Math Insight

1. If you don't observe the neuron's response (and don't observe the rat location, either, of course), and you know (or assume) that the rat is either in room 1 or room 2, what is the probability that the rat is in room 1?
   

   Let $X$ be a random variable indicating the rat's location. Then, $X=1$ is the event that the rat is in room 1. You have just estimated the probability of event $X=1$: $P(X=1) =＿$.

   Similarly, $X=2$ be the event that the rat is in room 2. What is $P(X=2)$?
   .

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   Hint

   If we assume that the rat must be in either room 1 or room 2, and we don't have any further information from the neuron response, the best we can do is estimate the probability using the relative frequencies based on how much time the rat tends to spend in each room.

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2. We will view these probabilities as the prior distribution of the random variable $X$, as the prior distribution captures our knowledge of the rat's location before recording anything from the rat's brain. If we denote the prior distribution as $f_X(x)$, then we can express our results as
   $f_X(1) = $
   
   $f_X(2) = $
   .

   Sketch the prior distribution of the rat's location.

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3. As mentioned above, you observed that the neuron has a different likelihood of firing a spike depending on the rat's location. These observations can be interpreted as conditional probabilities. You can now estimate the probability that the neuron spiked conditioned on each of the two events $X=1$ and $X=2$.

   Let $R$ denote the number of spikes recorded from the neuron in a 10 ms time window. We assume that you record either 0 or 1 spikes. Therefore, we will be working with two more events. If the neuron spiked once, we say we had the event $R=1$; if the neuron did not spike, we say we had the event $R=0$.

   If we know that the rat is in room 1, what is the probability that the neuron fires a spike in a 10 ms window (i.e., the probability of the event that $R=1$)?
   We denote the conditional probability of the event that $R=1$ given the event $X=1$ as $P(R=1 \,|\, X=1)$, which you estimated as $P(R=1\,|\,X=1)=＿$.

   Similarly, what is the conditional probability of the event that $R=1$ given the event $X=2$? $P(R=1 \,|\, X=2) = $
   .

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   Hint

   Recall that the neuron has a 10% chance of firing if $X=1$ and a 2% chance of firing if $X=2$. Just convert these percentages to fractions to estimate the conditional probabilities.

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4. We can summarize the neuron's properties by plotting how its firing probability depends on the location of the rat.
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   In this simple example, the neuron functions like a place cell, in the sense that it is more likely fire when the rat is in a certain location, $X=1$.

5. For completeness, we define two more conditional probabilities for the event that the neuron does not spike, i.e., $R=0$.

   $P(R=0 \,|\, X=1) = $
   
   $P(R=0 \,|\, X=2) = $
   

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   Hint

   Since we assume the neuron fired either zero or one spikes, the probability of zero spikes plus the probability of one spike must add up to one. The probabilities must sum to one both for the case when the rat is in room 1 and, separately, for the case when the rat is in room 2.

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6. We can combine these probabilities to determine probabilities of the four possible outcomes of the experiment where we measure one neuron and look which room the rat is in.

   1. The rat is room 1 and the neuron spikes (the event $X=1$ and the event $R=1$).
   2. The rat is room 1 and the neuron does not spike (the event $X=1$ and the event $R=0$).
   3. The rat is room 2 and the neuron spikes (the event $X=2$ and the event $R=1$).
   4. The rat is room 2 and the neuron does not spike (the event $X=2$ and the event $R=0$).

   We just need to multiply the probability of the rat being in the room times the probability that the neuron spiked condition on the rat being in that room. These four probabilities are the following.

   1. $P(R=1, X=1) = P(X=1) P(R=1 \,|\, X=1) = $
      
   2. $P(R=0, X=1) =$
      
   3. $P(R=1, X=2) =$
      
   4. $P(R=0, X=2) =$
      

   Summarize the results in a contingency table.

   	$R=0$	$R=1$	Total
   $X=1$
   $X=2$
   Total

   In the bottom row, you calculated the probability that a neuron fires a spike
   $P(R=1) = $
   
   as well as the probability that a neuron does not fire a spike
   $P(R=0) = $
   .

We state the question in this way: if we record a spike from the neuron in a 10 ms window (without observing the rat's location), what's the probability the rat is in room 1 and what's the probability it is in room 2? In other words, we want to know the conditional probabilities $P(X=1 \,|\, R=1)$ and $P(X=2 \,|\, R=1)$ (as well as maybe $P(X=1 \,|\, R=0)$ and $P(X=2 \,|\, R=0)$).

1. Calculating these probabilities is an application of Bayes' Theorem. But, rather than just state Bayes' Theorem, let go over the procedure to obtain Bayes' theorem in order to see what that means in this context.

   If we measure a spike from the neuron, then the probability that an event from the second column of the above contingency table is
   . That's because, if we recorded that spike, we know that we had to be in the second column of the contingency table, so the total probability of that column must be 1.

   Given that we know that we are in the second column (i.e., that we measured that a spike occurred), we want to know likelihood we were in the first row versus in the second row (i.e., that the rat was in room 1 rather than room 2). To answer this question, we just need to rescale the probabilities so that the total probability for second column is 1. In other words, we need to divide the values in the second column by its total $P(R=1) = $
   .

   	Given that $R=1$
   $X=1$
   $X=2$
   Total

2. What you calculated in that rescaled table were the conditional probabilities. Without realizing it, you used Bayes' Theorem.

   Recall that the contingency contained values like $P(R=1,X=1) = P(R=1 \,|\, X=1) P(X=1)$. Then, when dividing by $P(R=1)$, you calculated Bayes' theorem as

   • $\displaystyle P(X=1 \,|\, R=1) = \frac{P(R=1 \,|\, X=1) P(X=1)}{P(R=1)} =$
     
     (the probability of being in room 1 conditioned on the presence of a spike), and
   • $\displaystyle P(X=2 \,|\, R=1) = \frac{P(R=1 \,|\, X=2)P(X=2)}{P(R=1)} =$
     
     (the probability of being in room 2 conditioned on the presence of a spike).

   (The division by $P(R=1)$ is the rescaling due to the fact we are only considering cases where the neuron spiked. You could think of estimating the probability $P(X=1 \,|\, R=1)$ by counting all the times the rat was in room 1 with the neuron spiking and dividing, i.e., rescaling, by the total number of times that the neuron spiked.)

3. These conditional probabilities are exactly the results we were looking for: the posterior distribution of the rat's location given that we read the rat's mind and measured a spike. If we denote this distribution of $X$, conditioned on $R=1$, as $f_{X|R=1}(x)$, we can express our results as
   $f_{X|R=1}(1) = $
   
   $f_{X|R=1}(2) = $
   

   Plot the posterior distribution of $X$.

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   Bayes' Theorem shows how we calculate the posterior distribution from the prior distribution. We simply need to know how the spiking probability of a neuron depends on the rat's location (which is easy to determine experimentally). Armed with this tool, when we measure a spike from the neuron, we can read the rat's mind to determine these probabilities for being in the different rooms.

   decode_linear_track_two_neurons.m