# Notes and Solutions for: The Mathematics of Financial

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Notes and Solutions for: The Mathematics of Financial Derivatives
by Paul Wilmott, Sam Howison, and Jeﬀ Dewynne
John L. Weatherwax∗ December 7, 2015
[email protected]
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Please Do Not Redistribute Without Permission from the Author
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Introduction
Here you’ll ﬁnd some notes that I wrote up as I worked through this excellent book. For some of the problems I used MATLAB to perform any needed calculations. The code snippets for various exercises can be found at the following location:
http://waxworksmath.com/Authors/N_Z/Wilmott/MathOfFinancialDerivatives/wilmott.html
I’ve worked hard to make these notes as good as I can, but I have no illusions that they are perfect. If you feel that that there is a better way to accomplish or explain an exercise or derivation presented in these notes; or that one or more of the explanations is unclear, incomplete, or misleading, please tell me. If you ﬁnd an error of any kind – technical, grammatical, typographical, whatever – please tell me that, too. I’ll gladly add to the acknowledgments in later printings the name of the ﬁrst person to bring each problem to my attention.
Acknowledgements
Special thanks to (most recent comments are listed ﬁrst): Felix Huber and Peter Hogeveen for helping improve these notes and solutions. All comments (no matter how small) are much appreciated.
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Chapter 1 (Options and Markets)
Exercise 1 (stock splits)
If a company issues a stock split (doubling the number of stock shares) I would expect that the value of each share of stock to divide by two and the number of stock shares that a person owned to double keeping the net value of all the stock constant. I would also expect the options values to also divide their value by two.
As an example, of a stock split consider the value of a Rolls-Royce call (the option to buy on Rolls-Royce stock at a given price at a future date). According to Figure 1.1 from the book its value with a March expiration date is 11p. If Rolls-Royce issues a one-for-one stock split I would expect that the value of the March call option (for one share) to go to 11p/2 = 5.5p.
For a two-for-one issue, I would expect the value of the underlying security to drop to one third its original price, since the owner of one share ends up owning three shares after the issue. In the same way I would expect that the value of options on the underlying would be reduced by a factor of three.
Exercise 2 (dividends)
I would assume that the price of the stock S before the dividend is issued must implicitly include the value of the dividend in the stocks evaluation. Thus after the dividend issue the stock price would decrease by an amount D to S − D.
Exercise 3 (the direction of uncertainty)
As the volatility of the underlying security increases it becomes more likely that the underlying will obtain its strike price. Since it is more likely to be exercised there is more value to the option.
Exercise 4 (zero-sum game)
What is meant by the statement “options transactions are a zero sum game” is that there is no loss or gain in the immediate purchase of an option. It can be sold immediately after its purchase for the same price. Thus options have an intrinsic value that does not change as they are purchased.
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Chapter 2 (Asset Price Random Walks)

Exercise 1 (stochastic derivatives)

For this problem, we require Ito’s lemma for a function f (S), when S is by a stochastic process that satisﬁes dS = µSdt + σSdX, with dX the random variable. Here we are using the notation that a capital letter represents a random variable and a lower case letter represents a deterministic variable. With that background Ito’s lemma (for a function f (S) with no explicit time dependence) is given by

df = (σSdX + µSdt) df + 1 σ2S2 d2S dt dS 2 dS2

= σS df dX + µS df + 1 σ2S2 d2S dt

(1)

dS

dS 2 dS2

Part (a): Here our function f is given by f (S) = AS, so df /dS = A, and d2f /dS2 = 0 so Ito’s lemma then gives

df = σSAdX + µSAdt = σf (S)dX + µf (S)dt .
Part (b): Here our function f is given by f (S) = Sn, so df /dS = nSn−1 and d2f /dS2 = n(n − 1)Sn−2 and Ito’s lemma becomes
df = σS(nSn−1)dX + µSnSn−1 + 1 σ2S2n(n − 1)Sn−2 dt 2
= nσSndX + µnSn + n(n − 1) σ2Sn dt 2
= nσf (S)dX + µn + n(n − 1)σ2 f (S)dt . 2

Exercise 2 (verifying a stochastic integral with Ito’s lemma)
The Equation 2.12 is a stochastic integral given by t X(τ )dX(τ ) = 12(X(t)2 − X(t0)2) − 21 (t − t0) .
t0
Note that this diﬀers from what one would expect from non-stochastic calculus in the term linear in t. The diﬀerential of the left hand side of this

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expression is given by XdX. The diﬀerential of the right hand side is given

symbolically by

d 1 X2 − 1 dt .

