Thread: Heavy duty MLE

I will be impressed if someone can get this. I will also be impressed if someone can explain why this problem is of great interest.

Two correlated random variables and are Bivariate normal

~ Where is the Variance Covariance matrix.

Given samples the likelihood function is:

(Assume )

Some useful information:



Find the MLE and asymptotic variance.

I'm working out the answer now. If any one is curious I will post my solution.

I assume you want the MLE of a.

Yes I do. It's tedious to calculate, but can you see why this problem is interesting?

Well, there are formulas to estimate a covariance matrix... so it should be possible to narrow it to a huh ?

Did I hear someone thinking maximization of a cubic?

Originally Posted by Anonymous1

Did I hear someone thinking maximization of a cubic?

Look, if you know the answer then say so and/or post it. If this was meant to be a challenge question to members then it should have been posted in the subforum whose link is given below and the rules of that subforum (plainly stated in the stickies) followed:

http://www.mathhelpforum.com/math-he...enge-problems/

Been working on this for a while now so I thought I'd share .

Here is a data correlation project that involves finding the solution to the above estimator, which entailed solving a cubic.

My instructor gave us this problem to point out that can run into trouble for "multiple hump" functions, cubics. This is because there may be various local maximums.

To make sure I found the proper solution to: the maximum, I used the naive estimator, as my initial guess along with 10-step Newton's.

I hope someone finds this interesting. Cause I do!

Code:

u1= load('UNITED.txt');
u2= load('STATES.txt');
u3= load('HONG.txt');
u4= load('kONG.txt');

m1= sum(abs(u1.^2));    u1= u1./sqrt(m1);
m2= sum(abs(u2.^2));    u2= u2./sqrt(m2);
m3= sum(abs(u3.^2));    u3= u3./sqrt(m3);
m4= sum(abs(u4.^2));    u4= u4./sqrt(m4);

k= [20, 50, 100, 200, 300, 400];
a= sum(u1.*u2);

aHatUS0= zeros(100,length(k));  aHatUS= zeros(100,length(k));
aHatHK0= zeros(100,length(k));  aHatHK= zeros(100,length(k));
for t= 1:100
    for i= 1:length(k)
        R= randn(length(u1),k(i));
        v1= R'*u1;  v2= R'*u2; v3= R'*u3;  v4= R'*u4;
        aHatUS0(t,i)= (1/k(i)).*sum(v1.*v2);
        aHatHK0(t,i)= (1/k(i)).*sum(v3.*v4);
        
        xn1= aHatUS0;   xn2= aHatHK0;   n= length(v1);
        for j=1:10
            a1= xn1;    a2= xn2;
            
            f1= n*a1.^3 - a1.^2 .* sum (v1.*v2) + a1.*(-n+sum(v1.^2+v2.^2)) - sum (v1.*v2);
            fp1= 3*n*a1.^2 - 2.*a1.* sum (v1.*v2) -n + sum(v1.^2+v2.^2);
            f2= n*a2.^3 - a2.^2 .* sum (v3.*v4) + a2.*(-n+sum(v3.^2+v4.^2)) - sum (v3.*v4);
            fp2= 3*n*a2.^2 - 2.*a2.* sum (v3.*v4) -n + sum(v3.^2+v4.^2);
            
            xn1= a1 - f1./fp1;  xn2= a2 - f2./fp2;
        end 
        aHatUS(t,i)=xn1(1);
        aHatHK(t,i)=xn2(1);
    end
end

aHatUS0Var= (m1*m2+a^2)*k.^(-1);   aHatUSVar= (1-a^2)^2./((1+a^2)*k);
aHatHK0Var= (m3*m4+a^2)*k.^(-1);   aHatHKVar= (1-a^2)^2./((1+a^2)*k); 

mSEaHatUS0= aHatUS0Var + (mean(aHatUS0-a)).^2;
mSEaHatHK0= aHatHK0Var + (mean(aHatHK0-a)).^2;
mSEaHatUS= aHatUSVar + (mean(aHatUS-a)).^2;
mSEaHatHK= aHatHKVar + (mean(aHatHK-a)).^2;


figure; hold on;    grid on;
plot(k,mSEaHatUS0,'g');
plot(k,aHatUS0Var,'b');
plot(k,mSEaHatHK0,'r');
plot(k,aHatHK0Var,'y');

title('Figure 1: Empirical MSEs and Theoretical Variances')
xlabel('k'); ylabel('MSE & Variance');
h=legend('Empirical MSE of aUS0_{hat}','Theoretical Variance of aUS0_{hat}','Empirical MSE of aHK0_{hat}','Theoretical Variance of aHK0_{hat}',4);
set(h,'Location','NorthEast')
hold off;

figure; hold on; grid on;
plot(k,mSEaHatUS,'g');
plot(k,aHatUSVar,'b');
plot(k,mSEaHatHK,'r');
plot(k,aHatHKVar,'y');

title('Figure 1: Empirical MSEs and Theoretical Variances')
xlabel('k'); ylabel('MSE & Variance');
h=legend('Empirical MSE of aUS_{hat}','Theoretical Variance of aUS_{hat}','Empirical MSE of aHK_{hat}','Theoretical Variance of aHK_{hat}',4);
set(h,'Location','NorthEast')
hold off;

Note that the final theoretical variance was: