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Financial Engineering
Zvi Wiener
[email protected]
tel: 02-588-3049
Zvi Wiener
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slide 1
Main Books
Shimko D. Finance in Continuous Time, A
Primer. Kolb Publishing Company, 1992,
ISBN 1-878975-07-2
 Wilmott P., S. Howison, J. Dewynne, The
Mathematics of Financial Derivatives, A
Student Introduction, Cambridge University
Press, 1996, ISBN 0-521-49789-2

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Useful Books
Duffie D., Dynamic Asset Pricing Theory.
 Duffie D., Security Markets, Stochastic Models.
 Neftci S., An Introduction to the Mathematics of
Financial Derivatives.
 Steele M., Invitation to Stochastic Differential
Equations and Financial Applications.
 Karatzas I., and S. Shreve, BM and Stochastic
Calculus.

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Primary Asset Valuation
Discrete-time random walk:
W(t+1) = W(t) + e(t+1);
W(0) = W0;
e~i.i.d. N(0,1)
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Primary Asset Valuation
N(0,1)
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Primary Asset Valuation
Discrete-time random walk refinement:
W(t+) = W(t) + e(t+);
W(0) = W0;
e~i.i.d. N(0, )
This process has the same expected drift and
variance over n periods as the initial process
had in one period.
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Primary Asset Valuation
t=0
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0.25
0.5
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0.75
N(0,1)
1
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Primary Asset Valuation
Set   dt
W(t+ dt) = W(t) + e(t+ dt);
W(0) = W0;
e~i.i.d. N(0, dt)
Define dW(t) = W(t+dt) - W(t) white noise,
(dt)a = 0 for any a > 1
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Primary Asset Valuation
Needs["Statistics`NormalDistribution`"]
nor[mu_,sig_]:=Random[NormalDistribution[mu,sig]];
tt=NestList[ (#+nor[0, 0.1])&, 0, 300];
ListPlot[tt,PlotJoined->True,PlotLabel->"Random Walk"];
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Primary Asset Valuation
Random
Walk
0.5
50
100
150
200
250
300
-0.5
-1
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Main Properties
1.
E[dW(t)] = 0
2.
E[dW(t) dt] = E[dW(t)] dt = 0
3.
E[dW(t)2] = dt
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Main Properties (cont.)
4.
Var[dW(t)2] = E[dW(t)4] - E2[dW(t)2] =
3 dt2 - dt2 = 0
5.
E[(dW(t)dt)2] = E [dW(t)2] (dt)2 = 0
6.
Var[dW(t)dt] =
E[(dW(t)dt)4] - E2[dW(t)dt] = 0
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E[dW(t)] = 0
By definition the mean of the normally
distributed variable is zero.
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E[dW(t) dt] = E[dW(t)] dt = 0
The expectation of the product of a random
variable (dW) and a constant (dt) equals the
constant times the expected value of the
random variable.
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E[dW(t)2] = dt
For any distribution with zero mean the
expected value of the squared random variable
is the same as its variance.
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2
Var[dW(t) ]
4
E[dW(t) ]
-
=
2
2
E [dW(t) ]
=
3 dt2 - dt2 = 0
The fourth central moment of the standard
normal distribution is 3, and (dt)2 = 0.
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E[(dW(t)dt)2] = E [dW(t)2] (dt)2 = 0
Follows from properties 2 and 3.
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Var[dW(t)dt] =
E[(dW(t)dt)4] - E2[dW(t)dt] = 0
Follows from properties 2 and 5.
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Important Property
if
Var[f(dW)] = 0
then
E[f(dW)] = f(dW)
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Multiplication Rules
Rule 1.
(dW(t))2 = dt
Rule 2.
(dW(t)) dt = 0
Rule 3.
dt2 = 0
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W(t) is called a standard Wiener process,
or a Brownian motion.
W (0)  W0
t
W (t )  W0 
dW
(

)

