Problems: Quote for six-month American style euro currency options on plain vanilla, Max[S-K,0]and 〖Max[S-K,0]〗^0.5. Exchange rate S_0=\$1.3721 /euro
Six-month continuously compounded inter-bank rates: r=0.4472%,r_f=1.2840%.
Assumptions:The exchange rate for euro follows an iid log normal price changes and volatility is constant.
Methodology:Binomial Model is used to price American currency options on euros.
We calculate the payoffs at time T and discount payoffs to the prior time step. Under the risk neutral probability measure,
c_(t-1)=(q×c_u+(1-q)×c_d)/R
Since these two options are American styles, we need to check for optimal early exercise at each node of the binomial tree. For these two currency options, we check Max[S-K,c_(t-1),0] and Max[〖Max[S-K,0]〗^0.5,c_(t-1) ] . Matlab is the software used to implement the binomial model.
--Parameters:1. Time steps n
As the number of time steps n increases, we can apply CRR model and the binomial model approaches real world price changes. We choose n=80,h=T/n=0.5/80=0.00625.
2. u and dWe choose CRR model to define u and d for binomial model.
u=e^(σ√h) ; d=1/u=e^(-σ√h)
Where h is the length of the binomial times step and σ is the annualized log price change volatility.
3. Annualized log price change volatility σThe daily log changes and daily squared log changes for two year exchange rates from 11/5/2008 – 11/5/2010 are as follows. We consider the volatility to be constant since 05/01/2009. Thus, we choose the historical prices from 05/01/2009-11/5/2010 and apply the volatility of the daily log changes as an estimate. Then the annualized log price change volatility equals the square root of the trading days in one year (252 days) times the daily log price change volatility.
σ=√252×σ_daily
4. Risk Neutral Measure QOptions on currencies can be regarded as an asset providing a yield at the foreign risk-free rate of interest. Thus, the risk neutral probability measure Q :
q=(e^((r-r_f ) )-d)/(u-d) ;
1-q= 〖u- e〗^((r-r_f ) )/(u-d)
5. Strike Price KSet strike price K from \$1.3000/euro to \$1.5000/euro with \$0.005 per euro increments.
Reference: 1. John C. Hull.Options, Futures and Other Derivatives, 7th edition. Prentice Hall. 2008.
2. Base SAS 9.2 Procedures Guide. SAS Publishing. 2009

`**********************AUTHOR(DAPANGMAO)----HCHAO8@GMAIL.COM***********************************;****************(1) CONSTRUCT € TO \$ EXCHANGE RATIO VECTOR******************;data vector; attrib value informat=dollar10.4 format=dollar10.4 ; StrikeS=1.3000;   StrikeE=1.5000; Increment=0.005; do value=StrikeS to StrikeE by  Increment;  drop StrikeS StrikeE Increment; output; end;run;**************(2) BUILD THE FUNCTION TO EVALUATE OPTION PRICES********;proc fcmp outlib = myfunc.finance.subrout;subroutine mysubr(T, n, r, rf, s0, sigma, c1[*], c2[*]);/*Two vectors are output arguments*/  outargs c1, c2;/*Inside calculation from input arguments*/ dt=T/n; length=dim(c1); u=exp(sigma*sqrt(dt));       d=1/u; q=(exp((r-rf)*dt)-d)/(u-d);/*Announce 4 arrays -- 1 vector and 3 matrixes */ array k[1]/ nosymbols; array s[1] /nosymbols; array x1[1] /nosymbols; array x2[1] /nosymbols;  n=n+1;/*The sizes of the arrays are specified*/ call dynamic_array(s, n, n); call dynamic_array(x1, n, n); call dynamic_array(x2, n, n); call dynamic_array(k, length);/*Read the exchange ratio into function*/ rc=read_array('vector', k);/*Assign values to S matrix */ call zeromatrix(s); S[1,1]=S0;  do _col=2 to n ;    do _row=1 to n;        S[_row,_col]=S0*u**(_col-_row)*d**(_row-1);        if _row gt _col then S[_row, _col]=0;    end; end;/*Generate final option vectors */  do i=1 to length;   x=k[i];   call zeromatrix (x1);   call zeromatrix (x2);   do j=1 to n;      x1[j,n]=max(S[j,n]-x,0);         x2[j,n]=max(S[j,n]-x,0)**0.5;        end;   do _col=(n-1) to 1 by -1;          do _row=1 to (n-1) by 1;              h=exp(-r*dt)*(q*x1[_row,_col+1]+(1-q)*x1[_row+1,_col+1]);              h2=exp(-r*dt)*(q*x2[_row,_col+1]+(1-q)*x2[_row+1,_col+1]);              x1[_row,_col]=max(S[_row,_col]-X,h,0);                  x2[_row,_col]=max( max(S[_row,_col]-x ,0) **0.5, h2);            end;                   c1[i]=x1[1,1];     c2[i]=x2[1,1];  end; end;endsub;run;quit;***********SOME INITIAL VALUES*****************;/*T=1/2;*//*n=80;*//*r=0.004472;    *//*rf=0.01284;*//*S0=1.3721;*//*Sigma=0.1074*/**************(3) MEASURE THE LENGTH OF PRICING VECTOR**************;proc sql; select count(*) into: vecnum from vector;quit;**************(4) USE THE SUBROUTINE TO GENERATE TWO VECTORS********************;options cmplib = (myfunc.finance);data final;array c1[&vecnum] _temporary_;array c2[&vecnum] _temporary_;call mysubr(0.5, 80, 0.004472, 0.01284, 1.3721, 0.1074, c1, c2);   /*subroutine mysubr(T, n, r, rf, s0, sigma, c1[*], c2[*]);*/do i=1 to dim(c1); c1value= c1[i]; c2value=c2[i]; output;end;run;****************TEST PASSED ----------- END****************************;`