I am currently looking at Binomial Option Pricing. I have written the code below, which works fine, when you enter the variables in one at a time. However, entering each set of values is very tedious, and I need to be able to analyse a large set of data. I have created arrays for each of the variables. But, I keep getting the error; A(I) = B, the number of elements in B must equal I. The function is shown below.
function C = BinC(S0,K,r,sig,T,N);
% PURPOSE:
% To return the value of a European call option using the Binomial method
%-------------------------------------------------------------------------
% INPUTS:
% S0 - The initial price of the underlying asset
% K - The strike price
% r - The risk free rate of return, expressed as a decimal
% sig - The volatility of the underlying asset, expressed as a decimal
% T - The time to maturity, expressed as a decimal
% N - The number of steps
%-------------------------------------------------------------------------
dt = T/N;
u = exp(sig*sqrt(dt));
d = 1/u;
p = (exp(r*dt) - d)/(u - d);
S = zeros(N+1,1);
% Price of underlying asset at time T
for n = 1:N+1
S(n) = S0*(d^(N+1-n))*(u^(n-1));
end
% Price of Option at time T
for n = 1:N+1
C(n) = max(S(n)- K, 0);
end
% Backtrack to get option price at time 0
for i = N:-1:1
for n = 1:i
C(n) = exp(-r*dt)*(p*C(n+1) + (1-p)*C(n));
end
end
disp(C(1))
After importing my data, I entered this in to the command window.
for i=1:20
w(i)= BinC(S0(i),K(i),r(i),sig(i),T(i),N(i));
end
When I enter w, all I get back is w = []. I have no idea how I can make A(I) = B. I apologise, if this is a very silly question, but I am new to Matlab and in need of help. Thanks
Your function computes an entire vector C, but displays only C(1). This display is deceptive: it makes you think the function is returning a scalar, but it's not: it's returning the entire vector C, which you try to store into a scalar location.
The solution is simple: Change your function definition to this (rename the output variable):
function out = BinC(S0,K,r,sig,T,N);
Then at the last line of the function, remove the disp, and replace it with
out = C(1);
To verify all of this (compare with your non-working example), try calling it by itself at the command line, and examine the output.
Related
Suppose I have a CuArray with random zeros and ones and I want to get a random index of CuArray corresponding to value one. For instance,
m = 100;
A = CuArray(rand([0, 1], m));
i = rand(1:m);
while A[i]!=1
i = rand(1:m);
end
Is there a function so that I can not use while looping?
Your construction of A has the following equivalent representation:
using Distributions
n_ones = rand(Binomial(m, 0.5))
one_inds = shuffle(1:m)[1:n_ones]
A = zeros(Int, m)
A[one_inds] .= 1
That is, you first choose the number of ones you are going to set (from a binomial distribution, since you have m independent choices), and then select without repetition that many indices (by just taking the init of all indices, shuffled).
Written this way, choosing a random index of a one is just
rand(one_inds)
I want to create an array that has incremental random steps, I've used this simple code.
t_inici=(0:10*rand:100);
The problem is that the random number keeps unchangable between steps. Is there any simple way to change the seed of the random number within each step?
If you have a set number of points, say nPts, then you could do the following
nPts = 10; % Could use 'randi' here for random number of points
lims = [0, 10] % Start and end points
x = rand(1, nPts); % Create random numbers
% Sort and scale x to fit your limits and be ordered
x = diff(lims) * ( sort(x) - min(x) ) / diff(minmax(x)) + lims(1)
This approach always includes your end point, which a 0:dx:10 approach would not necessarily.
If you had some maximum number of points, say nPtsMax, then you could do the following
nPtsMax = 1000; % Max number of points
lims = [0,10]; % Start and end points
% Could do 10* or any other multiplier as in your example in front of 'rand'
x = lims(1) + [0 cumsum(rand(1, nPtsMax))];
x(x > lims(2)) = []; % remove values above maximum limit
This approach may be slower, but is still fairly quick and better represents the behaviour in your question.
