 # Monte Carlo Method Problems

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## Transcribed Text

Mathematics of Finance Background The script MonteCarlo2.m on Blackboard generates 10,000 year long daily price paths for a stock with initial price \$100, expected annual return r = 10% (=0.1) and annual volatility σ = 30%. The matrix C of size 10,000 x 252 stores all those paths in its rows. The script then calculates an approximation to E[ln(ST )] where ST is the final price of the stock. It does so by simply taking the mean of the last column of C: LogLastPrice = log(C(:, 252)); ←− C(:, 252) is the 252nd column; “:” stands for “any row” The Task Use the above script as a template (if you’re not working with Matlab) or modify its code to 1. Generate 100 (ideally 10,000 or more) 3 month long price paths for a stock with initial price \$50, expected annual return r = 10% (=0.1) and annual volatility σ = 30%. Assume that there are 21 trading days in a month. 2. Suppose that you want to sell a put option on the stock in part 1 with the strike price K = \$50 and expiration date in 3 months. You need to find how much to charge your customers for the option. You could proceed in two steps • Find the expected payoff of the put option E[|K − ST | +] where ST is the stock’s price in 3 months (i.e., in 63 days). You would use here the same method MonteCarlo2.m uses to find E[ln(ST )]. • Finde the price by discounting the expected payoff obtained above to the present, i.e., with the factor1 e −r(T −t) . • In order to make the obtained price robust, run your script several times and take the average of the results. How much would you charge for the option? 1 It seems improbable that using the same discount rate as the expected return of the stock would lead to a correct result; surprisingly, we will learn later that it does. MonteCarlo2.m script s = 100; %Stock's price now mu = 0.1; %Annual return rate dt = 1/252; %Time increment in years sigma = 0.3; %Annual volatility (risk) C = []; %The matrix in which each row will hold a price path for q=1:10000 M =[]; M = normrnd(1+mu*dt, sigma*sqrt(dt), 1, 252); %252 (i.e. the whole year) %of draws from normal %distribution with expected %value=1+mu*dt and std= % sigma*sqrt(dt) DP = cumprod(M); %Consecutive cumulative products of the above draws TS = s*DP; %To obtain the time series of price we need to multiply %by the initial price C = [C;TS]; end %Set the range on the plot and plot %plot(C'); ... %set(gca,'YLim',[50, 150]); set(gca,'XLim',[0, 253]); LogLastPrice = log(C(:, 252)); mean(LogLastPrice) %This line will print the simulated expected value %of the log of the price S_T

## Solution Preview

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function Solution()
%Q1
Npath= 100; % number of simulated paths
mu=0.1;%annual return for stock
sigma=0.3;
S0=50;
dt=1/(21*12); %time increment in years
T=1/4;
Nt=21*3;
t=0:dt:T;
DW=zeros(Npath,Nt+1); %initialize the increments of brownian motions to zero
%every row contains values for the a path and is of size Nt+1
% DW(q,i)= W((i+1)*dt)-W(i*dt) where q is the path
%since W((i+1)*dt)-W(i*dt) ~ N(0,dt)=sqrt(dt)*N(0,1) and they are
%independent
DW(:,2:Nt+1)=normrnd(0,1,Npath,Nt)*sqrt(dt);
W=cumsum(DW,2); % cumulative sum of each row
S=S0*exp((mu-sigma^2/2)*repmat(t,Npath,1)+W);...
\$78.00 for this solution

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