I am trying to use the Metropolis Hastings algorithm with a random walk sampler to simulate samples from a function $$ in matlab, but something is wrong with my code. The proposal density is the uniform PDF on the ellipse 2s^2 + 3t^2 ≤ 1/4. Can I use the acceptance rejection method to sample from the proposal density?
N=5000;
alpha = #(x1,x2,y1,y2) (min(1,f(y1,y2)/f(x1,x2)));
X = zeros(2,N);
accept = false;
n = 0;
while n < 5000
accept = false;
while ~accept
s = 1-rand*(2);
t = 1-rand*(2);
val = 2*s^2 + 3*t^2;
% check acceptance
accept = val <= 1/4;
end
% and then draw uniformly distributed points checking that u< alpha?
u = rand();
c = u < alpha(X(1,i-1),X(2,i-1),X(1,i-1)+s,X(2,i-1)+t);
X(1,i) = c*s + X(1,i-1);
X(2,i) = c*t + X(2,i-1);
n = n+1;
end
figure;
plot(X(1,:), X(2,:), 'r+');
You may just want to use the native implementation of matlab mhsample.
Regarding your code, there are a few things missing:
- function alpha,
- loop variable i (it might be just n but it is not suited for indexing since it starts at zero).
And you should always allocate memory in matlab if you want to fill it dynamically, i.e. X in your case.
To expand on the suggestions by #max, the code appears to work if you change the i indices to n and replace
n = 0;
with
n = 2;
X(:,1) = [.1,.1];
It would probably be better to assign X(:,1) to random values within your accept region (using the same code you use later), and/or include a burn-in period.
Depending upon what you are going to do with this, it may also make things cleaner to evaluate the argument to sin in the f function to keep it within 0 to 2 pi (likely by shifting the value by 2 pi if it exceeds those bounds)
Related
How can I simulate this question using MATLAB?
Out of 100 apples, 10 are rotten. We randomly choose 5 apples without
replacement. What is the probability that there is at least one
rotten?
The Expected Answer
0.4162476
My Attempt:
r=0
for i=1:10000
for c=1:5
a = randi(1,100);
if a < 11
r=r+1;
end
end
end
r/10000
but it didn't work, so what would be a better way of doing it?
Use randperm to choose randomly without replacement:
A = false(1, 100);
A(1:10) = true;
r = 0;
for k = 1:10000
a = randperm(100, 5);
r = r + any(A(a));
end
result = r/10000;
Short answer:
Your problem follow an hypergeometric distribution (similar to a binomial distribution but without replacement), if you have the necessary toolbox you can simply use the probability density function of the hypergeometric distribution:
r = 1-hygepdf(0,100,10,5) % r = 0.4162
Since P(x>=1) = P(x=1) + P(x=2) + P(x=3) + P(x=4) + P(x=5) = 1-P(x=0)
Of course, here I calculate the exact probability, this is not an experimental result.
To get further:
Noticed that if you do not have access to hygepdf, you can easily write the function yourself by using binomial coefficient:
N = 100; K = 10;
n = 5; k = 0;
r = 1-(nchoosek(K,k)*nchoosek(N-K,n-k))/nchoosek(N,n) % r = 0.4162
You can also use the binomial probability density function, it is a bit more tricky (but also more intuitive):
r = 1-prod(binopdf(0,1,10./(100-[0:4])))
Here we compute the probability to obtain 0 rotten apple five time in a row, the probabily increase at every step since we remove 1 good juicy apple each time. And then, according to the above explaination, we take 1-P(x=0).
There are a couple of issues with your code. First of all, implicitly in what you wrote, you replace the apple after you look at it. When you generate the random number, you need to eliminate the possibility of choosing that number again.
I've rewritten your code to include better practices:
clear
n_runs = 1000;
success = zeros(n_runs, 1);
failure = zeros(n_runs, 1);
approach = zeros(n_runs, 1);
for ii = 1:n_runs
apples = 1:100;
a = randperm(100, 5);
if any(a < 11)
success(ii) = 1;
elseif a >= 11
failure(ii) = 1;
end
approach(ii) = sum(success)/(sum(success)+sum(failure));
end
figure; hold on
plot(approach)
title("r = "+ approach(end))
hold off
The results are stored in an array (called approach), rather than a single number being updated every time, which means you can see how quickly you approach the end value of r.
