Finance-Theory · Unit 12 · Video 7 · Interactive Practice

Cardano to Bachelier: Fair Games, Random Walks and the Heat Equation

IKey Formulas

Formula	Name	What it says
$E_t[W_{t+1}] = W_t$	Martingale (fair game)	Given everything known at time $t$ , expected wealth next period equals wealth today: expected winnings are zero
$E_t[W_{t+1}] = p(W_t + 1) + (1-p)(W_t - 1) = W_t + (2p - 1)$	One $1 bet on a coin	Fair only at $p = \tfrac{1}{2}$ ; at $p = 0.6$ the expected gain is $0.2$ , twenty cents per dollar bet
$\dfrac{1}{c}\dfrac{\partial \mathcal{P}}{\partial t} - \dfrac{\partial^2 \mathcal{P}}{\partial x^2} = 0$	Heat equation (Bachelier, 1900)	$x$ = price move from the start, $t$ = time, $\mathcal{P}$ = probability density of the move (not a put price)
$\mathcal{P}(x,t) = \dfrac{1}{\sqrt{4\pi c t}}\, e^{-x^2/(4ct)}, \quad \operatorname{Var} = 2ct$	Law of Gauss	Starting from a single point, the spread of possible moves widens with $\sqrt{t}$

Key Insight: Cardano's equal conditions became the martingale; many fair bets add up to a random walk, and in continuous time to Brownian motion; Bachelier's option value is the average payoff over the spreading bell curve, so pricing an option in his model means solving the heat equation.

IIFair Game or Tilted Coin?

A $1 coin bet is a martingale only when the expected next wealth lands exactly on today's wealth.

Chance of winning p 0.60

IIIFrom Random Walk to Brownian Motion

Smaller, more frequent fair steps turn a jagged walk into Brownian motion, while the spread after time $t$ stays $\sqrt{t}$ .

Steps per unit time n 10

IVRadiation of Probability

Random-walk outcomes pile up the way heat spreads from a hot spot: a bell curve with variance $2ct$ .

Time t (steps) 4

💡 In Bachelier's fair-game model an option's value is the average of its payoff over this bell curve; whether that average payoff is the option's price is the question Black, Scholes and Merton take up.

VQuiz Questions

Problem 1 · A Tilted Coin

Given: Your wealth today is $20. You bet $1 on a coin that you win with probability $p = 0.65$ . Find your expected wealth next period, and Cardano's verdict on the bet.

Expected wealth next period

Cardano's verdict

✅ Correct! The expected gain is

2p - 1 = 0.30

, thirty cents per dollar bet, tilted in your favour.

❌ That is the fair-game answer. Wealth stays put in expectation only at

p = \tfrac{1}{2}

; here the coin favours you.

❌ Not quite. The gain is not

p

itself: you also lose the dollar with probability

1 - p = 0.35

, so the expected gain is

0.65 - 0.35

❌ Close, but…

p - \tfrac{1}{2} = 0.15

is only how far the coin is tilted; each unit of tilt moves both the win and the loss, so the expected gain is

2p - 1

❌ Check the expected wealth. Weight each outcome by its probability:

0.65 \times 21 + 0.35 \times 19

❌ Check the verdict. Departing from equality in your opponent's favour makes you a fool; in your own favour, unjust. Which way does

p = 0.65

tilt?

❌ Wrong direction. A fool accepts a game tilted in the opponent's favour; at

p = 0.65

the coin favours you.

❌ Not a fair game. Equal conditions would need

p = \tfrac{1}{2}

; here expected wealth rises above today's wealth.

Show solution

With $W_t = 20$ (in dollars), the two outcomes are 21 with probability $0.65$ and 19 with probability $0.35$ :

E_t[W_{t+1}] = 0.65(21) + 0.35(19) = 13.65 + 6.65 = 20.30

Equivalently $E_t[W_{t+1}] = W_t + (2p - 1) = 20 + 0.30$ . Expected wealth is $20.30, so this is not a martingale: the bet carries an expected gain of 30 cents per dollar. The tilt is in your favour, and Cardano's rule calls that unjust.

Problem 2 · Fair Without Being Fifty-Fifty

Given: You roll one die. A six wins $6, a five wins $2, and any of one to four loses $2. Your wealth today is $10. Is your wealth a martingale?

✅ Correct! The two larger wins exactly offset the more frequent losses, so expected winnings are zero.

❌ Not quite. How often you lose is only half the story; the martingale condition is about the expected value, and the wins are larger than the loss.

❌ Check the probabilities.

(6 + 2 - 2)/3 = 2

treats the three outcomes as equally likely, but the loss covers four faces out of six.

❌ Check the losing faces.

