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Vectorization of a function

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Francesco Rossi
Francesco Rossi on 30 Sep 2019
Edited: the cyclist on 30 Sep 2019
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!

  2 Comments

Geoff Hayes
Geoff Hayes on 30 Sep 2019
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
Francesco Rossi on 30 Sep 2019
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.

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Accepted Answer

the cyclist
the cyclist on 30 Sep 2019
Edited: the cyclist on 30 Sep 2019
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?

  2 Comments

the cyclist
the cyclist on 30 Sep 2019
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
the cyclist on 30 Sep 2019
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).

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More Answers (1)

meghannmarie
meghannmarie on 30 Sep 2019
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));

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