1. Simulation

2. Downloads  Today’s work is in: matlab_lec03.m  Datasets we need today: data_msft.m

3. 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

5. 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)

6. 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;

8. 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

10. 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.25)+1*(y<=.25); >>hist(x,[0:.25:4]);

12. 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  Let z be the mean of j binomial rv’s  Note that z itself is a rv, in particular, when j=1, y=z

13. 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)*1; z(i,k)=mean(y); end;

14. 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?

16. 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);

18. 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?

20. A better process  R=exp(mu+sigma*X(t)) (R is log- normal)  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?

21. Calibration Issue: Jensen’s Inequality  Jensen’s Inequality: E[f(X)] ≠ f(E[X])  Stein’s Lemma: E[exp(X)]=exp(m+.5*s 2 ) 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*s 2  With non-normal processes (ie jumps), things will be more complicated

22. Calibration Issue: Interval length  The “right” way to simulate is:  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

23. 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');

24. 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))]);

26. 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

27. 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)

28. 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