[projects] example code for mutual information

projects/README

@@ -7,12 +7,18 @@ Put your solution into the `code/` subfolder.
 Don't forget to add the project files to git (`git add FILENAMES`).
 
 
+Upload projects to Ilias
+------------------------
+
+Simply upload ALL zip files into one folder or Uebungseinheit.
+Provide an additional file that links project names to students.
+
+
 Projects
 --------
 
 1) project_activation_curve
 medium
-Write questions
 
 2) project_adaptation_fit
 OK, medium
@@ -34,7 +40,6 @@ OK, medium-difficult
 
 7) project_ficurves
 OK, medium
-Maybe add correlation test or fit statistics
 
 8) project_lif
 OK, difficult
@@ -42,7 +47,6 @@ no statistics
 
 9) project_mutualinfo
 OK, medium
-Example code is missing
 
 10) project_noiseficurves
 OK, simple-medium

projects/project_mutualinfo/code/mi.m

@@ -0,0 +1,8 @@
+function I = mi(nxy)
+    pxy = nxy / sum(nxy(:));
+    px = sum(nxy, 2) / sum(nxy(:));
+    py = sum(nxy, 1) / sum(nxy(:));
+    pi = pxy .* log2(pxy./(px*py));
+    pi(nxy == 0) = 0.0;
+    I = sum(pi(:));
+end

projects/project_mutualinfo/code/mutualinfo.m

@@ -0,0 +1,90 @@
+%% load data:
+x = load('../data/decisions.mat');
+presented = x.presented;
+reported = x.reported;
+
+%% plot data:
+figure()
+plot(presented, 'ob', 'markersize', 10, 'markerfacecolor', 'b');
+hold on;
+plot(reported, 'or', 'markersize', 5, 'markerfacecolor', 'r');
+hold off
+ylim([0.5, 2.5])
+
+p1 = sum(presented == 1);
+p2 = sum(presented == 2);
+r1 = sum(reported == 1);
+r2 = sum(reported == 2);
+figure()
+bar([p1, p2, r1, r2]);
+set(gca, 'XTickLabel', {'p1', 'p2', 'r1', 'r2'});
+
+%% histogram:
+nxy = zeros(2, 2);
+for x = [1, 2]
+    for y = [1, 2]
+        nxy(x, y) = sum((presented == x) & (reported == y));
+    end
+end
+figure()
+bar3(nxy)
+set(gca, 'XTickLabel', {'p1', 'p2'});
+set(gca, 'YTickLabel', {'r1', 'r2'});
+
+%% normalized histogram:
+pxy = nxy / sum(nxy(:));
+figure()
+imagesc(pxy)
+
+px = sum(nxy, 2) / sum(nxy(:));
+py = sum(nxy, 1) / sum(nxy(:));
+
+%% mutual information:
+miv = mi(nxy);
+
+%% permutation:
+np = 10000;
+mis = zeros(np, 1);
+for k = 1:np
+    ppre = presented(randperm(length(presented)));
+    prep = reported(randperm(length(reported)));
+    pnxy = zeros(2, 2);
+    for x = [1, 2]
+        for y = [1, 2]
+            pnxy(x, y) = sum((ppre == x) & (prep == y));
+        end
+    end
+    mis(k) = mi(pnxy);
+end
+alpha = sum(mis>miv)/length(mis);
+fprintf('signifikance: %g\n', alpha);
+bins = [0.0:0.025:0.4];
+hist(mis, bins)
+hold on;
+plot([miv, miv], [0, np/10], '-r')
+hold off;
+xlabel('MI')
+ylabel('Count')
+
+%% maximum MI:
+n = 100000;
+pxs = [0:0.01:1.0];
+mis = zeros(length(pxs), 1);
+for k = 1:length(pxs)
+    p = rand(n, 1);
+    nxy = zeros(2, 2);
+    nxy(1, 1) = sum(p<pxs(k));
+    nxy(2, 2) = length(p) - nxy(1, 1);
+    mis(k) = mi(nxy);
+%nxy(1, 2) = 0;
+%nxy(2, 1) = 0;
+%mi(nxy)
+end
+figure();
+plot(pxs, mis);
+hold on;
+plot([px(1), px(1)], [0, 1], '-r')
+hold off;
+xlabel('p(x=1)')
+ylabel('Max MI=Entropy')
+

projects/project_mutualinfo/mutualinfo.tex

@@ -31,21 +31,31 @@
 
     \part Use that probability distribution to compute the mutual
     information 
+
     \[ I[x:y] = \sum_{x\in\{1,2\}}\sum_{y\in\{1,2\}} P(x,y)
-    \log_2\frac{P(x,y)}{P(x)P(y)}\] 
+       \log_2\frac{P(x,y)}{P(x)P(y)}\]
     that the answers provide about the actually presented object.
 
     The mutual information is a measure from information theory that is
     used in neuroscience to quantify, for example, how much information
     a spike train carries about a sensory stimulus.
 
-    \part What is the maximally achievable mutual information (try to
+    \part What is the maximally achievable mutual information?
-    find out by generating your own dataset which naturally should
+
-    yield maximal information)?
+    Show this numerically by generating your own datasets which
+    naturally should yield maximal information. Consider different
+    distributions of $P(x)$.
+
+    Here you may encounter a problem when computing the mutual
+    information whenever $P(x,y)$ equals zero. For treating this
+    special case think about (plot it) what the limit of $x \log x$ is
+    for $x$ approaching zero. Use this information to fix the
+    computation of the mutual information.
 
-    \part Use bootstrapping (permutation test) to compute the $95\%$
+    \part Use a permutation test to compute the $95\%$ confidence
-    confidence interval for the mutual information estimate in the
+    interval for the mutual information estimate in the dataset from
-    dataset from {\tt decisions.mat}.
+    {\tt decisions.mat}. Does the measured mutual information indicate
+    signifikant information transmission?
 
   \end{parts}