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index cb77c74..c9d4fdc 100644
Binary files a/pointprocesses/lecture/pointprocessscetchB.pdf and b/pointprocesses/lecture/pointprocessscetchB.pdf differ
diff --git a/projects/README b/projects/README
index 16e9758..0f688a4 100644
--- a/projects/README
+++ b/projects/README
@@ -1,10 +1,21 @@
 How to make a new project
 -------------------------
-Copy `project_template/` to your `project_NAME/` and adapt according to your needs.
-Rename `template.tex` to `NAME.tex` and write questions.
-Put data that are needed for the project into the `data/` subfolder.
-Put your solution into the `code/` subfolder.
-Don't forget to add the project files to git (`git add FILENAMES`).
+
+- Copy `project_template/` to your `project_NAME/` and adapt according to your needs.
+- Rename `template.tex` to `NAME.tex` and write questions.
+- Put code needed for the project into the project's root directory.
+- Put data that are needed for the project into the `data/` subfolder.
+- Put your solution into the `solution/` subfolder.
+- Put code that is needed to generate some data for the project, 
+  but which is not part of the project, 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
@@ -12,7 +23,7 @@ Projects
 
 1) project_activation_curve
 medium
-Write questions
+also normalize activation curve to maximum.
 
 2) project_adaptation_fit
 OK, medium
@@ -34,7 +45,6 @@ OK, medium-difficult
 
 7) project_ficurves
 OK, medium
-Maybe add correlation test or fit statistics
 
 8) project_lif
 OK, difficult
@@ -42,7 +52,6 @@ no statistics
 
 9) project_mutualinfo
 OK, medium
-Example code is missing
 
 10) project_noiseficurves
 OK, simple-medium
diff --git a/projects/project_activation_curve/solution/ivcurve.m b/projects/project_activation_curve/solution/ivcurve.m
new file mode 100644
index 0000000..0431d4f
--- /dev/null
+++ b/projects/project_activation_curve/solution/ivcurve.m
@@ -0,0 +1,16 @@
+function [vsteps, peakcurrents] = ivcurve(vsteps, time, currents, tmax)
+
+peakcurrents = zeros(1, length(vsteps));
+for k = 1:length(peakcurrents)
+    c = currents((time>0.0)&(time<tmax), k);
+    minc = min(c);
+    maxc = max(c);
+    if abs(minc) > maxc
+        peakcurrents(k) = minc;
+    else
+        peakcurrents(k) = maxc;
+    end
+end
+
+end
+
diff --git a/projects/project_activation_curve/solution/main.m b/projects/project_activation_curve/solution/main.m
new file mode 100644
index 0000000..f2bd03f
--- /dev/null
+++ b/projects/project_activation_curve/solution/main.m
@@ -0,0 +1,50 @@
+%% plot data:
+x = load('../data/WT_01.mat');
+wtdata = x.data;
+plotcurrents(wtdata.t, wtdata.I);
+
+x = load('../data/A1622D_01.mat');
+addata = x.data;
+plotcurrents(addata.t, addata.I);
+
+%% I-V curve:
+[wtsteps, wtpeaks] = ivcurve(wtdata.steps, wtdata.t, wtdata.I, 100.0);
+
+[adsteps, adpeaks] = ivcurve(addata.steps, addata.t, addata.I, 100.0);
+
+figure();
+plot(wtsteps, wtpeaks, '-b');
+hold on;
+plot(adsteps, adpeaks, '-r');
+hold off;
+
+%% reversal potential:
+wtE = reversalpotential(wtsteps, wtpeaks);
+adE = reversalpotential(adsteps, adpeaks);
+
+%% activation curve:
+wtg = wtpeaks./(wtsteps - wtE);
+adg = adpeaks./(adsteps - adE);
+
+wtinfty = wtg(wtsteps<40.0)/mean(wtg((wtsteps>=20.0)&(wtsteps<=40.0)));
+adinfty = adg(adsteps<40.0)/mean(adg((adsteps>=20.0)&(adsteps<=40.0)));
+wtsteps = wtsteps(wtsteps<40.0);
+adsteps = adsteps(adsteps<40.0);
+
+figure();
+plot(wtsteps, wtinfty, '-b');
+hold on;
+plot(adsteps, adinfty, '-r');
+
+%% boltzmann fit:
+bf = @(p, v) 1.0./(1.0+exp(-p(1)*(v - p(2))));
+p = lsqcurvefit(bf, [1.0, -40.0], wtsteps, wtinfty);
+wtfit = bf(p, wtsteps);
+p = lsqcurvefit(bf, [1.0, -40.0], adsteps, adinfty);
+adfit = bf(p, adsteps);
+
+plot(wtsteps, wtfit, '-b');
+plot(wtsteps, adfit, '-r');
+hold off;
+
+
diff --git a/projects/project_activation_curve/solution/plotcurrents.m b/projects/project_activation_curve/solution/plotcurrents.m
new file mode 100644
index 0000000..e2c510c
--- /dev/null
+++ b/projects/project_activation_curve/solution/plotcurrents.m
@@ -0,0 +1,13 @@
+function plotcurrents(time, currents)
+
+figure();
+hold on;
+for k = 1:size(currents, 2)
+    plot(time, currents(:, k))
+end
+hold off;
+xlabel('Time [ms]')
+ylabel('Current')
+
+end
+
diff --git a/projects/project_activation_curve/solution/reversalpotential.m b/projects/project_activation_curve/solution/reversalpotential.m
new file mode 100644
index 0000000..686b667
--- /dev/null
+++ b/projects/project_activation_curve/solution/reversalpotential.m
@@ -0,0 +1,4 @@
+function E = reversalpotential(vsteps, currents)
+    p = polyfit(vsteps((vsteps>=20.0)&(vsteps<50.0)), currents((vsteps>=20.0)&(vsteps<50.0)), 1);
+    E = -p(2)/p(1);
+end
\ No newline at end of file
diff --git a/projects/project_mutualinfo/mutualinfo.tex b/projects/project_mutualinfo/mutualinfo.tex
index 9dda5f7..44e6f88 100644
--- a/projects/project_mutualinfo/mutualinfo.tex
+++ b/projects/project_mutualinfo/mutualinfo.tex
@@ -11,6 +11,49 @@
 
