Feynman-Kac and the Black-Scholes PDE

Skill positionModule 3 of 5 · Topic 2 of 2

3.2 Feynman-Kac and the Black-Scholes PDE

—Lab score— / 100Total XP—Streak—

Two ways to the same price

There are two ways to price an option. One is to average the payoff over random paths, as in the last topic. The other is to solve a partial differential equation (PDE), an equation that ties how the price changes with time to how it changes with the stock price. The Feynman-Kac theorem says the two are the same: the average of a payoff over random paths is the solution of a PDE. For a stock with $dS=rS\,dt+\sigma S\,dW$ under $Q$ , the option value $V(t,S)$ solves the Black-Scholes PDE, with the payoff $g$ as the end condition.

\frac{\partial V}{\partial t}+\frac12\sigma^{2}S^{2}\frac{\partial^{2}V}{\partial S^{2}}+rS\frac{\partial V}{\partial S}-rV=0,\qquad V(T,S)=g(S)

(1)

Read the PDE like a sentence. $\partial V/\partial t$ is time decay. The term with $\partial^{2}V/\partial S^{2}$ is the reward for curvature: randomness adds value to an option, and this term exists because $(dW)^{2}=dt$ . The term with $\partial V/\partial S$ is the drift of the stock and $-rV$ is discounting. For a European call the solution is the Black-Scholes formula below, where $N$ is the standard normal distribution function. A put follows from put-call parity.

C=S\,N(d_1)-Ke^{-rT}N(d_2),\quad d_1=\frac{\ln(S/K)+(r+\sigma^{2}/2)T}{\sigma\sqrt T},\quad d_2=d_1-\sigma\sqrt T,\qquad C-P=S-Ke^{-rT}

(2)

An at-the-money callTake

S=K=100

r=5\%

\sigma=20\%

and

T=1

. Then

d_1=(0+0.07)/0.2=0.35

and

d_2=0.15

, with

N(d_1)=0.6368

and

N(d_2)=0.5596

. The price is

C=100\times0.6368-95.12\times0.5596\approx10.45

. The number

N(d_1)=0.6368

is also the call's delta, the shares you hold to hedge it.

Feynman-Kac: the average payoff over random paths solves a PDE, and the PDE solves for that average.
Black-Scholes PDE terms: time decay, reward for curvature, drift and discounting.
Put-call parity: $C-P=S-Ke^{-rT}$ , so one price gives the other.

Lab · what's graded

Computing d115 pts
Pricing the call15 pts
Put-call parity15 pts
Testing a function in the PDE15 pts
Which functions solve the PDE15 pts
Read the PDE10 pts
The Feynman-Kac idea15 pts

0/ 7 answered

1Numeric answer15 ptsNot answered

For $S=K=100$ , $r=5\%$ , $\sigma=20\%$ and $T=1$ , what is $d_1=\dfrac{\ln(S/K)+(r+\sigma^{2}/2)T}{\sigma\sqrt T}$ ?

d =

accepted: ± 0.005

2Numeric answer15 ptsNot answered

Use $N(d_1)=0.6368$ and $N(d_2)=0.5596$ with $S=K=100$ , $r=5\%$ and $T=1$ . What is $C=S\,N(d_1)-Ke^{-rT}N(d_2)$ ? Give two decimals.

c =

accepted: ± 0.03

3Numeric answer15 ptsNot answered

A call is worth $C=12$ with $S=K=100$ , $r=5\%$ and $T=1$ . Use $C-P=S-Ke^{-rT}$ to find the put value $P$ . Give two decimals.

p =

accepted: ± 0.02

4Numeric answer15 ptsNot answered

Put $V=S^{2}$ into the left side of the PDE, $\partial_t V+\tfrac12\sigma^{2}S^{2}\partial_{SS}V+rS\,\partial_S V-rV$ . What number do you get at $S=10$ , $\sigma=0.2$ and $r=0.05$ ? (It is not $0$ , so $S^{2}$ is not a price.)

v =

accepted: ± 0.01

5Choice15 ptsNot answered

Which of these are prices that satisfy the Black-Scholes PDE? Pick all that apply.

6Match10 ptsNot answered

Match each term of the PDE with its meaning.

Pick the partner of each item on the left.

