## Annuities With Inflation

```Suppose we have a certain amount of savings p at age 65, invested in
a fund that yields 10% annual return, and we wish to make withdrawals
in amounts that increase at 4% each year (to keep up with inflation)
in such a way that our savings are exhausted at age 85.  What should
be the size of our withdrawals?

The precise answer depends on how we compound the interest/inflation,
and how we schedule our withdrawals, but the general method is the
same, regardless.  Suppose we want to do our calculations on an
annual basis.  An inflation rate of 4% means our withdrawals must
increase by a factor of I = 1.04 each year.  Also, an investment yield
of 10% means our principle would increase by a factor of Y = 1.10
each year, if we made no withdrawals.

Beginning with a principal amount of p[0] we would start by with-
drawing w[0] dollars, and then let the remaining principal gather
interest for 1 year.  At the end of the first year our principle is

p[1] = (p[0]-w[0])*Y                     (1)

At that point we withdraw w[1] dollars, and then let the remaining
principle gather interest for a year, at the end of which time we
have
p[2] = (p[1]-w[1])*Y                     (2)

Of course, to keep up with inflation, we have w[1] = I*w[0].  Also,
from equation (1) we have

w[0] = p[0] - p[1]/Y                     (3)

Therefore, w[1] equals I*(p[0] - p[1]/Y), so we can substitute into
equation (2) to give

p[2] = Y*(p[1] - I*(p[0] - p[1]/Y))       (4)

which simplifies to

p[2] = (Y+I)*p[1] - (YI)*p[0]              (5)

Obviously this same relation applies to each successive step, so
we have the recurrence

p[k+2]  =  (Y+I) p[k+1]  -  (YI) p[k]            (6)

The roots of the characteristic polynomial are just Y and I, so the
general term is of the form

p[n]  =   A Y^n  +  B I^n                       (7)

where the coefficients A and B are determined by the initial
conditions.  We require

A + B   =  p[0]
YA + IB  =  p[1]  = (p[0]-w[0])*Y

Solving for A and B gives

Ip[0] - Y(p[0]-w[0])                 -Yw[0]
A  =  --------------------          B  =  ---------
I - Y                           I - Y

so the general formula for the principle after n years is

(Y-I)p[0] - Yw[0]           Yw[0]
p[n]  =  ----------------- Y^n  +  --------- I^n           (8)
Y - I                 Y - I

This value reaches zero when

[ Yw[0] - (Y-I)p[0] ] Y^n   =   [ Yw[0] ] I^n

so we have
(Y-I)p[0]
(I/Y)^n  =  1  -  -----------                    (9)
Yw[0]

Solving for w[0] gives the result

1 - (I/Y)
w[0]  =  p[0] ------------                     (10)
1 - (I/Y)^n

As an example, suppose our initial principal is p[0] = \$200,000, the
inflation rate is I = 1.04, the investment yield is Y = 1.10, and we
intend to run out of money in exactly 20 years.  The above equation
says our initial withdrawal should be \$16,178.31 (increased by 4%
every year thereafter).

Conversely, we might ask how many years our money would last if we
start by withdrawing \$20,000 the first year.  Solving (9) for n gives

/     / Y-I \  p[0] \
ln ( 1 - ( ----- ) ----  )
\     \  Y  /  w[0] /
n  =  ---------------------------
ln(I/Y)

which indicates we would run out of money in 14.06 years.

There's actually a much simpler way to derive this formula.  Let
I=1.04 be the inflation rate, and Y=1.10 be the investment yield.
If x is our initial payout, then it's present value is obviously
just x.  The payout one year from now will be Ix, which has a present
investment value of x(I/Y).  In general, the present value of the
payout k years from now is x(I/Y)^k.  If we want to run out of money
in exactly n=20 years, we want the sum of the present values of the 20
payouts at year 0 through 19 to equal our initial principle P.  Thus,

P  =  x [ 1 + (I/Y) + (I/Y)^2 + ... + (I/Y)^(n-1) ]

1 - (I/Y)^n
=   x -------------
1 - (I/Y)

which is identical to the result given previously.
```