1. Section 9.4 Suppose $10 is the cost of a widget today, but one year from today the same widget costs $12.50. In other words, the purchasing power of $10 today is essentially equivalent to the purchasing power of $12.50 one year later. If we let r be the rate at which the cost of a widget has grown, then we can write 10(1 + r) = 12.50. Such a rate of increase in cost is mathematically the same concept as the rate of increase in the value of an investment; however, from an investor/consumer perspective, an increase in the value of an investment is desirable, whereas an increase in the cost of an item is typically not desirable. The rate of increase r in one period (year) is called the rate of inflation. In the previous illustration, the rate of inflation is r = 12.50/10  1 = 0.25.

2. In general, the rate of inflation for a period is determined by If amount A is invested at effective rate i, then without inflation, the value of the investment after n periods is However, with inflation at rate r per period, the purchasing power of this investment (i.e., the value in terms of dollars at time 0) after n periods is [old cost of X](1 + r) = [new cost of X]. A(1 + i) n. A(1 + i) n . (1 + r) n If we let i / be the real rate of interest and call i the nominal rate of interest (where “nominal” has a different meaning here than in the past), then the real value of the investment isand we must have that 1 + i 1 + i / = . 1 + r A(1 + i / ) n, It is now easy to see that i  r i / =  and i = i / + r + i / r. 1 + r

3. Although it is often true that i > r, there can be cases where this is not true. In a situation where i < r, then i / will be negative. Suppose we want the present value of a series of n yearly payments of R beginning one year from now, with (nominal) interest rate i. If the payments are not affected by inflation, then the present value is a – n|i R If the inflation rate is r, and the payments need to be adjusted for inflation, then the present value can be written as R(1 + r)  + (1 + i) R(1 + r) 2  + … + (1 + i) 2 R(1 + r) n  = (1 + i) n R(1 + r) 1 – ————— i – r 1 + r —— 1 + i n or can be written as

4. If the inflation rate is r, and the payments need to be adjusted for inflation, then the present value can be written as R(1 + r)  + (1 + i) R(1 + r) 2  + … + (1 + i) 2 R(1 + r) n  = (1 + i) n R(1 + r) 1 – ————— i – r 1 + r —— 1 + i n or can be written as R  + (1 + i / ) R  + … + (1 + i / ) 2 R  = (1 + i / ) n a – n|i / R

5. 1. Chapter 9 Exercises Bond X has face value F and coupon rate r where the first semiannual coupon is paid six months from today, and it is expected that the inflation rate for the six-month period will be 0.044. (a)Find the purchasing power of the coupon to be paid six months from today. Fr  1.044 (b)If Bond Y has face value F where the first semiannual coupon is paid six months from today, find the coupon rate necessary in order that the purchasing power of the first semiannual coupon be Fr. 1.044r TIPS (Treasury Inflation Protection Securities) are a type of security where both the maturity value and coupons are adjusted for future changes in the inflation rate.

6. We have assumed that the rate of inflation r is the same for each year (period). This may often be a reasonable assumption for the “near” future, but as time passes an actual rate of inflation can be calculated. One approach (but not the only approach) to estimate the rate of inflation for a previous year is to use the percent increase in the CPI (Consumer Price Index) for that year. 2. Chapter 9 Exercises Look at Example 9.2 in the textbook, and verify the entries in Table 9.1.

7. 3.The nominal rate of interest for a single deposit invested for 10 years is 8%, and the rate of inflation is 5%. Let A = value of the investment at the end of 10 years with no inflation B = value of the investment at the end of 10 years in terms of dollars at time 0 C = value of the investment at the end of 10 years computed at the real rate of interest Find the ratios A/B, A/C, and B/C. A = X(1.08) 10, B =, C = (1.08) 10 X  (1.05) 10 0.03 X 1 +  1.05 X(1 + i / ) 10 = 10 A/B = A/C = B/C = (1.05) 10 = 1.629 1