\partial V/\partial t

\tfrac12\sigma^{2}S^{2}\,\partial^{2}V/\partial S^{2}

rS\,\partial V/\partial S

-rV

7Explain in words15 ptsNot answered

In one or two sentences, explain why pricing by averaging over random paths and pricing with the Black-Scholes PDE give the same number.

Graded on:The average payoff over paths is a function of time and priceThat function satisfies the PDE0 / 400 characters

Math modeKeyboard or LaTeX inputTask 1 of 7 · Ctrl ⏎ to check

Feynman-Kac and the Black-Scholes PDE

Skill positionModule 3 of 5 · Topic 2 of 2

3.2 Feynman-Kac and the Black-Scholes PDE

—Lab score— / 100Total XP—Streak—

Two ways to the same price

\frac{\partial V}{\partial t}+\frac12\sigma^{2}S^{2}\frac{\partial^{2}V}{\partial S^{2}}+rS\frac{\partial V}{\partial S}-rV=0,\qquad V(T,S)=g(S)

(1)

C=S\,N(d_1)-Ke^{-rT}N(d_2),\quad d_1=\frac{\ln(S/K)+(r+\sigma^{2}/2)T}{\sigma\sqrt T},\quad d_2=d_1-\sigma\sqrt T,\qquad C-P=S-Ke^{-rT}

(2)

An at-the-money callTake

S=K=100

r=5\%

\sigma=20\%

and

T=1

. Then

d_1=(0+0.07)/0.2=0.35

and

d_2=0.15

, with

N(d_1)=0.6368

and

N(d_2)=0.5596

. The price is

C=100\times0.6368-95.12\times0.5596\approx10.45

. The number

N(d_1)=0.6368

is also the call's delta, the shares you hold to hedge it.

Feynman-Kac: the average payoff over random paths solves a PDE, and the PDE solves for that average.
Black-Scholes PDE terms: time decay, reward for curvature, drift and discounting.
Put-call parity: $C-P=S-Ke^{-rT}$ , so one price gives the other.

Lab · what's graded

Computing d115 pts
Pricing the call15 pts
Put-call parity15 pts
Testing a function in the PDE15 pts
Which functions solve the PDE15 pts
Read the PDE10 pts
The Feynman-Kac idea15 pts

0/ 7 answered

1Numeric answer15 ptsNot answered

For $S=K=100$ , $r=5\%$ , $\sigma=20\%$ and $T=1$ , what is $d_1=\dfrac{\ln(S/K)+(r+\sigma^{2}/2)T}{\sigma\sqrt T}$ ?

d =

accepted: ± 0.005

2Numeric answer15 ptsNot answered

Use $N(d_1)=0.6368$ and $N(d_2)=0.5596$ with $S=K=100$ , $r=5\%$ and $T=1$ . What is $C=S\,N(d_1)-Ke^{-rT}N(d_2)$ ? Give two decimals.

c =

accepted: ± 0.03

3Numeric answer15 ptsNot answered

A call is worth $C=12$ with $S=K=100$ , $r=5\%$ and $T=1$ . Use $C-P=S-Ke^{-rT}$ to find the put value $P$ . Give two decimals.

p =

accepted: ± 0.02

4Numeric answer15 ptsNot answered

v =

accepted: ± 0.01

5Choice15 ptsNot answered

Which of these are prices that satisfy the Black-Scholes PDE? Pick all that apply.

6Match10 ptsNot answered

Match each term of the PDE with its meaning.

Pick the partner of each item on the left.

\partial V/\partial t

\tfrac12\sigma^{2}S^{2}\,\partial^{2}V/\partial S^{2}

rS\,\partial V/\partial S

-rV

7Explain in words15 ptsNot answered

In one or two sentences, explain why pricing by averaging over random paths and pricing with the Black-Scholes PDE give the same number.

Graded on:The average payoff over paths is a function of time and priceThat function satisfies the PDE0 / 400 characters

Math modeKeyboard or LaTeX inputTask 1 of 7 · Ctrl ⏎ to check

Feynman-Kac and the Black-Scholes PDE

Two ways to the same price

Lab · what's graded

Log in to run your code.

Feynman-Kac and the Black-Scholes PDE

Two ways to the same price

Lab · what's graded

Log in to run your code.