Another good habit is including clear at the beginning of any script, which reduces the possibility of an error occurring due to variables stored in the workspace.
I am currently working on an assignment where I need to create two different controllers in Matlab/Simulink for a robotic exoskeleton leg. The idea behind this is to compare both of them and see which controller is better at assisting a human wearing it. I am having a lot of trouble putting specific equations into a Matlab function block to then run in Simulink to get results for an AFO (adaptive frequency oscillator). The link has the equations I'm trying to put in and the following is the code I have so far:
function [pos_AFO, vel_AFO, acc_AFO, offset, omega, phi, ampl, phi1] = LHip(theta, eps, nu, dt, AFO_on)
t = 0;
% syms j
% M = 6;
% j = sym('j', [1 M]);
if t == 0
omega = 3*pi/2;
theta = 0;
phi = pi/2;
ampl = 0;
else
omega = omega*(t-1) + dt*(eps*offset*cos(phi1));
theta = theta*(t-1) + dt*(nu*offset);
phi = phi*(t-1) + dt*(omega + eps*offset*cos(phi*core(t-1)));
phi1 = phi*(t-1) + dt*(omega + eps*offset*cos(phi*core(t-1)));
ampl = ampl*(t-1) + dt*(nu*offset*sin(phi));
offset = theta - theta*(t-1) - sym(ampl*sin(phi), [1 M]);
end
pos_AFO = (theta*(t-1) + symsum(ampl*(t-1)*sin(phi* (t-1))))*AFO_on; %symsum needs input argument for index M and range
vel_AFO = diff(pos_AFO)*AFO_on;
acc_AFO = diff(vel_AFO)*AFO_on;
end
https://www.pastepic.xyz/image/pg4mP
Essentially, I don't know how to do the subscripts, sigma, or the (t+1) function. Any help is appreciated as this is due next week
You are looking to find the result of an adaptive process therefore your algorithm needs to consider time as it progresses. There is no (t-1) operator as such. It is just a mathematical notation telling you that you need to reuse an old value to calculate a new value.
omega_old=0;
theta_old=0;
% initialize the rest of your variables
for [t=1:N]
omega[t] = omega_old + % here is the rest of your omega calculation
theta[t] = theta_old + % ...
% more code .....
% remember your old values for next iteration
omega_old = omega[t];
theta_old = theta[t];
end
I think you forgot to apply the modulo operation to phi judging by the original formula you linked. As a general rule, design your code in small pieces, make sure the output of each piece makes sense and then combine all pieces and make sure the overall result is correct.
How to generate 64 or 128 QAM signal in MATLAB without making use of signal processing/Communication tool box. Is there any general method for creating higher order qams?
Let's say you want to generate a QAM for M=2^b. Then you need constellation points that form a lattice of M_I = 2^b_I by M_Q = 2^b_Q points where M = M_I*M_Q and therefore b_I+b_Q==b. Distributing evenly, we get b_I = floor(b/2); and b_Q = ceil(b/2);.
This established, you set up the lattice by forming 1-D grids for real and imaginary part and then connect them to a 2-D lattice. Like this:
b = 6;
M = 2^b; % b = 6: 64-QAM, b = 7: 128-QAM, etc.
b_I = floor(b/2);
M_I = 2^b_I;
b_Q = ceil(b/2);
M_Q = 2^b_Q;
x_I = -(M_I-1)/2:(M_I-1)/2;
x_Q = -(M_Q-1)/2:(M_Q-1)/2;
x_IQ = repmat(x_I,[M_Q,1]) + 1i*repmat(x_Q.',[1,M_I]);
x_IQ = x_IQ / sqrt((M_I^2+M_Q^2-2)/12);
Note that in R2016b, the second-to last line can be simplified to x_IQ = x_I + 1i*x_Q.'; Moreover, the last line ensures that your constellation to has average energy per symbol of 1 (note: for even b, the normalization can be simplified to sqrt((M-1)/6)).
Now, x_IQ is an array that contains all possible constellation points.
