I need to code the Monte Carlo algorithm for approximating delta to Matlab and calculate confidence intervals:
but for some reason my code doesn't work, any ideas why?
randn('state', 100)
%Problem and method parameters
S=10; E=9; sigma=0.1; r=0.06; T=1;
Dt=1e-3; N=T/Dt; M=2^17;h=10^(-4);
delta = zeros(M,1);
for i = 1:M
Sfinal = S*exp((r-0.5*sigma^2)*T+sigma*sqrt(T).*randn(M,1));
S_h = (S+h)*exp((r-0.5*sigma^2)*T+sigma*sqrt(T).*randn(M,1));
delta(i) = exp(-r*T).*(max(Sfinal-E,0)-max(S_h-E,0))/h;
end
aM=mean(delta);
bM=std(delta);
conf=[aM-1.96*bM/sqrt(M),aM+1.96*bM/sqrt(M)]
The error message is
"Unable to perform assignment because the left and right sides have a different number of elements."
Any help is appreciated!
You do not need to explicitly write the for loop since you have already vectorized it. In other words, Sfinal and S_h are vectors of length M, and their i-th entries correspond to S_i and S^h_i in the image. Since the right hand side of the delta expression evaluates to a vector of length M, which contains all the values of delta, you should assign that vector directly to delta, not delta(i).
One more thing: the pseudo-code in the image seems to suggest that the same random number should be used for calculating both S_i and S^h_i. This is not the case in your code, since you call randn separately for calculating Sfinal and S_h. I think you should generate the random samples once, save them, and then use it for both calculations.
Here's the code:
randn('state', 100)
%Problem and method parameters
S=10; E=9; sigma=0.1; r=0.06; T=1;
Dt=1e-3; N=T/Dt; M=2^17;h=10^(-4);
xi = randn(M,1);
Sfinal = S*exp((r-0.5*sigma^2)*T+sigma*sqrt(T).*xi);
S_h = (S+h)*exp((r-0.5*sigma^2)*T+sigma*sqrt(T).*xi);
delta = exp(-r*T).*(max(Sfinal-E,0)-max(S_h-E,0))/h;
aM=mean(delta);
bM=std(delta);
conf=[aM-1.96*bM/sqrt(M),aM+1.96*bM/sqrt(M)]
Related
The problem:
If a large number of fair N-sided dice are rolled, the average of the simulated rolls is likely to be close to the mean of 1,2,...N i.e. the expected value of one die. For example, the expected value of a 6-sided die is 3.5.
Given N, simulate 1e8 N-sided dice rolls by creating a vector of 1e8 uniformly distributed random integers. Return the difference between the mean of this vector and the mean of integers from 1 to N.
My code:
function dice_diff = loln(N)
% the mean of integer from 1 to N
A = 1:N
meanN = sum(A)/N;
% I do not have any idea what I am doing here!
V = randi(1e8);
meanvector = V/1e8;
dice_diff = meanvector - meanN;
end
First of all, make sure everytime you ask a question that it is as clear as possible, to make it easier for other users to read.
If you check how randi works, you can see this:
R = randi(IMAX,N) returns an N-by-N matrix containing pseudorandom
integer values drawn from the discrete uniform distribution on 1:IMAX.
randi(IMAX,M,N) or randi(IMAX,[M,N]) returns an M-by-N matrix.
randi(IMAX,M,N,P,...) or randi(IMAX,[M,N,P,...]) returns an
M-by-N-by-P-by-... array. randi(IMAX) returns a scalar.
randi(IMAX,SIZE(A)) returns an array the same size as A.
So, if you want to use randi in your problem, you have to use it like this:
V=randi(N, 1e8,1);
and you need some more changes:
function dice_diff = loln(N)
%the mean of integer from 1 to N
A = 1:N;
meanN = mean(A);
V = randi(N, 1e8,1);
meanvector = mean(V);
dice_diff = meanvector - meanN;
end
For future problems, try using the command
help randi
And matlab will explain how the function randi (or other function) works.
Make sure to check if the code above gives the desired result
As pointed out, take a closer look at the use of randi(). From the general case
X = randi([LowerInt,UpperInt],NumRows,NumColumns); % UpperInt > LowerInt
you can adapt to dice rolling by
Rolls = randi([1 NumSides],NumRolls,NumSamplePaths);
as an example. Exchanging NumRolls and NumSamplePaths will yield Rolls.', or transpose(Rolls).
