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Simulation

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

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Simulation

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  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

  4. 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)

  5. 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;

  6. 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

  7. 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]);

  8. 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

  9. 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;

  10. 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?

  11. 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);

  12. 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?

  13. 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?

  14. 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

  15. 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

  16. 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');

  17. 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))]);

  18. 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

  19. 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)

  20. 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

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