 
 %%%%%%%%%%%%%% Questions %%%%%%%%%%%%%%%%%%%%%%%%%
+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. It quantifies the
+dependence of an output $y$ (e.g. a spike train) on some input $x$
+(e.g. a sensory stimulus).
+
+The probability of each of $n$ input values $x = {x_1, x_2, ... x_n}$
+is given by the corresponding probabilty distribution $P(x)$.  The entropy
+\begin{equation}
+  \label{entropy}
+  H[x] = - \sum_{x}  P(x) \log_2 P(x)
+\end{equation}
+is a measure for the surprise of getting a specific value of $x$. For
+example, if from two possible values '1' and '2', the probability of
+getting a '1' is close to one ($P(1) \approx 1$) then the probability
+of getting a '2' is close to zero ($P(2) \approx 0$). For this case
+the entropy, the surprise level, is almost zero, because both $0 \log
+0 = 0$ and $1 \log 1 = 0$. It is not surprising at all that you almost
+always get a '1'. The entropy is largest for equally likely outcomes
+of $x$. If getting a '1' or a '2' is equally likely then you will be
+most surprised by each new number you get, because you can not predict
+them.
+
+Mutual information measures information transmitted between an input
+and an output. It is computed from the probability distributions of
+the input, $P(x)$, the output $P(y)$ and their joint distribution
+$P(x,y)$:
+\begin{equation}
+  \label{mi}
+  I[x:y] = \sum_{x}\sum_{y} P(x,y) \log_2\frac{P(x,y)}{P(x)P(y)}
+\end{equation}
+where the sums go over all possible values of $x$ and $y$.  The mutual
+information can be also expressed in terms of entropies. Mutual
+information is the entropy of the outputs $y$ reduced by the entropy
+of the outputs given the input:
+\begin{equation}
+  \label{mientropy}
+  I[x:y] = E[y] - E[x|y]
+\end{equation}
+
+The following project is meant to explore the concept of mutual
+information with the help of a simple example.
+
 \begin{questions}
   \question A subject was presented two possible objects for a very
   brief time ($50$\,ms). The task of the subject was to report which of
@@ -19,40 +62,56 @@
   object was reported by the subject. 
   
   \begin{parts}
-    \part Plot the data appropriately.
+    \part Plot the raw data (no sums or probabilities) appropriately.
+
+    \part Compute and plot the probability distributions of presented
+    and reported objects.
 
     \part Compute a 2-d histogram that shows how often different
     combinations of reported and presented came up.
 
     \part Normalize the histogram such that it sums to one (i.e. make
     it a probability distribution $P(x,y)$ where $x$ is the presented
-    object and $y$ is the reported object). Compute the probability
-    distributions $P(x)$ and $P(y)$ in the same way.
-
-    \part Use that probability distribution to compute the mutual
-    information 
-    \[ I[x:y] = \sum_{x\in\{1,2\}}\sum_{y\in\{1,2\}} P(x,y)
-    \log_2\frac{P(x,y)}{P(x)P(y)}\] 
-    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
-    find out by generating your own dataset which naturally should
-    yield maximal information)?
-
-    \part Use bootstrapping (permutation test) to compute the $95\%$
-    confidence interval for the mutual information estimate in the
-    dataset from {\tt decisions.mat}.
-
+    object and $y$ is the reported object).
+    
+    \part Use the computed probability distributions to compute the mutual
+    information \eqref{mi} that the answers provide about the
+    actually presented object.
+
+    \part Use a permutation test to compute the $95\%$ confidence
+    interval for the mutual information estimate in the dataset from
+    {\tt decisions.mat}. Does the measured mutual information indicate
+    signifikant information transmission?
+    
   \end{parts}
+  
+  \question What is the maximally achievable mutual information?
+  
+  \begin{parts}
+    \part   Show this numerically by generating your own datasets which
+    naturally should yield maximal information. Consider different
+    distributions of $P(x)$.
 
