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This page contains a graphical simulation of the stochastic
differential equation,
dS
S
= mdt + sdX,
(1)
where dt is an infinitesimal time increment and dX is a
random variable drawn from N(0,dt),
a normal distribution with zero mean
and standard deviation Ödt (dX is a Wiener process).
This equation is often used to model stock prices and then
the drift, m, is the expected rate
of return on the asset, such as the interest on a risk-free bank
deposit (e.g. m = 0.15 for a 15% interest rate). The
volatility, s, represents the standard deviation on the
return dS/S and the random variable dX is included to model
stock market uncertainty. See [2] or
[5] for details.
It must be emphasized that (1) is
only an intuitive model.
Stock prices behave in a very complicated way and depend on a
great many factors: they neither know nor care anything about
differential equations and so, as the man said
(www.brunel.ac.uk/~icsrsss/sounds/cannot.wav"),
we have to use our intuition.
NOTE: for reasons I have not yet figured out this applet doesn't
work properly in Netscape running under GNU/Linux.
To use the java applet follow these steps:
Enter the initial condition, S(0) = S0, and the stopping
time, T. (Press ``enter'' each time.)
Enter the drift m, and the volatility s.
(Press ``enter'' each time.)
Select the colour of the graph, the accuracy (No. of time steps)
and press ``Plot''.
Note that to change either S0 or T you must press ``clear'' and
start again-otherwise the axes will not be correct.
You must press the ``enter'' key after changing a TextField value,
otherwise Java doesn't realise that you have altered anything
(i.e. you must generate an event on the component).
where dX ~ N(0,dt), we can calculate expectations (with respect to
N(0,dt)) using
E(dX) = 0 and E(dX2) = dt. This gives:
E(dW)
=
E(AdX + B dt) = A E(dX) + B dt = B dt,
E(dW2)
=
E(A2 dX2 + 2AB dX dt + B2 dt2) = A2 dt + B2 dt2,
Var(dW)
=
E(dW2) - E(dW)2 = A2 dt.
Now, what about (1)? We can fit this into these calculations
by choosing A = sS and B = mS, thus:
E(dS)
=
mS dt,
E(dS2)
=
s2 S2 dt + m2 S2 dt2,
Var(dS)
=
s2 S2 dt.
Convinced? you shouldn't be! These are in fact correct but it is
far from obvious why they are. In our first set of calculations we
tacitly assumed that A and B were deterministic (i.e.
non-random), and this allowed us to take them ``outside'' of the
expectaion operator. In the second set of calculations A and B
both depend on S, a stochastic process (and therefore random), so
how can we justify taking A and B (i.e. S) outside of the
expectation operator?
The answer to this riddle lies with the manner in which we interpret
(1). The key is that dS is the increment away from
S(t) during the small time dt. Thus, when we take the expectations
S(t) is already known, and, therefore, non-random.
Strictly speaking, the expectaions above are conditional on the
information S(t) being known at time t.
Denote by N(m, s2) the normal distribution with mean m
and variance s2. If X is a random variable with such a
normally distributed logarithm: lnX ~ N(m, s2),
then X is said to be lognormally distributed.
To determine the probability density function, q(x), for the
lognormal distribution we define Y : = lnX, then
Y ~ N(m, s2), and the probability measure for
N(m, s2) is,
p(y) dy =
1
sÖ(2p)
exp
æ ç
è
-
(y-m)2
2s2
ö ÷
ø
dy for y Î (-¥, ¥).
Now, using x = ey, so that dy = dx/x, we can find the probability
measure for X by the change-of-measure formula,
p(y) dy = p(lnx)
dx
x
= q(x) dx.
Filling in the details we therefore have,
p(y) dy
=
1
sÖ(2p)
exp
æ ç
è
-
(y-m)2
2s2
ö ÷
ø
dy,
=
1
xsÖ(2p)
exp
æ ç
è
-
(lnx-m)2
2s2
ö ÷
ø
dx,
=
q(x) dx.
Therefore, if lnX ~ N(m,s2), then X itself
has the probability density function,
q(x) =
1
xsÖ(2p)
exp
æ ç
è
-
(lnx - m)2
2s2
ö ÷
ø
, for x Î (0,¥).
This is the density function for the lognormal distribution, and we denote
it by eN(m,s2).
After some integrations one can show that the moment
generating function for this distribution is,
M(t) =
¥ å
n = 0
tn
n!
exp
æ ç
è
n
2
(2m+ ns2)
ö ÷
ø
.
The centered moments now follow by differentiation ``at t = 0'',
and give the following expectations with respect to eN(m,s2):
E(X)
=
em+ s2/2,
(2)
E(X2)
=
e2m+ 2s2,
(3)
:
E(Xm)
=
emm + m2s2/2.
