I have some basic questions regarding multivariate model. In the ARFIT toolbox, the demo file ardem.m shows the working of a 2nd order bivariate (v1,v2) AR model. The coefficient matrices
A1 = [ 0.4 1.2; 0.3 0.7 ]
A2 = [ 0.35 -0.3; -0.4 -0.5 ]
are concatenated into
A = [ A1 A2 ]
Then a transpose of A is taken. So the result is a 2*4 matrix.
My question is that there should be only 4 coefficients viz. 2 for v1 variable and 2 for v2 variable but why are there 8 coefficients? If the equation format is
v(k,:) = a11*v1(k-1)+a12*v1(k-2) + a21*v2(k-1)+ a22*v2(k-2)
where a11 = 0.4, a12=1.2, a21=0.3 and a22=0.7.
I think I am missing somewhere in understanding. Can somebody please explain what is the correct representation?
The matrices A1 and A2 contain transfer coefficients that describe the contribution of states at times k-1 and k-2, respectively, to the state at time k. Since this is a bivariate process, we are following two variables which can influence each other, and both A1 and A2 are 2 x 2. Writing v1 = v(k,1) and v2 = v(k,2):
v1(k) = A1(1,1)*v1(k-1) + A1(1,2)*v2(k-1) + A2(1,1)*v1(k-2) + A2(1,2)*v2(k-2)
and similarly for v2(k). Then collectively A1 and A2 contain 8 elements. If the two processes were independent then A1 and A2 would be diagonal and would collectively contain only 4 nonzero elements.
By the way this is not really a Matlab question so I don't think this is the right forum for this question.
Related
I have posted earlier also regarding KL divergence query but unfortunately did not received a reply may be due to the complexity of my question so I have tried to explain my problem by using simple example.
I have one reference sensor signal and 1 measured value of some sensor.
I want to find out the error or difference between Ref and Measured sensor signal Values.
SO I am using KL divergence.
First I normalized my reference and sensor signal histogram and then applied KL divergence.
My data is too much and complicated means it contains a lot of zeroes and negative values and also 0.001 like these values also.
I was applying KL divergence but unfortunately was not being able to get some good results so I was wondering that may be I did not able to get the good concept of KL divergence or I am doing wrong at some point in The code.
It will nice of people if someone help me out in this. I shall be grateful.
I have also seen a post explaining KL divergence and had a look at it but some concepts remained missing so I am posting my example along with query.
Am i on right way or there is some fault in my concepts .
Thanks a lot in advance.
ref = [5 6 7 5 8 7 8 9 -2 -3 -4];
measured_sensor = [3 3 4 5 7 8 9 9 -1 -2 -3];
%normalized histograms for
C= hist( ref);
C1 = C ./ sum(C);
D = hist(measured_sensor);
D1 = D ./ sum(D);
figure(1)
ax11=subplot(321);
bar(C1)
ax12=subplot(322);
bar(D1)
d = zeros(size(C1));
goodIdx = C1>0 & D1>0;
d1 = sum(C1(goodIdx) .* log(C1(goodIdx) ./ D1(goodIdx)))
d2 = sum(D1(goodIdx) .* log(D1(goodIdx) ./ C1(goodIdx)))
d(goodIdx) = d1 + d2
Mean Based Gaussian Hysterisis (Means Error Finding)
ref = [5 6 7 5 8 7 8 9 -2 -3 -4];
measured_sensor = [3 3 4 5 7 8 9 9 -1 -2 -3];
sig_diff = ref - measured_sensor ;
m = mean(sig_diff)
deviation = std(sig_diff);
pos = sig_diff(sig_diff>0)
neg = sig_diff(sig_diff<0)
m_pos = mean(pos)
m_neg = mean (neg)
hysterisis = abs( m_pos)+ abs(m_neg)
figure(6)
ax11=subplot(321);
histfit(sig_diff)
hold on
plot([m m],[0 5000],'r')
plot([m-deviation m-deviation],[0 5000],'r')
plot([m+deviation m+deviation],[0 5000],'r')
hold off
The error value or Hysterisis value that I am getting with mean based Gaussian Distribution is 3.2500.
So I am expecting the error values from KL divergence near to 3.2500 value or in the range with some tolerance is also accepted.
Actually I am confused about these techniques that which one is giving me more precise version of error and with which technique i will get the best results.
I am working on MATLAB on my own, and was doing problem 9 on Project Euler
It states
" A Pythagorean triplet is a set of three natural numbers, a < b < c, for which,
a2 + b2 = c2
For example, 32 + 42 = 9 + 16 = 25 = 52.