 0
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Major Properties of W
1. W(t) is continuous in t.
2. W(t) is nowhere differentiable.
3. W(t) is a process of unbounded variation.
4. W(t) is a process of bounded quadratic variation.
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Major Properties of W
5. The conditional distribution of W(u) given
W(t), for u > t, is normal with mean W(t)
and variance (u-t).
6. The variance of a forecast W(u) increases
indefinitely as u .
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Brownian Motion - BM
The standard BM is useful since many general
stochastic processes can be written in terms of
W.
X(t+1) = X(t) + (X(t),t) + (X(t),t) e(t+1)
X(0) = X0,
e~i.i.d. N(0,1)
generalized drift
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heteroscedastisity
(changing variance)
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Brownian Motion - BM
Choose a shorter time interval :
X(t+) = X(t) + (X(t),t)  + (X(t),t) e(t+ )
X(0) = X0,
e~i.i.d. N(0, )
generalized drift
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heteroscedastisity
(changing variance)
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Brownian Motion - BM
As we let dt we see that
dX(t) = (X(t),t) dt + (X(t),t) dW(t)
X(0) = X0
generalized univariate Wiener process, (diffusion).
dX = (X,t) dt + (X,t) dW,
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X(0) = X0
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Interpretation
How can we interpret the fact that
dX =  dt +  dW
the random variable dX has local mean dt
and local variance 2dt.
A discrete analogy is X =  +  z.
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Arithmetic BM
dX =  dt +  dW
Let (X,t) = , and (X,t) =  two constants.
Then X follows an arithmetic Brownian
Motion with drift  and volatility .
This is an appropriate specification for a
process that grows at a linear rate and exhibits
an increasing uncertainty.
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Arithmetic BM
dX =  dt +  dW
1. X may be positive or negative.
2. If u > t, then Xu is a future value of the
process relative to time t. The distribution of
Xu given Xt is normal with mean
Xt+  (u-t) and standard deviation (u-t)1/2.
3. The variance of a forecast Xu tends to
infinity as u does (for fixed t and Xt).
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Arithmetic BM
dX =  dt +  dW
X


time
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Arithmetic BM
dX =  dt +  dW
X


time
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Arithmetic BM
dX =  dt +  dW
Is appropriate for variables that can be
positive and negative, have normally
distributed forecast errors, and have forecast
variance increasing linearly in time.

Example: net cash flows.
 Is inappropriate for stock price.

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Geometric BM
dX = Xdt + XdW
Let (X,t) = X, and (X,t) = X.
Then X follows an geometric Brownian Motion.
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Geometric BM
dX = Xdt + XdW
This is an appropriate specification for a process
that

grows exponentially at an average rate of 

has volatility proportional to the level of the
variable.

It also exhibits an increasing uncertainty.
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Geometric BM
dX = Xdt + XdW
1. If X(0) > 0, it will always be positive.
2. X has an absorbing barrier at X = 0. Thus if
X hits zero (a zero probability event) it will
remain there forever.
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Geometric BM
dX = Xdt + XdW
3. The conditional distribution of Xu given Xt
is lognormal. The conditional mean of ln(Xt)
is ln(Xt) +  (u-t) - 0.5 2(u-t) and conditional
standard deviation of ln(Xt) is (u-t)1/2.
ln(Xt) is normally distributed.
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Geometric BM
dX = Xdt + XdW
4. The conditional expected value of Xu is
Et[Xu] = Xtexp[ (u-t)]
5. The variance of a forecast Xu tends to
infinity as u does (for fixed t and Xt).
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Geometric BM
dX = Xdt + XdW
X
time
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Geometric BM
dX = Xdt + XdW
X
time
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Geometric BM
dX = Xdt + XdW
Is often used to model security values, since
the proportional changes in security price are
independent and identically normally
distributed (sometimes).
 Example: currency price, stocks.
 Is inappropriate for dividends, interest
rates.

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Mean Reverting Process

dX = (-X)dt + X dW
Ornstein-Uhlenbeck when  = 1

Let (X,t) = (-X), and (X,t) = X , where
0 - speed of adjustment
 - long run mean
 - volatility
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Mean Reverting Process

dX = (-X)dt + X dW
This is an appropriate specification for a process
that has a long run value but may be beset by
short-term disturbances.
We assume that , , and  are positive for
simplicity.
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Mean Reverting Process

dX = (-X)dt + X dW
1. If X(0) > 0, it will always be positive.
2. As X approaches zero, the drift is positive
and volatility vanishes.
3. As u becomes infinite, the variance of a
forecast Xu is finite.
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Mean Reverting Process

dX = (-X)dt + X dW
4. If  = 0.5, the distribution of Xu given Xt for
u > t is non-central chi-squared, the mean of
the distribution is:
(Xt- ) exp[-(u-t)] + 
the variance of the distribution is (CIR 85):
 2  (u t ) 2 (u t )  2
 ( u t ) 2




