Question

Money is invested in a savings account with a nominal interest rate of 2.4% convertible monthly for three years. The rate of inflation is 1.5% for the first year, 2.8% for the second year, and 3.4% for the third year. Find the percentage of purchasing power lost during the time the money is invested; that is, find p so that if you could purchase exactly u units at the time the money was invested, three years later you could purchase u(1 ? 0.01p).

Answer 1

Answer:

real rate of return -0.48%

if before I could purchase X units now I purchase X*(1 - 0.0048)

Explanation:

We solve using the fishcer model:

$\frac{1+r_n}{1+ \theta } = 1+r_e$

As we have more than one inflation period we multiply each other as it was a succession of interest

$\frac{1.024^3}{1.015 \times 1.028 \times 1.034} - 1 = r_e$

real rate -0.004777527 = -0.48%