Finance-Theory · Unit 4 · Video 4 · Interactive Practice

Inflation vs. Time Value: Wealth and the Price Index

IKey Formulas

Formula	Name	What it says
$\text{Wealth } W_t \;\Leftrightarrow\; \text{Price Index } I_t$	The pairing at a date	$W_t$ counts your dollars; $I_t$ is the price of the basket you actually consume
$\text{Increase in Cost of Living} \;\equiv\; \dfrac{I_{t+k}}{I_t} \;=\; (1+\pi)^k$	Cost-of-living ratio	How much dearer the same basket is $k$ periods later, at the constant rate $\pi$ per period
$I_{t+k} = I_t\,(1+\pi)^k$	The index $k$ periods on	The same statement rearranged: the later price of the basket, grown $k$ times at $\pi$
$\dfrac{W_t}{I_t}$	Wealth in baskets	What those dollars reach — and a standard of living is measured by consumption, not by a count of dollars

Key Insight: Impatience runs one way only — a later cashflow is discounted because people prefer money now, and prices play no part in it. Inflation is the other statement, about prices, and it has no fixed sign: $\pi > 0$ makes the basket dearer so each dollar buys less, $\pi < 0$ makes it cheaper so each dollar buys more. Here $\pi$ is a variable name for the inflation rate, exactly as $r$ names an interest rate — not the constant from geometry.

IIInflation Runs Both Ways

A barrel more than doubled over two years, then fell back within months — the same dollar, a different amount of heat.

Measured against

💡 Impatience never runs backwards: through every one of these months, a dollar due next winter was still worth less than a dollar today.

IIIThe Cost of Living Over $k$ Periods

At a constant rate $\pi$ per period, the price of the basket compounds, and $I_{t+k}/I_t$ collapses to $(1+\pi)^k$ .

Inflation π 3.0% Periods k 5.0

IVTwo Tracks: What You Hold, What It Costs

Wealth alone cannot say whether you are better off; the price of the basket has to be read beside it.

💡 Challenge: with the basket at $125, find the wealth at $t+5$ that leaves your standard of living exactly unchanged.

VQuiz Questions

Problem 1 · Four Periods of Inflation

Given: the cost of living grows at the constant rate $\pi = 3\%$ per period — find the cost-of-living ratio $I_{t+4}/I_t$ .

✅ Correct!

(1.03)^4 = 1.1255

: the basket costs 12.55% more after four periods.

❌ Close, but…

1.12

1 + 4\pi

. Each period's 3% is charged on a basket that is already dearer, so the ratio is

(1+\pi)^4

, not

1 + 4\pi

❌ Not quite.

0.8885 = 1/(1.03)^4

— that is what one dollar comes to buy, the reciprocal of the cost-of-living ratio.

❌ Not quite. The ratio over

k

periods is

(1+\pi)^k

with

\pi = 0.03

and

k = 4

Show solution

The increase in the cost of living over $k$ periods is the ratio of the two index levels:

\frac{I_{t+4}}{I_t} = (1+\pi)^4 = (1.03)^4 = 1.1255

Step by step: $(1.03)^2 = 1.0609$ , and $(1.0609)^2 = 1.1255$ .

The same basket costs 12.55% more, not 12% — the extra 0.55% is the inflation charged on the inflation. A dollar therefore buys $1/1.1255 = 0.8885$ of what it bought at date $t$ .

Problem 2 · When the Basket Gets Cheaper

Given: a household's basket costs $120 at date $t$ and $108 at date $t+3$ — which statement is correct?

✅ Correct!

108/120 = 0.90 < 1

, so

\pi

is negative — the basket got cheaper and the same dollar reaches further.

❌ Close, but… the ratio is right. A ratio below 1 means the basket is cheaper at

t+3

, so each dollar buys more, not less.

❌ Not quite. You divided the earlier index by the later one. The cost-of-living ratio is

I_{t+3}/I_t = 108/120

, the later level over the earlier one.

❌ Not quite. That is the other claim. Impatience discounts later cashflows whatever prices do; inflation is a statement about prices and has no fixed sign.

❌ Not quite. Compute

I_{t+3}/I_t

first, then read its size against 1.

