using vectors as inputs for a function

1 visualización (últimos 30 días)
Locks
Locks el 9 de Mayo de 2013
Hi,
I have this function here:
function HestonCallPriceF
x0 = [1.1768 0.082259 0.83746 -0.54589]; %parameters: psi m xi rho
PC = 1; F = 100; K = 100; T = 0.2; r = 0.04; V0 = 0.02;
CallPriceF(PC, F, K, T, r, V0, x0)
%Delta
epsD = abs(F)*eps^(1/6);
Delta=(CallPriceF(PC, F+epsD, K, T, r, V0, x0)-CallPriceF(PC, F-epsD, K, T, r, V0, x0))/(2*epsD)
%Vega
epsV = abs(V0)*eps^(1/6);
Vega=(CallPriceF(PC, F, K, T, r, V0+epsV, x0)-CallPriceF(PC, F-epsD, K, T, r, V0-epsV, x0))/(2*epsV)
Which is working fine for those inputs:
PC = 1; F = 100; K = 100; T = 0.2; r = 0.04; V0 = 0.02;
Which I have used above. Indest of having this inputs, I would like the function to compute me the delta and vega for each row of a matrix and therefore I tried to replace those iputs:
PC = 1; F = 100; K = 100; T = 0.2; r = 0.04; V0 = 0.02;
with this here:
PC=data9(1,3); F=data9(1,1); K=data9(1,2); T=data9(1,5); r=data9(1,7); V0=(data9(1,8)/100)^2;
For some reason this is not working.
I would like to store the deltas and vegas in two additional vectors which I then add to the existing matrix called data9
is there a way to do this or where is the mistake I am making?
  1 comentario
Jordan Monthei
Jordan Monthei el 9 de Mayo de 2013
why not have a loop that goes through each element of your input vectors one at a time and calculates the delta vega with the output then being placed in an output vector?

Iniciar sesión para comentar.

Respuesta aceptada

Matt Tearle
Matt Tearle el 9 de Mayo de 2013
Editada: Matt Tearle el 9 de Mayo de 2013
"this is not working" What is the error message you're getting?
Is there any reason to have HestonCallPriceF be a function? It has no inputs or outputs. Why not just comment out the function declaration line and run it as a script? Then you can at least see what's going on a bit easier, without having to go fully into debug mode.
In fact, the problem may well be that it's a function -- I don't see anywhere that data9 is defined within the function. If data9 is a variable in your base workspace and you're expecting it to be available within HestonCallPriceF, then that's your problem right there.
  3 comentarios
Matt Tearle
Matt Tearle el 9 de Mayo de 2013
It sounds like you're calling HestonCallPriceF without any inputs. (MATLAB doesn't care about this until you try to use an input that you haven't provided. In this case, that happens on line 4, where you reference size(data9,1).) If you have data9 in the base workspace, you need to call HestonCallPriceF with that as an input explicitly:
>> HestonCallPriceF(data9)
Locks
Locks el 9 de Mayo de 2013
it's working now, thanks for your help!!

Iniciar sesión para comentar.

Más respuestas (0)

Categorías

Más información sobre Entering Commands en Help Center y File Exchange.

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by