(4)
The variance is then,
Var(X) = E(X2) - E(X)2 = e2m+s2( es2 - 1).
(5)
These calculations are useful when choosing parameters in the
binomial tree option pricing technique (see, for example,
www.brunel.ac.uk/~icsrsss/finance/options/binomial).
The reason for this is demonstrated in
Section 6, but to make sense of that section
we first need Ito's lemma.
Equation (1) is one of the few Stochastic
Differential Equations to have a closed-form solution. It is:
S(t) = exp
æ ç
è
(m-
1
2
s2)t + sX(t)
ö ÷
ø
,
(6)
and this can be confirmed with Ito's lemma.
Ito's lemma is often referred to as Taylor's series for
stochastic calculus. This is not quite right, unless one interprets
the ``differences'' in Taylor's series as a kind of shorthand for
definite integrals. Stochastic differential equations don't exist
(at least not in the way classical differential equations exist), and
expressions such as (1) are only a shorthand
for Ito integrals. Equation (1) actually means,
S(t) = S(t) +
ó õ
t
t
mS dt+
ó õ
t
t
sS dX,
for t £ t and where the rightmost integral is taken in the sense
of Ito. Neftci, [3], is an excellent source for this
stuff, but the other popular books,
[5,4,1] for example, also outline
Ito calculus in treatments of various depths.
All we are attempting here is a summary treatment, so here goes.
Define a process W by,
dW = A(W,t) dt + B(W,t) dX,
where dX ~ N(0,dt) (normal: mean zero, variance dt) is a
Wiener process and A and B are known, well-behaved, functions.
Given a function f(W,t) Ito's lemma gives us the process followed
by f and, in essence, is the following result
(probably more famous in the
stock exchanges than the universities!),
df =
æ ç
è
¶f
¶t
+ A(W,t)
¶f
¶W
+
B(W,t)2
2
¶2 f
¶W2
ö ÷
ø
dt+B(W,t)
¶f
¶W
dX.
(7)
To confirm that (6) is indeed the solution
of (1) we use Ito's lemma with S = f, A = 0
and B = 1, giving W = X and,
dS
=
æ ç
è
¶f
¶t
+
1
2
¶2 f
¶X2
ö ÷
ø
+
¶f
¶X
dX,
=
æ ç
è
(m-
1
2
s2) S +
1
2
s2 S
ö ÷
ø
dt+ sS dX,
=
mS dt + sS dX.
Easy! Well, not really. We didn't actually solve the
stochastic differential equation, we only verified that we had
a solution. In general this is about as good as it gets. You have to
get the solution by luck, guesswork, intuition, ...2, and then
use Ito's lemma to verify that it really is a solution. Stochastic
differential equations are much harder to solve than their
classical counterparts.
In the next section we look a little harder at the solution of
(1) and confirm that S is lognormally distributed.
Starting with (1) let f(S,t) be a function of
S and t, then Ito's lemma, in the form,
df = sS
¶f
¶S
dX+
æ ç
è
mS
¶f
¶S
+
s2 S2
2
¶2 f
¶S2
+
¶f
¶t
ö ÷
ø
dt,
can be used to find the process (i.e. the stochastic differential
equation) followed by f.
For example, let f = lnS, then
¶f
¶S
=
1
S
,
¶2 f
¶S2
=
-1
S2
and
¶f
¶t
= 0.
Using these in the Ito formula gives,
df =
æ ç
è
m-
1
2
s2
ö ÷
ø
dt + sdX.
Since dX ~ N(0,dt) we can perform calulations similar to
those in Section 3 and conclude that,
df ~ N( (m-s2/2)dt, s2 dt).
Integrating from t < t to t and setting k: = t-t we get,
f(t) - f(t) ~ N( (m-s2/2)k, s2 k),
or, translating the mean,
f(t) ~ N( f(t) + (m-s2/2)k, s2 k).
Therefore lnS(t) is normally distributed with
mean = lnS(t) + (m-s2/2)k,
(using f(t) = lnS(t)) and,
variance = s2 k.
In other words, S(t) is lognormally distributed.
Some (conditional on information at t)
statistics for S(t) therefore follow from equation
(4) once we make the simultaneous
replacements:
m
¬
lnS(t) +
æ ç
è
m-
s2
2
ö ÷
ø
k,
s2
¬
s2 k.
Thus:
E(S(t)|S(t))
=
S(t) emk,
E(S(t)2 |S(t))
=
S(t)2 e2mk + s2 k,
Var(S(t)|S(t))
=
S(t)2 e2mk(es2 k - 1).
These results are useful in constructing a Binomial Tree
(see e.g. www.brunel.ac.uk/~icsrsss/finance/options/binomial)
of lognormally distributed asset prices. In this case one would take
t = ti, t = ti-1 and let k be the timestep.