\tfrac{1}{6}(6) + \tfrac{1}{6}(2) - \tfrac{1}{6}(2) = 1

counts the loss on one face only; it happens on four, with probability

\tfrac{4}{6}

Show solution

With $W_t = 10$ (in dollars), next period's wealth is 16 with probability $\tfrac{1}{6}$ , 12 with probability $\tfrac{1}{6}$ and 8 with probability $\tfrac{4}{6}$ :

E_t[W_{t+1}] = \tfrac{1}{6}(16) + \tfrac{1}{6}(12) + \tfrac{4}{6}(8) = \tfrac{16 + 12 + 32}{6} = \tfrac{60}{6} = 10 = W_t

The expected gain per roll is $\tfrac{1}{6}(6) + \tfrac{1}{6}(2) - \tfrac{4}{6}(2) = 0$ . The martingale condition asks only that expected wealth next period equal wealth today, so this lopsided game is still a fair game.

Problem 3 · Bachelier's Spreading Bell Curve

Given: In Bachelier's model the price move $x$ after $t$ days has density $\mathcal{P}(x,t) = \frac{1}{\sqrt{4\pi c t}}\, e^{-x^2/(4ct)}$ with $c = 0.09$ ( $x$ in dollars, $t$ in days; illustrative). Find the standard deviation of the move after 50 days, and the number of days until it reaches $6.

Standard deviation after 50 days

Days until the standard deviation is $6

✅ Correct! Doubling the spread takes four times as long, because the spread grows with

\sqrt{t}

❌ That is the variance.

2ct = 9

is in squared dollars; the standard deviation is its square root.

❌ Missing a factor of 2. The variance of this density is

2ct

, not

ct

: the

4ct

in the exponent is twice the variance.

❌ Not quite.

ct = 4.5

is neither the variance (

2ct

) nor the standard deviation (

\sqrt{2ct}

❌ The spread is not linear in time. Doubling the time doubles the variance, so the standard deviation grows only by

\sqrt{2}

; doubling the standard deviation needs four times the time.

❌ Check which quantity grows linearly. Setting

2ct = 6

treats the standard deviation as the variance; the variance must reach

6^2 = 36

❌ Check the standard deviation. Compute the variance

2ct

first, then take its square root.

❌ Check the time. A standard deviation of 6 means a variance of 36; solve

2ct = 36

Show solution

Step 1: Variance after 50 days.

\operatorname{Var} = 2ct = 2(0.09)(50) = 9, \qquad \sqrt{9} = 3

The standard deviation of the price move is $3.

Step 2: Time for a standard deviation of 6. The variance must be $6^2 = 36$ :

2(0.09)\,t = 36 \;\Rightarrow\; t = \frac{36}{0.18} = 200

It takes 200 days: four times 50 days to double the spread, since the standard deviation $\sqrt{2ct}$ grows with the square root of time.

Problem 4 · One Equation, Two Worlds

Given: A hot spot is applied at the middle of a long rod. After 9 minutes the temperature profile along the rod is a bell curve with standard deviation 3 cm. The same heat equation governs Bachelier's price moves. Find the standard deviation after 36 minutes, and what plays the role of the temperature in Bachelier's reading.

Spread after 36 minutes

Bachelier's counterpart of the temperature at position x, time t

✅ Correct! Four times the time gives twice the spread, in the rod and in the price alike; the temperature profile reads as the density of price moves.

❌ The spread grows with $\sqrt{t}$ , not $t$ . Multiplying the time by 4 multiplies the variance by 4 and the standard deviation by 2.

❌ Not quite.

3 \times 16 = 48

squares the time ratio; the standard deviation scales with the square root of the ratio,

\sqrt{4} = 2

❌ The spread does not stay fixed. The variance

2ct

grows in proportion to time, so the bell curve keeps widening as heat (or probability) spreads out.

❌ Not quite. The temperature is a quantity spread along

x

at each time; its counterpart is how probability is spread over the possible price moves

x

, not a single expected return.

❌ Not quite.

\mathcal{P}

is a probability density, not an option price; in Bachelier's model the option's value is the average of its payoff over that density.

❌ Check the spread. The variance

2ct

grows in proportion to time; the standard deviation is its square root.

❌ Check the analogy. In the heat equation

\frac{1}{c}\frac{\partial \mathcal{P}}{\partial t} - \frac{\partial^2 \mathcal{P}}{\partial x^2} = 0

, the quantity spreading along

x

is what the temperature represents.

Show solution

Spread. Starting from a single point the solution has variance $2ct$ , so the standard deviation is $\sqrt{2ct}$ , proportional to $\sqrt{t}$ :

\sigma_{36} = \sigma_9 \sqrt{\tfrac{36}{9}} = 3 \times 2 = 6 \text{ cm}

Analogy. In physics $\mathcal{P}(x,t)$ is the temperature at position $x$ along the rod; in Bachelier's thesis the same equation describes $\mathcal{P}(x,t)$ , the probability density of the price having moved by $x$ after time $t$ . Heat spreading from a hot spot is probability "radiating" from today's price, which is why Poincaré found Fourier's reasoning applicable almost without change.

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