-\end{questions}
+    \part Compare the maximal mutual information with the corresponding
+    entropy \eqref{entropy}.
+  \end{parts}
 
+  \question What is the minimum possible mutual information?
 
+  This is the mutual information between an output is independent of the
+  input.
 
+  How is the joint distribution $P(x,y)$ related to the marginls
+  $P(x)$ and $P(y)$ if $x$ and $y$ are independent? What is the value
+  of the logarithm in eqn.~\eqref{mi} in this case? So what is the
+  resulting value for the mutual information?
+  
+\end{questions}
 
+Hint: 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.
 
 \end{document}
diff --git a/projects/project_mutualinfo/solution/mi.m b/projects/project_mutualinfo/solution/mi.m
new file mode 100644
index 0000000..dc69d7f
--- /dev/null
+++ b/projects/project_mutualinfo/solution/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
diff --git a/projects/project_mutualinfo/solution/mutualinfo.m b/projects/project_mutualinfo/solution/mutualinfo.m
new file mode 100644
index 0000000..0e7a95e
--- /dev/null
+++ b/projects/project_mutualinfo/solution/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')
+
diff --git a/projects/project_noiseficurves/noiseficurves.tex b/projects/project_noiseficurves/noiseficurves.tex
index 3baa199..a2df3df 100644
--- a/projects/project_noiseficurves/noiseficurves.tex
+++ b/projects/project_noiseficurves/noiseficurves.tex
@@ -9,49 +9,50 @@
 
 \input{../instructions.tex}
 
-\begin{questions}
-  \question You are recording the activity of a neuron in response to
-  constant stimuli of intensity $I$ (think of that, for example,
-  as a current $I$ injected via a patch-electrode into the neuron).
-
-  Measure the tuning curve (also called the intensity-response curve) of the
-  neuron.  That is, what is the mean firing rate of the neuron's response
-  as a function of the constant input current $I$?
-
-  How does the intensity-response curve of a neuron depend on the
-  level of the intrinsic noise of the neuron?
-
-  How can intrinsic noise be usefull for encoding stimuli?
-
-  The neuron is implemented in the file \texttt{lifspikes.m}.  Call it
-  with the following parameters:\\[-7ex]
-    \begin{lstlisting}
-trials = 10;
-tmax = 50.0;
-current = 10.0;  % the constant input current I
-Dnoise = 1.0;    % noise strength
-spikes = lifspikes(trials, current, tmax, Dnoise);
-    \end{lstlisting}
-    The returned \texttt{spikes} is a cell array with \texttt{trials}
-    elements, each being a vector of spike times (in seconds) computed
-    for a duration of \texttt{tmax} seconds.  The input current is set
-    via the \texttt{current} variable, the strength of the intrinsic
-    noise via \texttt{Dnoise}. If \texttt{current} is a single number,
-    then an input current of that intensity is simulated for
-    \texttt{tmax} seconds. Alternatively, \texttt{current} can be a
-    vector containing an input current that changes in time. In this
-    case, \texttt{tmax} is ignored, and you have to provide a value
-    for the input current for every 0.0001\,seconds.
-
-    Think of calling the \texttt{lifspikes()} function as a simple way
-    of doing an electrophysiological experiment. You are presenting a
-    stimulus with a constant intensity $I$ that you set. The neuron
-    responds to this stimulus, and you record this response. After
-    detecting the timepoints of the spikes in your recordings you get
-    what the \texttt{lifspikes()} function returns. In addition you
-    can record from different neurons with different noise properties
-    by setting the \texttt{Dnoise} parameter to different values.
+You are recording the activity of neurons that differ in the strength
+of their intrinsic noise in response to constant stimuli of intensity
+$I$ (think of that, for example, as a current $I$ injected via a
+patch-electrode into the neuron).
+
+We first characterize the neurons by their tuning curves (also called
+intensity-response curve).  That is, what is the mean firing rate of
+the neuron's response as a function of the constant input current $I$?
+
+In the second part we demonstrate how intrinsic noise can be useful
+for encoding stimuli on the example of the so called ``subthreshold
+stochastic resonance''.
+
+The neuron is implemented in the file \texttt{lifspikes.m}.  Call it
+with the following parameters:\\[-7ex]
+\begin{lstlisting}
+  trials = 10;
+  tmax = 50.0;
+  current = 10.0;  % the constant input current I
+  Dnoise = 1.0;    % noise strength
+  spikes = lifspikes(trials, current, tmax, Dnoise);
+\end{lstlisting}
+The returned \texttt{spikes} is a cell array with \texttt{trials}
+elements, each being a vector of spike times (in seconds) computed for
+a duration of \texttt{tmax} seconds.  The input current is set via the
+\texttt{current} variable, the strength of the intrinsic noise via
+\texttt{Dnoise}. If \texttt{current} is a single number, then an input
+current of that intensity is simulated for \texttt{tmax}
+seconds. Alternatively, \texttt{current} can be a vector containing an
+input current that changes in time. In this case, \texttt{tmax} is
+ignored, and you have to provide a value for the input current for
+every 0.0001\,seconds.
+
+Think of calling the \texttt{lifspikes()} function as a simple way of
+doing an electrophysiological experiment. You are presenting a
+stimulus with a constant intensity $I$ that you set. The neuron
+responds to this stimulus, and you record this response. After
+detecting the timepoints of the spikes in your recordings you get what
+the \texttt{lifspikes()} function returns. In addition you can record
+from different neurons with different noise properties by setting the
+\texttt{Dnoise} parameter to different values.
 
