Vectorization of a function

I have dotted all the variable that are defined as a vector, but matlab gives me an error.

Do you see where is my problem?

function [C] = BlackScholesCall(S,K,t,r,sigma)
% Calculates the price of a call option
% INPUT   S 1x1 ... Current stock price (underlying)
%         K 1x1 ... Strike price
%         t 1x1 ... Time to maturity
%         r 1x1 ... Risk-free interest rate
%         sigma 1x1 ... standard deviation (volatility of the underlying)
% OUTPUT  C 1x1 ... The price of a call option
% USAGE   BlackScholesCall(S,K,t,r,sigma)

d1=(log(S/K.))+(r+(1/2)*sigma^2)*t)/(sigma*sqrt(t));
d2=d1-sigma*sqrt(t.);
C=(S*normcdf(d1))-(K.*(exp(-r*.t))*normcdf(d2));
end

Code to call the function:

S = 22
K = 20:25
t = 0.1:0.1:0.6
r = 0.02
sigma = 0.25
C = BlackScholesCall(S, K, t, r, sigma)

Thank you very much!

[Geoff Hayes]

Francesco - please copy and paste the full error message to this question. In the function header, you have

t 1x1 ... Time to maturity

which implies that t is a scalar...but in your input to this function, you define t as an array

t = 0.1:0.1:0.6

Which should it be - a scalar or an array? Perhaps this is the problem...you are passing in an array but the code is expecting a scalar? Are you the author of BlackScholesCall?

[Francesco Rossi]

Error in solution: Line: 12 Column: 13

Invalid expression. When calling a function or indexing a variable, use parentheses. Otherwise, check for mismatched delimiters.

This is the error message I got if I run the function.

Thank you for spotting the error in the description, it should be an array, as defined later.

Answer 1

[the cyclist]

The problem is that you don't seem to have a grasp on how the "." syntax actually works. (Sorry if that sounds harsh!)

For example, you have

log(S/K.)

I'm not sure what you intend there -- just stating that K is a vector? -- but that's just not how it works. You don't just "dot the vectors". Oversimplifying a bit -- you dot the operations not the variables.

I would start by reading this documentation.

Your code is just a bit tricky to fix, because you are also trying to use two vectors of different lengths, and I'm guessing you actually want all combinations of K and t to generate results. So, you actually have two-dimensional input. Is that right?

[the cyclist]

Oh, maybe not as tricky as I thought. Are the values of K and t paired, such that each pair of inputs gives one value of C? In that case, this should work:

d1=(log(S./K))+(r+(1/2)*sigma^2)*t./(sigma*sqrt(t));
d2=d1-sigma*sqrt(t);
C=(S*normcdf(d1))-(K.*(exp(-r*t)).*normcdf(d2));

in place of what you had. Notice how I moved some of your dots to be associated with vector operations, not the variables themselves.

[the cyclist]

Still easier than I expected if you do not want K and t paired, but there just happened to be the same number of inputs for each.

Due to "implicit expansion", you can send in a row vector for K, and a column vector of t, and get all the combinations. Just change your input to this:

t = (0.1:0.1:0.6)'

(in addition to making the changes I suggested to your code).

Answer 2

[meghannmarie]

I think you have some of your dot operators wrong:

d1 = (log(S./K) +(r+(1/2)*sigma^2)*t)/(sigma*sqrt(t));
d2 = d1-sigma*sqrt(t);
C = (S*normcdf(d1)) - (K.*(exp(-1.*t)).*normcdf(d2));