(2)

2

2

To evaluate the diﬀerential d( 12X2) we must invoke Ito’s lemma since the variable X(t) is random. Now Ito’s lemma for a function f (X) is

df = df dX + 1 d2f dX2 .

dx

2 dX2

In our case f (X) ≡ X22 , so ddXf = X, ddX2f2 = 1. Finally using the heuristic that dX2 = dt we see that Eq. 2 becomes

df = XdX + 1(1)dt − 1dt = XdX ,

2

2

proving the desired equivalence.

Exercise 3 (the density function for the log-normal random variable)
Considering S a stochastic variable with increments given by

dS = σSdX + µSdt ,

And a function f deﬁned by f (S) = log(S), Ito’s lemma simpliﬁes and we ob-

serve that f satisﬁes the following stochastic diﬀerential equation (see Page 28

of the text)

df = σdX + (µ − 1 σ2)dt . 2

In the above expression, as dX is the only random variable it follows that

the distribution of f can be obtained from the distribution of dX. To do

this imagine integrating the above from some ﬁxed point f0 to f . This can

be thought of as adding together a large number of independent identically

distributed random variables dX. Because dX is assumed to be normally

distributed with a mean of 0 and a variance t, we see that the variable f is normally distributed with mean (µ − 12 σ2)t and variance equal to σ2t. The fact that the variance is given by σ2t follows from the fact that a random

variable deﬁned as a sum of independent random variables has a variance

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given by the sum of the individual variances. Thus the probability density function (PDF) of f is given by

√2π(1σ√t) e . − 12 (f−(f0+σ(µ2−t 12 σ2)t)2

To compute the PDF of the random variable S given the PDF of the random variable f we use the following theorem involving transformations of random variables from probability theory

df pS(s) = pF (f (s)) ds ,

Here pS(s) is the PDF of the random variable S and pF (f ) is the PDF of

the random variable F . Since when f (S) = log(S), the derivative is given by

df ds

=

1 s

using

the

above

we

get

for

pS (s)

the

following

p (s) = √2π1tsσ e S − 12 (log(s)−(log(st0σ)2+(µ− 21 σ2)t)2 = σs√12πt e , − 12 (log(s/s0)−tσ(2µ− 12 σ2)t)2

as claimed in the book.

Exercise 4 (a function of a random variable that has no drift)

We assume that our random variable G is deﬁned in terms of X and t by a stochastic diﬀerential equation given by

dG = A(G, t)dX + B(G, t)dt .

Now lets consider a function of G say f (G). We will apply Ito’s lemma to f . Now Ito’s lemma applied to the function f is given by

df = df dG + 1 d2f dG2

dG

2 dG2

= (A(G, t)dX + B(G, t)dt) df + 1 (A(G, t)dX + B(G, t)dt)2 d2f

dG 2

dG2

= (A(G, t)dX + B(G, t)dt) df + 1 A(G, t)2dt d2f

dG 2

dG2

= A(G, t) df dX + B(G, t) df + 1A(G, t)2 d2f dt

dG

dG 2

dG2

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Where in the above we have used the “rule” that dX2 = dt. Now we can select f such that we eliminate the drift term by setting the coeﬃcient of dt to zero. Speciﬁcally we require

B(G, t) df + 1 A(G, t)2 d2f = 0 .

dG 2

dG2

With this speciﬁcation we ﬁnd that the evolution of f follows a random with

no drift given by

df = A(G, t) df dX . dG

Thus given a random variable that has a drift term we can construct a func-

tion of G such that this function has no drift.

