakland’s population = .023713305
-- get notes for the first part of the class-Fertility
TFR = sum fx
= n sum nfx
= avg family size
= completed family size
Gross reproductive rate => counts girl births only
GRR = sum fx(f)
= n sum nfx(f)
= avg # of girl babies
= ratio of girls in the next generation to 1 woman in this
generation.
= .4886.TFR
Growth Measure
Net reproductive rate/ration
NRR= Sum 1fx(f)*1Ln(f)
= Sum nfx(f)*nlx
= Ratio of girl babies in the next generation to 1girl baby in
this generation
Expe
Slightly more boys are born than girls. The sex ratio at birth
worldwide is roughly 1.05 boys to 1 girl.
In the long run:
r>0
r<0
r=0
NRR > 1 population grows
NRR<1 population decreases
NRR =1 population is constant.
Example:
Mid 1970’s
age 5Fx
10
.006
15
.048
20
.111
25
.108
30
.05
35
.017
40
.004
45
.003
----------tot. .338
x5
----------1.69 - TFR -> Below replacement.

** Draws graph** <img src=”chart1sept21.xls”>
Mid 1970’s
age 5Fx 5fx*5Lx
10
.006 .003
15
.048 .234
20
.111
25
.108
30
.05
35
.017
40
.004
child bearing
45
.003
------------------
(
maternity fuction avg)
*12.5 (weights)
*17.5
--42.55/1.647 = 25.84 = average age of
tot. .338 (.647*4886) = .8047
x5
----------1.69 - TFR -> Below replacement.
r= (K(t)-K(0))/(t*K(0))
Calculate weighted means
v =(1,2,7)
(1+2+7)/3 = 3.33
Weighted mean = multiply each number for a number = weight then
divide 3=
vector => x(n)(vector of n integers)- series of weights
Maternity function = *** formula?***
In 25.84 years, in the long run (ignoring migration) the
population is .8047 times as large as it was before.
K(t)=.8047*K(0)
NRR = ~ e^(r*25.84)
r = ~ (log (.8047))/(25.84)
How are Nrr & GRR related?\
GRR= sum 1fx(f)
NRR= sum 1fx(f) lx
NRR/GRR =(sum 1fx(f))/(sum 1fx(f) lx)
Now we have a relationship of the NRR to the growth rate but
also:
NRR/Grr =l(m)
m = mean age of childbearing.
NRR = l(m)*GRR
NRR is a growth measure (like r)
l(m) pure mortality
GRR is pure fertility
I have decomposed growth into two components:
One is a mortality measure and the second is a fertility measure.
We can see what of a change in NRR has to do with mortality and
how much has to do with fertility.
NRR is a function of fertility and female mortality
NRR = Proportion of femalle babies * SUM nfx * nLx = SUM
nfx(f)*nLx
Proportion of female babies = ffab = .4886 (by convention when
it’s not given)
nLx = measure of survivorship
nfx(f)*nLx = net maternity function
Alternative Formulation:
Unsing Parity-progression ratios (PPR)
Distribution of (life) children ever born by woman = Parity
Parity = 0 = nulliparous
w(j) = proportion of women in a completed cohort with parity j
SUM (0<=j<=15) w*(j) = 1
NRR = [0*w(0)+ 1*w(1) + 2*w(2)+....] * proportion of girl babies
(or .4886 by default)
= SUM j * w(j)] proportion of girls
Example with US women cohort of 1934
72 y.o. women
Parity Distribution:
Parity=j w(j)
j
w(j)
0
.079
6
.047
1
.097
7
.029
2
.233
8
.012
3
.233
9+
.008 (presume it’s exactly 9 because
.008 is such a small #)
4
.166
5
.090
Proportion of female babies = .4877
NRR = .4877 * [(0*.079) + (1*.097)+...] = 1.4860 (population is
growing because NRR >1)
GRR ~ 1.5
TFR ~ 3
PPR = a(j) = proportion of those women with at least j kids who
go on to have a t least j+1 kids
Example with US women cohort of 1934
72 y.o. women
Parity Distribution:
Parity=j w(j)
T(j)
0
.079
1
1
.097
.921
2
.233
3
.233
4
.166
5
.090
j
6
7
8
9+
w(j)
.047
.029
.012
.008
T(j) = proportion with at least j kids
SUM of all proportions at and above j kids.
a (j) = [T (j+1)]/Tj]
a (0) = T(1)/T(0) = .921/1 = .921
Chapter 5:
Projection
Simplified projection: Ignore Migration
Fertility, Mortality only
Female only = One sex model
Model Life Transitions
Example: One Period Marriage Transitions
Single, Married, Divorced, Widowed, Die 5 possibilities together
must be complete (must exhaust all possibilities)

We’re going to organize these arrows into a matrix, with each
arrow labeled by a probability:

** Homework: 3.12 and 5.1, 3, 4 and 5**
Projection Matrices
Transition Matrices - Marital status and transitioning statuses
From last time
matrice * vector = E x(1)
E(x(1)) = P*(x(0))
E(x(2)) = P*#(x(1)) = P*(P*(x(0))
E(x(t)) = (P^t)* x(0)
Projecting Populations Structured by Age
The Leslie Matrix
Project pop vector K(t) -> k(t+n)

Simplifications:
1. Width of age groups = length of projection period.
2. Females (1 sex)