To draw random symbols, use s = x_IQ(randi(M,NSymbols,1));
Note that for odd b this script generates a rectangular QAM (e.g., for 128-QAM it is a 16x8 constellation). There are other ways to do so by placing constellation points more symmetrical. There is no unique way of doing that though so if you need one you have to specify which. Though in practice one usually uses QAM with even b.
I found the answer! If it is 64,256 or square of any number, then it can be easily generated as below.
% matlab code to generate 64 qam
m = 0; n = 0;
for k = -7:2:7
m = m+1;
for l = -7:2:7
n = n+1;
x(m,n) = k+j*l;
end
n = 0;
end
So I am trying to go through a for loop that will increment .1 every time and will do this until the another variable h is less than or equal to zero. Then I am suppose to graph this h variable along another variable x. The code that I wrote looks like this:
O = 20;
v = 200;
g = 32.2;
for t = 0:.1:12
% Calculate the height
h(t) = (v)*(t)*(sin(O))-(1/2)*(g)*(t^2);
% Calculate the horizontal location
x(t) = (v)*(t)*cos(O);
if t > 0 && h <= 0
break
end
end
The Error that I keep getting when running this code says "Attempted to access h(0); index must be a positive integer or logical." I don't understand what exactly is going on in order for this to happen. So my question is why is this happening and is there a way I can solve it, Thank you in advance.
You're using t as your loop variable as well as your indexing variable. This doesn't work, because you'll try to access h(0), h(0.1), h(0.2), etc, which doesn't make sense. As the error says, you can only access variables using integers. You could replace your code with the following:
t = 0:0.1:12;
for i = 1:length(t)
% use t(i) instead of t now
end
I will also point out that you don't need to use a for loop to do this. MATLAB is optimised for acting on matrices (and vectors), and will in general run faster on vectorised functions rather than for loops. For instance, your equation for h could be replaced with the following:
O = 20;
v = 200;
g = 32.2;
t = 0:0.1:12;
h = v * t * sin(O) - 0.5 * g * t.^2;
The only difference is that you have to use the element-wise square (.^2) rather than the normal square (^2). This means that MATLAB will square each element of the vector t, rather than multiplying the vector t by itself.
In short:
As the error says, t needs to be an integer or logical.
But your t is t=0:0.1:12, therefore a decimal value.
O = 20;
v = 200;
g = 32.2;
for t = 0:.1:12
% Calculate the height
idx_t = 1:numel(t);
h(idx_t) = (v)*(t)*(sin(O))-(1/2)*(g)*(t^2);
% Calculate the horizontal location
x(idx_t) = (v)*(t)*cos(O);
if t > 0 && h <= 0
break
end
end
Look this question's answer for more options: Subscript indices must either be real positive integers or logical error
I have a square matrix that represents the frequency counts of co-occurrences in a data set. In other words, the rows represent all possible observations of feature 1, and the columns are the possible observations of feature 2. The number in cell (x, y) is the number of times feature 1 was observed to be x at the same time feature 2 was y.
I want to calculate the mutual information contained in this matrix. MATLAB has a built-in information function, but it takes 2 arguments, one for x and one for y. How would I manipulate this matrix to get the arguments it expects?
Alternatively, I wrote my own mutual information function that takes a matrix, but I'm unsure about its accuracy. Does it look right?
function [mutualinfo] = mutualInformation(counts)
total = sum(counts(:));
pX = sum(counts, 1) ./ total;
pY = sum(counts) ./ total;
pXY = counts ./ total;
[h, w] = size(counts);
mutualinfo = 0;
for row = 1:h
for col = 1:w
mutualinfo = mutualinfo + pXY(row, col) * log(pXY(row, col) / (pX(row)*pY(col)));
end;
end;
end
I don't know of any built-in mutual information functions in MATLAB. Perhaps you got a hold of one of the submissions from the MathWorks File Exchange or some other third-party developer code?
I think there may be something wrong with how you are computing pX and pY. Plus, you can vectorize your operations instead of using for loops. Here's another version of your function to try out:
function mutualInfo = mutualInformation(counts)
pXY = counts./sum(counts(:));
pX = sum(pXY,2);
pY = sum(pXY,1);
mutualInfo = pXY.*log(pXY./(pX*pY));
mutualInfo = sum(mutualInfo(:));
end