My first approach to this would be to generate N-2 samples, where N is the desired amount of samples randomly, sort them, and add the extrema:
N=50;
endpoint=100;
initpoint=0;
randsamples=sort(rand(1, N-2)*(endpoint-initpoint)+initpoint);
t_inici=[initpoint randsamples endpoint];
However not sure how "uniformly random" this is, as you are "faking" the last 2 data, to have the extrema included. This will somehow distort pure randomness (I think). If you are not necessarily interested on including the extrema, then just remove the last line and generate N points. That will make sure that they are indeed random (or as random as MATLAB can create them).
Here is an alternative solution with "uniformly random"
[initpoint,endpoint,coef]=deal(0,100,10);
t_inici(1)=initpoint;
while(t_inici(end)<endpoint)
t_inici(end+1)=t_inici(end)+rand()*coef;
end
t_inici(end)=[];
In my point of view, it fits your attempts well with unknown steps, start from 0, but not necessarily end at 100.
From your code it seems you want a uniformly random step that varies between each two entries. This implies that the number of entries that the vector will have is unknown in advance.
A way to do that is as follows. This is similar to Hunter Jiang's answer but adds entries in batches instead of one by one, in order to reduce the number of loop iterations.
Guess a number of required entries, n. Any value will do, but a large value will result in fewer iterations and will probably be more efficient.
Initiallize result to the first value.
Generate n entries and concatenate them to the (temporary) result.
See if the current entries are already too many.
If they are, cut as needed and output (final) result. Else go back to step 3.
Code:
lower_value = 0;
upper_value = 100;
step_scale = 10;
n = 5*(upper_value-lower_value)/step_scale*2; % STEP 1. The number 5 here is arbitrary.
% It's probably more efficient to err with too many than with too few
result = lower_value; % STEP 2
done = false;
while ~done
result = [result result(end)+cumsum(step_scale*rand(1,n))]; % STEP 3. Include
% n new entries
ind_final = find(result>upper_value,1)-1; % STEP 4. Index of first entry exceeding
% upper_value, if any
if ind_final % STEP 5. If non-empty, we're done
result = result(1:ind_final-1);
done = true;
end
end
I am working with Bank of America time series data for stock prices. I am trying to store the forecasted value for a specific step ahead (in this case 1:20 steps) in an array. I then need to subtract each value of the array from each value of the test array. Then I have to square each value of the array, sum all the squared values of the array, then divide by N (N = number of steps forecasted ahead).
I have the following so far. Also, the quantmod and fpp libraries are needed for this.
---------Bank of America----------
library(quantmod)
library(fpp)
BAC = getSymbols('BAC',from='2009-01-02',to='2014-10-15',auto.assign=FALSE)
BAC.adj = BAC$BAC.Adjusted
BAC.daily=dailyReturn(BAC.adj,type='log')
test = tail(BAC.daily, n = 20)
train = head(BAC.daily, n = 1437)
Trying to write a function to forecast, extract requisite value (point forecast for time i), then store it in an array where I can perform operations on that array (i.e. - add, multiply, exponentiate, sum the values of the array)
MSE = function(N){
for(i in 1:(N)){
x = forecast(model1, h = i)
y = x$mean
w = as.matrix(as.double(as.matrix(unclass(y))))
p = array(test[i,]-w[i,])
}
}
and we also have:
model1 = Arima(train, order = c(0,2,0))
MSE = function(N){
result = vector("list", length = (N))
for(i in 1:(N)){
x = forecast(model1, h = i)
point_forecast = as.double(as.matrix(unclass(x$mean)))
result[i] = point_forecast
}
result = as.matrix(do.call(cbind, result))
}
Neither of these functions have worked so far. When I run the MSE function, I get the following errors:
> MSE(20)
There were 19 warnings (use warnings() to see them)
> warnings()
Warning messages:
1: In result[i] = point_forecast :
number of items to replace is not a multiple of replacement length
2: In result[i] = point_forecast :
number of items to replace is not a multiple of replacement length
3: In result[i] = point_forecast :
number of items to replace is not a multiple of replacement length
4: In result[i] = point_forecast :
When I run MSE2 function, I get the following ouput:
MSE2(20)
[1] -0.15824
When putting a print statement inside, it printed out 'p' as a singular number, just like above (even though that had been run for i = 20). The x,y, and w variable in the MSE2 function act as vectors as far as storing the output, so I do not understand why p does not as well.
