[likelihood] improved solution to exercise 1

likelihood/exercises/mlestd.m

@@ -3,10 +3,11 @@ sigma =2.0;
 ns = [50, 1000];
 for k = 1:length(ns)
     n = ns(k);
+    fprintf('\nn=%d\n', n);
     % draw random numbers:
     x = randn(n,1)*sigma+mu;
-    fprintf('              mean of the data is %.2f\n', mean(x))
-    fprintf('standard deviation of the data is %.2f\n', std(x))
+    fprintf('                                mean of the data is %.2f\n', mean(x))
+    fprintf('                  standard deviation of the data is %.2f\n', std(x))
 
     % standard deviation as parameter:
     psigs = 1.0:0.01:3.0;
@@ -17,8 +18,14 @@ for k = 1:length(ns)
         p = exp(-0.5*((x-mu)/psig).^2.0)/sqrt(2.0*pi)/psig;
         lms(:,i) = p;
     end
-    lm = prod(lms, 1);          % likelihood
+    lm = prod(lms, 1);              % likelihood
     loglm = sum(log(lms), 1);   % log likelihood
+    maxlm = psigs(lm==max(lm));
+    if length(maxlm) > 1
+        maxlm = maxlm(1);
+    end
+    fprintf('      likelihood: standard deviation of the data is %.2f\n', maxlm)
+    fprintf('  log-likelihood: standard deviation of the data is %.2f\n', psigs(loglm==max(loglm)))
 
     % plot likelihood of standard deviation:
     subplot(2, 2, 2*k-1);