There exists exactly one Pythagorean triplet for which a + b + c = 1000.
Find the product abc."
Below is the code I wrote; however, it compiles, but does not produce and output. I was hoping to get some feedback on what's wrong, so I can fix it.
Thanks,
syms a;
syms b;
syms c;
d= 1000;
d= a + b + c ;
ab= a.^2 + b.^2;
ab= c.^2;
c
I propose a vectorized way (that is, without using loops) to solve the problem. It may seem relatively complicated, especially if you come from other programming languages; but for Matlab you should get used to this way of approaching problems.
Ingredients:
Vectorization;
Indexing;
Transpose;
Implicit singleton expansion;
hypot;
find.
Read up on these concepts if you are not familiar with them, and then try to solve the problem yourself (which of course is the whole point of Project Euler). As a hint, the code below proceeds along these lines:
Generate a 1×1000 vector containing all possible values for a and b.
Compute a 1000×1000 matrix with the values of c corresponding to each pair a, b
From that compute a new matrix such that each entry contains a+b+c
Find the row and column indices where that matrix equals 1000. Those indices are the desired a and b (why?).
You'll get more than one solution (why?). Pick one.
Compute the product of the obtained a and b and the corresponding c.
Once you have tried yourself, you may want to check the code (move the mouse over it):
ab = 1:1000; % step 1
cc = hypot(ab,ab.'); % step 2
sum_abc = ab+ab.'+cc; % step 3
[a, b] = find(sum_abc==1000); % step 4
a = a(1); b = b(1); % step 5
prod_abc = a*b*cc(a,b); % step 6
I have a data which is 1672x6. I have put some of them in the picture.
where x values are A1 A2 A3 A4 A5 A6 and y values are B1 B2 ....B1672.
I used the following code while generating decision trees:
vars = {'A1', ' A2 ','A3',' A4 ','A5',' A6'}
x = [A1 A2 A3 A4 A5 A6];
y = [B];
t = classregtree(x, y, 'method','classification', 'names',vars, ...
'categorical',[2 4], 'prune','off');
view(t)
and it generates super crazy trees like
I want to get the values which are greater than the values that I gave. When I say :
inst = [3 2.3 2 0 1 0];
prediction = eval(t, inst)
It only gives me the B value (like B271) which has that variable but I want to get all B variables which have greater values than inst variable such as A1>3 A2>2.3 A3>2 A4>0 A5>1 A6>0. How can I get them?
You seem to confuse two things: decision tree and finding desired rows
If you want to find all the rows that are greater than inst, the following is a simple code that prints all such rows.
for i = 1:size(B,1)
if all(a(i,:)>inst)==1
i,
end
end
However, decision tree is a totally different topic. In a decision tree, you have a set of conditions (A1 to A6 in your case) and many rows for training (B1 to B1672) and a consequence for each one of them. When a new test case is queried, the machine decides on best possible consequence out of all the consequences.
Some decision tree totorials: 1, 2 and wikipedia
In an attempt to speed up for loops (or eliminate all together), I've been trying to pass matrices into functions. I have to use sine and cosine as well. However, when I attempt to find the integral of a matrix where the elements are composed of sines and cosines, it doesn't work and I can't seem to find a way to make it do so.
I have a matrix SI that is composed of sines and cosines with respect to a variable that I have defined using the Symbolic Math Toolbox. As such, it would actually be even better if I could just pass the SI matrix and receive a matrix of values that is the integral of the sine/cosine function at every location in this matrix. I would essentially get a square matrix back. I am not sure if I phrased that very well, but I have the following code below that I have started with.
I = [1 2; 3 4];
J = [5 6; 7 8];
syms o;
j = o*J;
SI = sin(I + j);
%SI(1,1) = sin(5*o + 1)
integral(#(o) o.*SI(1,1), 0,1);
Ideally, I would want to solve integral(#(o) o*SI,0,1) and get a matrix of values. What should I do here?
Given that A, B and C are all N x N matrices, for the moment, let's assume they're all 2 x 2 matrices to make the example I'm illustrating more succinct to understand. Let's also define o as a mathematical symbol based on your comments in your question above.
syms o;
A = [1 2; 3 4];
B = [5 6; 7 8];
C = [9 10; 11 12];
Let's also define your function f according to your comments:
f = o*sin(A + o*B + C)
We thus get:
f =
[ o*sin(5*o + 10), o*sin(6*o + 12)]
[ o*sin(7*o + 14), o*sin(8*o + 16)]
Remember, for each element in f, we take the corresponding elements in A, B and C and add them together. As such, for the first row and first column of each matrix, we have 1, 5 and 9. As such, A + o*B + C for the first row, first column equates to: 1 + 5*o + 9 = 5*o + 10.