I appreciate any help in this matter, thank you.
Sincerely,
Mitchell Healy
Your question has two MSE functions: one in the first code block and one in the second code block.
Also, library(forecast) is needed to run Arima and forecast.
My understanding of what you are trying to do in the first paragraph is to compute the 20-step ahead forecast error. That is, what is the error in forecasts from model1 20 days ahead, based on your test data. This can be done in the code below:
model1 <- Arima(train, order = c(0,2,0))
y_fcst<-forecast(model1,h=20)$mean
errors<-as.vector(y_fcst)-as.vector(test)
MSE.fcst<-mean(errors^2)
However, I'm not sure what you're trying to do here: an ARIMA(0,2,0) model is simply modelling the differences in returns as a random walk. That is, this model just differences the returns twice and assumes this twice-differenced data is white noise. There's no parameters other than $\sigma^2$ being estimated.
Rob Hyndman has a blog post covering computing errors from rolling forecasts.
My solution to finding the MSE is below. I used log adjusted daily return data from Bank of America gathered through quantmod. Then I subsetted the data (which had length 1457) into training[1:1437] and testing[1438:1457].
The solution is:
forc = function(N){
forecast = matrix(data = NA, nrow = (N) )
for(i in 1:N){
fit = Arima(BAC.adj[(1+(i-1)):(1437+(i-1))], order = c(0,0,4))
x = forecast(fit, h = 1)
forecast[i,] = as.numeric(x$mean)
}
error = test - forecast
error_squared = error^2
sum_error_squared = sum(error_squared)
MSE = sum_error_squared/N
MSE
}
So, after trying a heavy cicle-based function for calculating the joint entropy of two sources of information, I found this useful MATLAB function, accumarray, and tried the following code:
function e = jointEntropy(fonte1, fonte2)
i = double(fonte1(:))+ 1;
j = double(fonte2(:)) + 1;
subs = [i j];
f = accumarray(subs, ones(length(fonte1), 1));
p = f / length(fonte1);
freq = f ~= 0;
prob = p(freq);
e = -sum(prob.*log2(prob));
end
, where fonte1 and fonte2 are the information sources, 1xN arrays. This worked just fine with examples in which both sources had only positive integer values, but then I tried to use it with audio files (i.e., arrays whose values ranged from -1 to 1) and it kept giving me an error. I tried adding 1 to each array (to range from 0 to 2), multiplying them by 100 and rounding the numbers, in order to obtain non-negative integers, but it still doesn't work.
Any idea/ alternative to this code would be greatly appreciated.
I need to iterate Newton-Raphson in MATLAB. It seems easy but I cannot figure out where I am wrong. The problem is:
For mmm=1:
1) If m=1 take c1=c1b and c2=1-c1 and do the loop for u1,2(i) and p1,2(i)
2)If m=2 take c1=c1+dc and c2=1-c1, and this time do the loop with new c1 and c2 for u1,2(i) and p1,2(i)
3) If m=3 take c1=(c1*st(1)-(c1-dc)*st(2))/(st(1)-st(2)) and do the loop for new c1 and c2.
Then increase the iteration number: mmm=2 ;
mmm keeps count of the number of N-R iterations. The first iteration has mmm=1, the second mmm=2, etc. (This particular run only do 2 iterations).
sumint are inside of the integrals.
I need to plot these figures in the code but MATLAB gives errors below. Please help me.