+\begin{questions}
+  \question Tuning curves
   \begin{parts}
     \part First set the noise \texttt{Dnoise=0} (no noise). Compute
     and plot the neuron's $f$-$I$ curve, i.e. the mean firing rate
@@ -64,37 +65,43 @@ spikes = lifspikes(trials, current, tmax, Dnoise);
 
     \part Compute the $f$-$I$ curves of neurons with various noise
     strengths \texttt{Dnoise}. Use for example $D_{noise} = 10^{-3}$,
-    $10^{-2}$, and $10^{-1}$.
+    $10^{-2}$, and $10^{-1}$. Depending on the resulting curves you
+    might want to try additional noise levels.
 
-    How does the intrinsic noise influence the response curve?
+    How does the intrinsic noise level influence the tuning curves?
 
     What are possible sources of this intrinsic noise?
 
     \part Show spike raster plots and interspike interval histograms
     of the responses for some interesting values of the input and the
     noise strength. For example, you might want to compare the
-    responses of the four different neurons to the same input, or by
-    the same resulting mean firing rate.
+    responses of the different neurons to the same input, or by the
+    same resulting mean firing rate.
 
     How do the responses differ?
+  \end{parts}
+  
+  \question Subthreshold stochastic resonance
+  
+  Let's now use as an input to the neuron a 1\,s long sine wave $I(t)
+  = I_0 + A \sin(2\pi f t)$ with offset current $I_0$, amplitude $A$,
+  and frequency $f$. Set $I_0=5$, $A=4$, and $f=5$\,Hz.
 
-    \part Let's now use as an input to the neuron a 1\,s long sine
-    wave $I(t) = I_0 + A \sin(2\pi f t)$ with offset current $I_0$,
-    amplitude $A$, and frequency $f$. Set $I_0=5$, $A=4$, and
-    $f=5$\,Hz.
-
-    Do you get a response of the noiseless ($D_{noise}=0$) neuron?
+  \begin{parts}
+    \part  Do you get a response of the noiseless ($D_{noise}=0$) neuron?
 
-    What happens if you increase the noise strength?
+    \part  What happens if you increase the noise strength?
 
-    What happens at really large noise strengths?
+    \part  What happens at really large noise strengths?
 
-    Generate some example plots that illustrate your findings.
+    \part  Generate some example plots that illustrate your findings.
 
-    Explain the encoding of the sine wave based on your findings
+    \part  Explain the encoding of the sine wave based on your findings
     regarding the $f$-$I$ curves.
 
- \end{parts}
+    \part Why is this phenomenon called ``subthreshold stochastic resonance''?
+
+  \end{parts}
 
 \end{questions}
 
diff --git a/projects/project_power_analysis/power_analysis.tex b/projects/project_power_analysis/power_analysis.tex
index 0957126..00dfcf8 100644
--- a/projects/project_power_analysis/power_analysis.tex
+++ b/projects/project_power_analysis/power_analysis.tex
@@ -1,6 +1,6 @@
 \documentclass[a4paper,12pt,pdftex]{exam}
 
-\newcommand{\ptitle}{EOD waveform}
+\newcommand{\ptitle}{Power analysis}
 \input{../header.tex}
 \firstpagefooter{Supervisor: Peter Pilz}{phone: 29 74835}%
 {email: peter.pilz@uni-tuebingen.de}