Xt
e
e

1 e

2
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Mean Reverting Process

dX = (-X)dt + X dW
X

time
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Mean Reverting Process

dX = (-X)dt + X dW
X

time
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Mean Reverting Process

dX = (-X)dt + X dW
SeedRandom[2]
tt=NestList[
(#+0.3(1-#)+0.1*#*nor[0,0.1])&,
1.01,130];
ListPlot[tt,PlotJoined->True, Axes->False];
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Mean Reverting Process

dX = (-X)dt + X dW

Is often used to model economic variables
and do not represent traded assets.

Example: interest rates, volatility.
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Ito’s lemma
Consider a real valued function f(X): RR.
Taylor series expansion:
1
2
f ( X  )  f ( X )  f X ( X )  f XX ( X )
2
1
3
3
 f XXX ( X )  o( )
6
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Ito’s lemma
If X is a “standard” variable, then 2 is o()
f ( X  )  f ( X )  f X ( X )  o()
f ( X  dX )  f ( X )  f X ( X )dX
df ( X )  f X ( X )dX
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Ito’s lemma
If X is a stochastic variable (following
diffusion) then the term dX2 does NOT
vanish.
dX  dt  dW
dX    (dt )
2
2
dX    dt
2
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2
2
 2dtdW   (dW )
2
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2
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Ito’s lemma
1
2
f ( X  dX )  f ( X )  f X ( X )dX  f XX ( X )dX 
2
1
2
df ( X )  f X ( X )dX  f XX ( X )dX 
2
1
2
df  f X dX  f XX dX 
2
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Ito’s lemma
If f = f(X,t) and dX = dt + dW, then


df  f X  0.5 f XX  f t dt  f X dW
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Financial Applications A
Suppose that a security with value V
guarantees $1dt every instant of time forever.
This is the continuous time equivalent of a
risk-free perpetuity of $1. If the risk-free
interest rate is constant r, what is the
(discounted) value of the security?
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Financial Applications A
1. V = V(t), there are NO stochastic variables.
dV = Vtdt
2. The expected capital gain on V is
ECG = E[dV] = Vtdt
3. The expected cash flows to V is ECF = 1 dt
4. The total return on V is
ECG + ECF = (Vt+1)dt
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Financial Applications A
5. Since there is no risk, the total return must
be equal to the risk-free return on V, or rVdt.
(Vt+1) dt = r V dt
6. Divide both sides by dt:
Vt = rV - 1
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Financial Applications A
Vt = rV - 1
DSolve[ V'[t]==r*V[t]-1, V[t], t ]
V(t) = c Exp[r t] + 1/r
given V(0) one can find c
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Financial Applications B
Suppose that X follows a geometric Brownian
motion with drift  and volatility . A security
with value V collects Xdt continuously forever.
V represents a perpetuity that grows at an
average exponential rate of , but whose risks in
cash flow variations are considered
diversificable. The economy is risk-neutral, and
the risk-free interest rate is constant at r. What is
the value of this security?
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Financial Applications B
1. V = V(X), since V is a perpetual claim, its
price does not depend on time.
dV = VxdX + 0.5 VxxdX2,
dX = Xdt + XdW,
dX2= 2X2dt
dV = [XVx+0.5 2X2Vxx]dt +XVxdW
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Financial Applications B
2. The expected capital gain:
ECG = E[dV] = [XVx+0.5 2X2Vxx]dt
since E[dW] = 0
3. The Expected cash flow:
ECF = X dt
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Financial Applications B
4. Total return:
TR = ECG + ECF =
[XVx+X+0.52X2Vxx]dt
5. But the return must be equal to the risk free
return on the same investment V.
rVdt = [XVx+X+0.52X2Vxx]dt
6. Thus the PDE:
rV = XVx+X+0.52X2Vxx
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Financial Applications B
rV = XVx+X+0.52X2Vxx
there are several ways to solve it. One can guess
that doubling X will double the price V.
If V is proportional to X, then V = X, Vx= , and
Vxx=0, then the equation becomes
r X= X+X
 = 1/(r- )
V(X) = X/(r- )
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