Relevant part of the code:
ii=101;
u = cell(2, 1);
ini_cond = [0,0];
for i = 1:2;
u{i} = zeros(1,ii);
u{i}(:, ii) = ini_cond(i) * rand(1, 1);
end
for i=1:ii;
fikness=fik*sin(pi.*x);
u{1}(i)=(c1-H1D*(x-0.5)+AD/2.*(x-0.5).^2)./(H1-0.5*fikness-A*(x-0.5));
u{2}(i)=(c2+H1D*(x-0.5)-AD/2.*(x-0.5).^2)./(1.-H1+0.5*fikness+A*(x-0.5));
end
p = cell(2, 1);
q = cell(2, 1);
for i = 1:2;
p{i} = zeros(1,ii);
q{i} = zeros(1,ii);
end
p{1}(1)=0.5*(1.-u{1}(1).^2);
q{1}(1)=0;
p{2}(1)=0.5*(1.-u{2}(1).^2);
q{2}(1)=0;
for i=2:101
q{1}(i)=q{1}(i-1)-dx*(u{1}(i-1)-ub{1}(i-1))./dt;
p{1}(i)=0.5*(1.-u{1}(i).^2)+q{1}(i);
q{2}(i)=q{2}(i-1)-dx*(u{2}(i-1)-ub{2}(i-1))./dt;
p{2}(i)=0.5*(1.-u{2}(i).^2)+q{2}(i);
end
st = zeros(2, length(t));
st(1,:)=p{1}(100)-p{2}(100);
m=m+1;
if m==3;
c1=(c1*st(1)-(c1-dc)*st(2))/(st(1)-st(2));
c2=1-c1;
end
for i = 1:2;
sumint{i} = zeros(1,length(t));
end
sumint = cell(2, 1);
sumint{1}(1)=0.5*(p{2}(1)-p{1}(1));
sumint{2}(1)=0.5*(p{2}(1)-p{1}(1)).*(-1/2);
for i=2:ii-1;
x=(i-1)*dx;
sumint{1}(i)=sumint{1}(i-1)+(p{2}(i)-p{1}(i));
sumint{2}(i)=sumint{2}(i-1)+(p{2}(i)-p{1}(i))*(x-1/2);
end
H1DDOT=-sumint{1}.*dx./rmass;
H1D=H1D+dt*H1DDOT;
H1=H1+dt*H1D;
ADDOT=sumint{2}*dx./rmomi;
AD=AD+dt*ADDOT;
A=A+dt*AD;
H1L=H1+A.*0.5;
H1R=H1-A.*0.5;
H2=1.-H1;
rat1=AD./ADinit;
rat2=ADDOT./AD;
u are the velocities p are the pressures c1,c2 are the camber effects H1DDOT and ADDOT are the second derivation of H1 and A. sum1 and sum2 are the inside of the integrals to define the values of H1DDOT and ADDOT. H1DDOT and ADDOT are functions of time.
As you can see from the message, the error is with this line:
sumint{2}(i)=sumint{2}(i-1)+(p{2}(i)-p{1}(i)).*(x-1/2);
Now, let's find out why:
sumint{2}(i) = ...
This part means you want to insert whatever is on the right side to the ith position of the array in cell sumint{2}. That means, the right side must be a scalar.
Is it?
Well, sumint{2}(i-1)+(p{2}(i)-p{1}(i)) is certainly a scalar, since you use a single value as index to all the vectors/arrays. The problem is the multiplication .*(x-1/2);.
From the code above, it is clear that x is a vector / array, (since you use length(x) etc). Multiplying the scalar sumint{2}(i-1)+(p{2}(i)-p{1}(i)), by the vector x, will give you back a vector, which as mentioned, will not work.
Maybe you want the ith value of x?
There are several other strange things going on in your code, for instance:
for i=1:101;
fikness=fik*sin(pi.*x);
u{1}=(c1-H1D*(x-0.5)+AD/2.*(x-0.5).^2)./(H1-0.5*fikness-A*(x-0.5));
u{2}=(c2+H1D*(x-0.5)-AD/2.*(x-0.5).^2)./(1.-H1+0.5*fikness+A*(x-0.5));
end
Why do you have a loop here? You are not looping anything, you are doing the same calculations 100 times. Again, I guess it should be x(i).
Update:
Your new error is introduced because you changed x into a scalar. Then u = zeros(1,length(x)) will only be a scalar, meaning u{1}(i) will fail for i ~= 1.