According to the Law of Large Numbers, the updated sample average after each roll should converge to the true mean, ExpVal (short for expected value), as the number of rolls (trials) increases. Notice that as NumRolls gets larger, the sample mean converges to the true mean. The image below shows this for two sample paths.
To get the sample mean for each number of dice rolls, I used arrayfun() with
CumulativeAvg1 = arrayfun(#(jj)mean(Rolls(1:jj,1)),[1:NumRolls]);
which is equivalent to using the cumulative sum, cumsum(), to get the same result.
CumulativeAvg1 = (cumsum(Rolls(:,1))./(1:NumRolls).'); % equivalent
% MATLAB R2019a
% Create Dice
NumSides = 6; % positive nonzero integer
NumRolls = 200;
NumSamplePaths = 2;
% Roll Dice
Rolls = randi([1 NumSides],NumRolls,NumSamplePaths);
% Output Statistics
ExpVal = mean(1:NumSides);
CumulativeAvg1 = arrayfun(#(jj)mean(Rolls(1:jj,1)),[1:NumRolls]);
CumulativeAvgError1 = CumulativeAvg1 - ExpVal;
CumulativeAvg2 = arrayfun(#(jj)mean(Rolls(1:jj,2)),[1:NumRolls]);
CumulativeAvgError2 = CumulativeAvg2 - ExpVal;
% Plot
figure
subplot(2,1,1), hold on, box on
plot(1:NumRolls,CumulativeAvg1,'b--','LineWidth',1.5,'DisplayName','Sample Path 1')
plot(1:NumRolls,CumulativeAvg2,'r--','LineWidth',1.5,'DisplayName','Sample Path 2')
yline(ExpVal,'k-')
title('Average')
xlabel('Number of Trials')
ylim([1 NumSides])
subplot(2,1,2), hold on, box on
plot(1:NumRolls,CumulativeAvgError1,'b--','LineWidth',1.5,'DisplayName','Sample Path 1')
plot(1:NumRolls,CumulativeAvgError2,'r--','LineWidth',1.5,'DisplayName','Sample Path 2')
yline(0,'k-')
title('Error')
xlabel('Number of Trials')
This Question is in continuation to a previous one asked Matlab : Plot of entropy vs digitized code length
I want to calculate the entropy of a random variable that is discretized version (0/1) of a continuous random variable x. The random variable denotes the state of a nonlinear dynamical system called as the Tent Map. Iterations of the Tent Map yields a time series of length N.
The code should exit as soon as the entropy of the discretized time series becomes equal to the entropy of the dynamical system. It is known theoretically that the entropy of the system is log_2(2). The code exits but the frst 3 values of the entropy array are erroneous - entropy(1) = 1, entropy(2) = NaN and entropy(3) = NaN. I am scratching my head as to why this is happening and how I can get rid of it. Please help in correcting the code. THank you.
clear all
H = log(2)
threshold = 0.5;
x(1) = rand;
lambda(1) = 1;
entropy(1,1) = 1;
j=2;
tol=0.01;
while(~(abs(lambda-H)<tol))
if x(j - 1) < 0.5
x(j) = 2 * x(j - 1);
else
x(j) = 2 * (1 - x(j - 1));
end
s = (x>=threshold);
p_1 = sum(s==1)/length(s);
p_0 = sum(s==0)/length(s);
entropy(:,j) = -p_1*log2(p_1)-(1-p_1)*log2(1-p_1);
lambda = entropy(:,j);
j = j+1;
end
plot( entropy )
It looks like one of your probabilities is zero. In that case, you'd be trying to calculate 0*log(0) = 0*-Inf = NaN. The entropy should be zero in this case, so you you can just check for this condition explicitly.
Couple side notes: It looks like you're declaring H=log(2), but your post says the entropy is log_2(2). p_0 is always 1 - p_1, so you don't have to count everything up again. Growing the arrays dynamically is inefficient because matlab has to re-copy the entire contents at each step. You can speed things up by pre-allocating them (only worth it if you're going to be running for many timesteps).