Show solution

The cost-of-living ratio is the later index over the earlier one:

\frac{I_{t+3}}{I_t} = \frac{108}{120} = 0.90

Since $0.90 < 1$ , the basket is cheaper at $t+3$ : the same dollars buy about 11% more of it ( $1/0.90 = 1.111$ ). Solving $0.90 = (1+\pi)^3$ gives

\pi = 0.90^{1/3} - 1 = -0.0345 = -3.45\%\ \text{per period}

Nothing here contradicts the time value of money. Impatience never runs backwards — a cashflow at $t+3$ is still discounted — but prices do run backwards, and here they did.

Problem 3 · Read Both Tracks

Given: a household holds wealth $W_t$ of $50,000 against a basket priced at $I_t$ = $100. Over five periods its wealth grows to $60,000 while inflation runs at $\pi = 4\%$ per period.

What is the later price of the basket, $I_{t+5}$ ?

How many baskets can the household consume at $t+5$ ?

✅ Correct! Wealth rose 20% but the basket rose 21.67%, so consumption slipped from 500 baskets to about 493 — more dollars, slightly less living.

❌ Close, but… $120 is

100 \times (1 + 5 \times 0.04)

. The index compounds:

100 \times (1.04)^5

❌ Check the index.

I_{t+5} = I_t (1+\pi)^5

with

I_t = 100

and

\pi = 0.04

❌ Not quite. That reads the wealth track alone: 20% more dollars buys 20% more baskets only if the basket's price stayed put.

❌ Check the consumption. Divide the later wealth by the later price of the basket:

W_{t+5}/I_{t+5}

Show solution

Step 1 — the price track.

I_{t+5} = I_t (1+\pi)^5 = 100 \times (1.04)^5 = 100 \times 1.21665 = 121.67

So the basket costs $121.67.

Step 2 — the wealth track, read in baskets.

\frac{W_t}{I_t} = \frac{50{,}000}{100} = 500 \qquad \frac{W_{t+5}}{I_{t+5}} = \frac{60{,}000}{121.67} = 493.2

Step 3 — compare. Wealth grew by the factor $60{,}000/50{,}000 = 1.20$ , the cost of living by $1.21665$ . Consumption therefore moves by

\frac{1.20}{1.21665} = 0.9863

a fall of about 1.4%, from 500 baskets to 493. Knowing the wealth at the later date was not enough on its own.

Problem 4 · Standing Still

Given: wealth $W_t$ of $40,000, a basket priced at $I_t$ = $200, and $\pi = 5\%$ per period — find the wealth $W_{t+2}$ that leaves the household's standard of living exactly unchanged.

✅ Correct! Wealth has to grow by the same factor as the basket,

(1.05)^2 = 1.1025

, and

40{,}000 \times 1.1025 = 44{,}100

❌ Close, but… $44,000 grows wealth by

1 + 2\pi = 1.10

. The basket grows by

(1.05)^2 = 1.1025

, so you would end up 200 baskets short by a hair.

❌ Not quite. That applies

\pi

once. Two periods of inflation compound:

(1+\pi)^2

❌ Not quite. Holding the same dollars while the basket gets dearer means consuming less — the standard of living falls.

❌ Not quite. Keep

W/I

fixed:

W_{t+2} = W_t \times I_{t+2}/I_t

Show solution

Step 1 — baskets at date $t$ .

\frac{W_t}{I_t} = \frac{40{,}000}{200} = 200 \ \text{baskets}

Step 2 — the basket's price two periods later.

I_{t+2} = 200 \times (1.05)^2 = 200 \times 1.1025 = 220.50

Step 3 — the wealth that still buys 200 baskets.

W_{t+2} = 200 \times 220.50 = 44{,}100

Equivalently, wealth must grow by exactly the cost-of-living ratio: $40{,}000 \times 1.1025 = 44{,}100$ , or $44,100. Anything less and the two tracks no longer offset; $44,000 would leave the household a fraction of a basket short.

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Inflation vs. Time Value: Wealth and the Price Index

IKey Formulas

IIInflation Runs Both Ways

IIIThe Cost of Living Over kkk Periods

IVTwo Tracks: What You Hold, What It Costs

VQuiz Questions

IIIThe Cost of Living Over $k$ Periods