Now if you want to integrate, just use the int command. This will find the exact integral, provided that the integral can be solvable in closed form. int also can handle matrices so it will integrate each element in the matrix. You can call it like so:
out = int(f,a,b);
This will integrate f for each element from the lower bound a to the upper bound b. As such, supposing our limits were from 0 to 1 as you said. Therefore:
out = int(f,0,1);
We thus get:
out =
[ sin(15)/25 - sin(10)/25 - cos(15)/5, sin(18)/36 - sin(12)/36 - cos(18)/6]
[ sin(21)/49 - sin(14)/49 - cos(21)/7, sin(24)/64 - sin(16)/64 - cos(24)/8]
Bear in mind that out is defined in the symbolic math toolbox. If you want the actual numerical values, you need to cast the answer to double. Therefore:
finalOut = double(out);
We thus get:
finalOut =
0.1997 -0.1160
0.0751 -0.0627
Obviously, this can generalize for any size M x N matrices, so long as they all share the same dimensions.
Caveat
sin, cos, tan and the other related functions have their units in radians. If you wish for the degrees equivalent, append a d at the end of the function (i.e. sind, cosd, tand, etc.)
I believe this is the answer you're after. Good luck!
I want to automatically calculate expansions of polynomials where there are variables (x1,x2,...) as well as coefficients (c1,c2, ...). My goal is to calculate p(1)*(c1*x1+c2*x2+...)^n+ ... + p(n)*(c1*x1+c2*x2+...)^n .
As you can notice the resulting expression can be written as F(x1,x2...)*g(c1,c2,...) [where F is a row matrix and g is column matrix], i.e. there is some multiplicative decoupling between the coefficients and the variables.
Right now I use the MATLAB symbolic toolbox and construct F and g by manually examining the resulting symbolic expansions. This is not very feasible as if n is big and c=(c1,c2,...) is too big there are too many terms and it is no longer possible manually. For instance for (c1*x1+c2*x2+c3) and n=2, what I want is following.
>> p=[2 5]
p =
2 5
>> syms c1 c2 c3
>> syms x1 x2
>> expression= p(1)*(c1*x1+c2*x2+c3)^2 + p(2)*(c1*x1+c2*x2+c3);
>> expand(expression)
ans =
2*c1^2*x1^2 + 4*c1*c2*x1*x2 + 4*c1*c3*x1 + 5*c1*x1 + 2*c2^2*x2^2 + 4*c2*c3*x2 + 5*c2*x2 + 2*c3^2 + 5*c3
>> F=[5*x1 5*x2 5 4*x1*x2 4*x1 4*x2 2*x1^2 2*x2^2 2]
F =
[ 5*x1, 5*x2, 5, 4*x1*x2, 4*x1, 4*x2, 2*x1^2, 2*x2^2, 2]
>> g=[c1 c2 c3 c1*c2 c1*c3 c2*c3 c1^2 c2^2 c3^2].'
g =
c1
c2
c3
c1*c2
c1*c3
c2*c3
c1^2
c2^2
c3^2
>> expand(F*g)
ans =
2*c1^2*x1^2 + 4*c1*c2*x1*x2 + 4*c1*c3*x1 + 5*c1*x1 + 2*c2^2*x2^2 + 4*c2*c3*x2 + 5*c2*x2 + 2*c3^2 + 5*c3
I have found the following question and it looks like there may be a way to do it automatically using conv etc. If one can come up with an automated solution (or at least some idea towards such automation) for the case where x=(x1,x2) and c=(c1,c2,c3) and n=2, the case depicted above; I guess I may be able to generalize it to higher dimensional cases.
Note: the ordering of terms in F or g does not matter, given that they are ordered in some structured way.
The coefficients from different terms don't overlap. The first term, p(1)*(c'*x)^1, has only terms of degree 1 in xi and ci, and so on. So it becomes a matter of computing the coefficients of one term at a time.
That, too, has a "simple" expression:
p(k)*(c'*x)^k = sum(i1,..,im>=0 with sum(i_)=k)
M(k;i1,..,im)*x1^i1*...*xm^im * c1^i1*...*cm^im
where the summation is such that the sum of all i equals k, and M is the multinomial coefficient.
For m=3, n=2, the i's would be in the order of your example: 110,101,011,200,020,002. M(2;110)=2 so the first term is `p(2)*M(2;110)*x1*x2 = 4*x1*x2'.
Your F and g terms are:
F(...) = p(k)*M(k;i1,..,im)*x1^i1*...*xm^im
g(...) = c1^i1*...*cm^im