I am trying to get the angle between every vector in a large array (1896378x4 -EDIT: this means I need 1.7981e+12 angles... TOO LARGE, but if there's room to optimize the code, let me know anyways). It's too slow - I haven't seen it finish yet. Here's the steps towards optimizing I've taken:
First, logically what I (think I) want (just use Bt=rand(N,4) for testing):
[ro,col]=size(Bt);
angbtwn = zeros(ro-1); %too long to compute!! total non-zero = ro*(ro-1)/2
count=1;
for ii=1:ro-1
for jj=ii+1:ro
angbtwn(count) = atan2(norm(cross(Bt(ii,1:3),Bt(jj,1:3))), dot(Bt(ii,1:3),Bt(jj,1:3))).*180/pi;
count=count+1;
end
end
So, I though I'd try and vectorize it, and get rid of the non-built-in functions:
[ro,col]=size(Bt);
% angbtwn = zeros(ro-1); %TOO LONG!
for ii=1:ro-1
allAxes=Bt(ii:ro,1:3);
repeachAxis = allAxes(ones(ro-ii+1,1),1:3);
c = [repeachAxis(:,2).*allAxes(:,3)-repeachAxis(:,3).*allAxes(:,2)
repeachAxis(:,3).*allAxes(:,1)-repeachAxis(:,1).*allAxes(:,3)
repeachAxis(:,1).*allAxes(:,2)-repeachAxis(:,2).*allAxes(:,1)];
crossedAxis = reshape(c,size(repeachAxis));
normedAxis = sqrt(sum(crossedAxis.^2,2));
dottedAxis = sum(repeachAxis.*allAxes,2);
angbtwn(1:ro-ii+1,ii) = atan2(normedAxis,dottedAxis)*180/pi;
end
angbtwn(1,:)=[]; %angle btwn vec and itself
%only upper left triangle are values...
Still too long, even to pre-allocate... So I try to do sparse, but not implemented right:
[ro,col]=size(Bt);
%spalloc:
angbtwn = sparse([],[],[],ro,ro,ro*(ro-1)/2);%zeros(ro-1); %cell(ro,1)
for ii=1:ro-1
...same
angbtwn(1:ro-ii+1,ii) = atan2(normedAxis,dottedAxis)*180/pi; %WARNED: indexing = >overhead
% WHAT? Can't index sparse?? what's the point of spalloc then?
end
So if my logic can be improved, or if sparse is really the way to go, and I just can't implement it right, let me know where to improve. THANKS for your help.
Are you trying to get the angle between every pair of vectors in Bt? If Bt has 2 million vectors that's a trillion pairs each (apparently) requiring an inner product to get the angle between. I don't know that any kind of optimization is going to help have this operation finish in a reasonable amount of time in MATLAB on a single machine.
In any case, you can turn this problem into a matrix multiplication between matrices of unit vectors:
N=1000;
Bt=rand(N,4); % for testing. A matrix of N (row) vectors of length 4.
[ro,col]=size(Bt);
magnitude = zeros(N,1); % the magnitude of each row vector.
units = zeros(size(Bt)); % the unit vectors
% Compute the unit vectors for the row vectors
for ii=1:ro
magnitude(ii) = norm(Bt(ii,:));
units(ii,:) = Bt(ii,:) / magnitude(ii);
end
angbtwn = acos(units * units') * 360 / (2*pi);
But you'll run out of memory during the matrix multiplication for largish N.
You might want to use pdist with 'cosine' distance to compute the 1-cos(angbtwn).
Another perk for this approach that it does not compute n^2 values but exaxtly .5*(n-1)*n unique values :)
I have two signals, let's call them 'a' and 'b'. They are both nearly identical signals (recorded from the same input and contain the same information) however, because I recorded them at two different 'b' is time shifted by an unknown amount. Obviously, there is random noise in each.
Currently, I am using cross correlation to compute the time shift, however, I am still getting improper results.
Here is the code I am using to calculate the time shift:
function [ diff ] = FindDiff( signal1, signal2 )
%FINDDIFF Finds the difference between two signals of equal frequency
%after an appropritate time shift is applied
% Calculates the time shift between two signals of equal frequency
% using cross correlation, shifts the second signal and subtracts the
% shifted signal from the first signal. This difference is returned.
length = size(signal1);
if (length ~= size(signal2))
error('Vectors must be equal size');
end
t = 1:length;
tx = (-length+1):length;
x = xcorr(signal1,signal2);
[mx,ix] = max(x);
lag = abs(tx(ix));
shifted_signal2 = timeshift(signal2,lag);
diff = signal1 - shifted_signal2;
end
function [ shifted ] = timeshift( input_signal, shift_amount )
input_size = size(input_signal);
shifted = (1:input_size)';
for i = 1:input_size
if i <= shift_amount
shifted(i) = 0;
else
shifted(i) = input_signal(i-shift_amount);
end
end
end
plot(FindDiff(a,b));
However the result from the function is a period wave, rather than random noise, so the lag must still be off. I would post an image of the plot, but imgur is currently not cooperating.
