Simulation
Downloads
 Today’s work is in: matlab_lec03.m
 Datasets we need today:
data_msft.m
Histograms: hist()
>>X=[2*ones(3,1); 3*ones(5,1); 7*ones(4,1)];
>>subplot(2,1,1);
>>hist(X); %draws histogram of X
>>subplot(2,1,2);
>>hist(X,[0:.25:10]); %draws histogram of X
%on the interval [0 10] with bins of size .25
 Default is a histogram with 20 bins,
from min(X) to max(X)
 hist(X,n) will keep default min and max
but make n bins
Uniform rv: rand()
 A uniform random variable (rv) has
equal probability of occurring at any
point on its support
>>T=1000;
>>X=rand(T,1); %creates a matrix of size
%(Tx1) of uniform rv’s on (0,1)
>>a=5; b=50;
>>Y=a+b*rand(T,1); %creates a matrix of
%size (Tx1) of unifrom rv’s on (a,a+b)
Using hist()
and rand()
>>subplot(3,1,1);
>>hist(X,[-.25:.025:1.25]); %draws
%histogram of X
>>subplot(3,1,2);
>>hist(Y,[.9*a:(b/30):1.1*(a+b)]); %draw
>>T=100000; Z=a+b*rand(T,1);
>>subplot(3,1,3);
>>hist(Z,[.9*a:(b/30):1.1*(a+b)]);
subplot;
Law of Large
Numbers (LLN)
 Note that E[X]=.5
>>X=rand(5,1); disp(mean(X));
>>X=rand(10,1); disp(mean(X));
>>clear Y;
for i=1:200;
Y(i)=mean(rand(i,1));
end;
>>plot(Y);
 How quickly does Y tend to .5? CLT will tell
us
Discrete rv’s
 Oftentimes you will need to simulate rv’s
with a small number of possible outcomes
 You can use the uniform rv to create
discrete rv’s (ie coinflips)
%x=1 with probability p and 0 with (1-p)
>>p=.25; if rand(1,1)<p; x=1; else; x=0; end;
%x=3 with p=.25, 2 with p=.5, 1 with p=.25
>>y=rand(20,1);
>>x=3.*(y>.75)+2*(y<=.75 & y>.25)+1*(y<=.25);
>>hist(x,[0:.25:4]);
Central Limit
Theorem (CLT)
 Tells us that means of many rv’s
converge to a normal rv
 This is why normals are so common in
nature!
 Let x=uniform rv
 Let y=0 if x<.3 and 1 if x>=.3
 Let z be the mean of j binomial rv’s
 Note that z itself is a rv, in particular,
when j=1, y=z
CLT (cont’d)
 Think of y as a biased coin flip
 Think of z as the mean of j coinflips
>>A=[1 5 10 25 50 100];
for k=1:6;
j=A(k);
for i=1:5000;
x=rand(j,1);
y=(x<.3)*0+(x>=.3)*1;
z(i,k)=mean(y);
end;
end;
CLT (con’d)
>>for k=1:6;
subplot(3,2,k); hist(z(:,k),50);
end;
subplot;
 We will now have 6 plots. Each will have a
histogram of rv z, which is a mean of j
binomial rv’s y.
 What do distributions with high j look
like?
 Ever wonder why distribution of human
heights looks like a normal rv?
Normal rv:
randn()
 Works same as uniform, but produces
a normal of mean 0, standard
deviation 1
>>X=randn(10000,1);
>>subplot(2,1,1); hist(X,50);
 To produce normal rv’s with different
mean or variance, just skew and shift
>>m=1.1; s=.16; X=m+s*randn(10000,1);
>>subplot(2,1,2); hist(X,50);
A simple
security process
 R(t)=mu+sigma*x(t) (x is normal, R is
normal)
 10% annual return and 30% annual
standard deviation are quite typical for
equity
>>T=10000; mu=1.1; sigma=.3;
>>x=randn(T,1);
>>R=mu+sigma*x;
>>subplot; hist(R,50);
 Do you notice anything “strange” about
this process or the histogram?
A better process
 R=exp(mu+sigma*X(t)) (R is lognormal)
 exp(x) is approximately 1+x, so if
want mean of process approximately
1.1, you need x to be approximately
.1
 Can this return be negative?
Calibration Issue:
Jensen’s Inequality
 Jensen’s Inequality: E[f(X)] ≠ f(E[X])
 Stein’s Lemma:
E[exp(X)]=exp(m+.5*s2) where X is
normal rv with mean m, standard
deviation s
 If you want R to have mean exp(m),
than make sure rv X inside of exponent
has mean m-.5*s2
 With non-normal processes (ie jumps),
things will be more complicated
Calibration Issue:
Interval length
 The “right” way to simulate is:

R(t )  exp m * dt  X (t ) * * dt

 X(t) is N(0,1)
 dt=1/T where T is the number simulations per period
 m is the mean per period, σ is the standard deviation
per period
 For example, if one period is one year and we are
simulating monthly, than T=12, m is the annual mean
(ie 10%), σ is the annual standard dev (ie 20%)
 When the length of the period (over which we measure
parameters) is equal to the simulation period, than T=1
and this reduces to what we saw earlier
Calibrate Microsoft
 Get daily microsoft data from CRSP or
course website
>>data_msft;
>>disp(mean(msft(:,4)));
>>disp(std(msft(:,4)));
%Microsoft returns have a daily mean of
%.097% and standard deviation of 2.21%
>>subplot(2,1,1); hist(msft(:,4),[-.2:.01:.2]);
>>axis([-.2 .2 0 800]); xlabel('Actual');
Simulate Microsoft
>>T=3022;
>>mu=.00097-.5*.0221^2; sigma=.0221;
>>x=randn(T,1);
>>r=exp(mu+sigma*x)-1;
>>subplot(2,1,2);
>>hist(r,[-.2:.01:.2]); axis([-.2 .2 0 800]);
>>disp(mean(r)); disp(std(r));
>>disp([skewness(r) skewness(msft(:,4))]);
>>disp([kurtosis(r) kurtosis(msft(:,4))]);
Compare
simulated to actual
 Mean, standard deviation, skewness
match well
 Kurtosis (extreme events) does not
match well
 Actual has much more mass in the tails
(fat tails)
 This is extremely important for option
pricing!
 CLT fails when tails are “too” fat
Next Week:
Modelling Volatility
 How to make tales fatter?
 Add jumps to log normal distribution to
make tails fatter
 Jumps also help with modeling default
 Make volatility predictable:
 Stochastic volatility, governed by state
variable
 ARCH process (2003 Nobel prize, Rob
Engle)
Optional
Homework (1)
 Create a function that will simulate
microsoft stock using a log-normal
process
 The function should take in
arguments mu (mean), sigma
(standard deviation), and T (number
of days)
 Its output should be a vector of daily
returns