Exercise 5 (Ito’s lemma for multidimensional functions)

We are told that the variables Si satisfy the stochastic diﬀerential equations

dSi = σiSidXi + µiSidt for i = 1, 2, . . . , n ,

where the random increments dXi satisfy

E[dXi] = 0 E[dXi2] = dt E[dXi dXj] = ρij dt ,

and we desire to compute the diﬀerential df of a multidimensional function

f = f (S1, S2, . . . , Sn). For such a function using calculus we can write its

diﬀerential as

n ∂f

1 n ∂2f

df = ∂Si dSi + 2 ∂Si∂Sj dSidSj . . .

i=1

i,j=1

Splitting the second sum in the above (the one over i and j) into a diagonal term and the non-diagonal terms we ﬁnd that

n ∂f

df =

∂Si dSi

i=1

+ 1 n ∂2f dS2 2 i=1 ∂Si2 i

1n

∂2f

+2

∂Si∂Sj dSidSj . . .

i,j=1;i=j

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To evaluate the diagonal sums in the above we use the two facts. The ﬁrst is that dSi = σiSidXi+µiSidt and the second is that dSi2 = σi2Si2dXi2 = σi2Si2dt, to ﬁrst order. To evaluate the non-diagonal terms we can expand the product
above as

dSidSj = (σiSidXi + µiSidt)(σjSjdXj + µjSjdt)
= σiσjSiSjdXidXj + σiµjSiSjdXidt + µiσjSiSjdtdXj + µiµjSiSjdt2 .
√ Since dXi is O( dt) the terms in the above expression are ordered with increasing powers of dt, as dt, dt3/2, dt3/2, and dt2. As always, we are concerned with the limit of this expression as dt → 0 we see that to leading order only the term σiσjSiSjdXidXj remains. With these substitutions we obtain for df the following

n ∂f

df =

∂Si (σiSidXi + µiSidt)

i=1

+ 1 n ∂2f σ2S2dt 2 i=1 ∂Si2 i i

1n

∂2f

+2

∂Si∂Sj σiσjSiSjdXidXj . . .

i,j=1;i=j

Using dXidXj → ρijdt and rearranging terms we have that Ito’s lemma for a multivariate function f is given by

n ∂f

df =

∂Si σi Si dXi

i=1

n
+

∂f µ S + 1 ∂2f σ2S2 + 1

n

∂2f σ σ S S ρ dt .

i=1 ∂Si i i 2 ∂Si2 i i

2

∂SiSj i j i j ij

i,j=1;i=j

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Chapter 3 (The Black-Scholes Model)

The derivation of ∆ for a European call
Here we perform the derivation of the expression for ∆ for a European call option. As such, recall that a European call has a value given by

C(S, t) = SN (d1) − Ee−r(T −t)N (d2) ,

with d1 and d2 functions of S and t given by

d1(S, t) = ln(S/E) +σ√(rT+−12 σt 2)(T − t) = lσo√g(TS/−Et) + (r +σ12 σ2) √T − t d2(S, t) = ln(S/E) +σ√(rT−−21 σt 2)(T − t) = lσo√g(TS/−Et) + (r −σ12 σ2) √T − t ,

with N(x) the cumulative distribution function for the standard normal and

is given by

N(x) = √1

x

e−

1 2

y2

dy

.

(3)

2π −∞

From the deﬁnition of ∆ we ﬁnd that

∆ ≡ ∂C = N (d1) + SN ′(d1) ∂d1 − Ee−r(T −t)N ′(d2) ∂d2 .

∂S

∂S

∂S

√ From the deﬁnitions of d1 and d2 we see that d1 − d2 = σ T − t and thus

that ∂d1 = ∂d2 . ∂S ∂S

Finally N ′(d) = √12π e− 21 d2. With these results the expression for N ′(d2) is given by

N ′(d2)

=

√1

e−

1 2

d22

=

√1

e−

1 2

(d1

σ

T

t

)2

=

√1

e−

1 2

d21

ed1σ T −t

e−

1 2

σ

2

(T

t)

=

N

′(d1)ed1σ T

−t

e−

1 2

σ2 (T

−t)

.

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