Is there a more accurate way to calculate lag other than cross correlation, or is there a way to improve the results from cross correlation?
Cross-correlation is usually the simplest way to determine the time lag between two signals. The position of peak value indicates the time offset at which the two signals are the most similar.
%// Normalize signals to zero mean and unit variance
s1 = (signal1 - mean(signal1)) / std(signal1);
s2 = (signal2 - mean(signal2)) / std(signal2);
%// Compute time lag between signals
c = xcorr(s1, s2); %// Cross correlation
lag = mod(find(c == max(c)), length(s2)) %// Find the position of the peak
Note that the two signals have to be normalized first to the same energy level, so that the results are not biased.
By the way, don't use diff as a name for a variable. There's already a built-in function in MATLAB with the same name.
Now there are two functions in Matlab:
one called finddelay
and another called alignsignals that can do what you want, I believe.
corr finds a dot product between vectors (v1, v2). If it works bad with your signal, I'd try to minimize a sum of squares of differences (i.e. abs(v1 - v2)).
signal = sin(1:100);
signal1 = [zeros(1, 10) signal];
signal2 = [signal zeros(1, 10)];
for i = 1:length(signal1)
signal1shifted = [signal1 zeros(1, i)];
signal2shifted = [zeros(1, i) signal2];
d2(i) = sum((signal1shifted - signal2shifted).^2);
end
[fval lag2] = min(d2);
lag2
It is computationally worse than cross-calculation which can be speeded up by using FFT. As far as I know you can't do this with euclidean distance.
UPD. Deleted wrong idea about cross-correlation with periodic signals
You can try matched filtering in frequency domain
function [corr_output] = pc_corr_processor (target_signal, ref_signal)
L = length(ref_signal);
N = length(target_signal);
matched_filter = flipud(ref_signal')';
matched_filter_Res = fft(matched_filter,N);
corr_fft = matched_filter_Res.*fft(target_signal);
corr_out = abs(ifft(corr_fft));
The peak of the matched filter maximum-index of corr_out above should give you the lag amount.
i am trying to calculate the inverse fourier transform of the vector XRECW. for some reason i get a vector of NANs.
please help!!
t = -2:1/100:2;
x = ((2/5)*sin(5*pi*t))./((1/25)-t.^2);
w = -20*pi:0.01*pi:20*pi;
Hw = (exp(j*pi.*(w./(10*pi)))./(sinc(w./(10*pi)))).*(heaviside(w+5*pi)-heaviside(w-5*pi));%low pass filter
xzohw = 0;
for q=1:20:400
xzohw = xzohw + x(q).*(2./w).*sin(0.1.*w).*exp(-j.*w*0.2*((q-1)/20)+0.5);%calculating fourier transform of xzoh
end
xzohw = abs(xzohw);
xrecw = abs(xzohw.*Hw);%filtering the fourier transform high frequencies
xrect=0;
for q=1:401
xrect(q) = (1/(2*pi)).*trapz(xrecw.*exp(j*w*t(q))); %inverse fourier transform
end
xrect = abs(xrect);
plot(t,xrect)
Here's a direct answer to your question of "why" there is a nan. If you run your code, the Nan comes from dividing by zero in line 7 for computing xzohw. Notice that w contains zero:
>> find(w==0)
ans =
2001
and you can see in line 7 that you divide by the elements of w with the (2./w) factor.
A quick fix (although it is not a guarantee that your code will do what you want) is to avoid including 0 in w by using a step which avoids zero. Since pi is certainly not divisible by 100, you can try taking steps in .01 increments:
w = -20*pi:0.01:20*pi;
Using this, your code produces a plot which might resemble what you're looking for. In order to do better, we might need more details on exactly what you're trying to do, or what these variables represent.
Hope this helps!