Math Insight

1. If you don't observe the neuron's response (and don't observe the rat location, either, of course), what is the probability that the rat is in room A?
   (If you don't have enough information, enter underdetermined.)

   Need help?Hide help

   Hint

   Given what is stated right here, do we know that the rat is definitely in room A or room B? For example, could he be halfway in between the two rooms?

   Hide help

2. 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 A or room B, what is the probability that the rat is in room A?
   

   Let $A$ be the event that the rat is in room A. You have just estimated the probability of event $A$: $P(A) =＿$.

   Let $B$ be the event that the rat is in room B. What is $P(B)$?
   .

   Need help?Hide help

   Hint

   If we assume that the rat must be in either room A or room B, 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.

   Hide help

3. The spiking data you measured can be interpreted as conditional probabilities. You measured the probability that the neuron spiked conditioned on each of the two events $A$ and $B$.

   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 $R=0$. (We labeled the events $R=0$ and $R=1$ differently that we had previous events like events $A$ and $B$. If we didn't like the equal signs in the events, we could have labeled the events $C$ and $D$ instead. But, the event labels $R=0$ and $R=1$ communicate more about the nature of the events.)

   If we know that the rat is in room A, 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 $A$ as $P(R=1 \,|\, A)$, which you estimated as $P(R=1\,|\,A)=＿$.

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

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

   $P(R=0 \,|\, A) = $
   
   $P(R=0 \,|\, B) = $
   

4. 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 A and the neuron spikes (the event $A$ and the event $R=1$).
   2. The rat is room A and the neuron does not spike (the event $A$ and the event $R=0$).
   3. The rat is room B and the neuron spikes (the event $B$ and the event $R=1$).
   4. The rat is room B and the neuron does not spike (the event $B$ 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, A) = P(A) P(R=1 \,|\, A) = $
      
   2. $P(R=0, A) =$
      
   3. $P(R=1, B) =$
      
   4. $P(R=0, B) =$
      

   Summarize the results in a contingency table.

   	$R=0$	$R=1$	Total
   $A$
   $B$
   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 A and what's the probability it is in room B? In other words, we want to know the conditional probabilities $P(A \,|\, R=1)$ and $P(B \,|\, R=1)$ (as well as maybe $P(A \,|\, R=0)$ and $P(B \,|\, R=0)$).

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 A rather than room B). 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 $R=1$
   $A$
   $B$
   Total

2. What you calculated in that rescaled table were the conditional probabilities:

   • the probability
     $\displaystyle P(A \,|\, R=1) = \frac{P(R=1,A)}{P(R=1)} =$
     
     of being in room $A$ conditioned on the presence of a spike, and
   • the probability
     $\displaystyle P(B \,|\, R=1) = \frac{P(R=1,B)}{P(R=1)} =$
     
     of being in room $B$ 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 $P(A \,|\, R=1)$ by counting all the times the rat was in room A with the neuron spiking and dividing (i.e., rescaling) by the total number of times that the neuron spiked.

   These conditional probabilities are exactly the results we were looking for. Now, 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.

3. We don't typically write out the contingency table and them rescale the columns to calculate the conditional probabilities. Instead, we just use Bayes' Theorem directly.

   Rewrite the above two equations as instances of Bayes' Theorem. You simply need to substitute $P(R=1, A) = P(R=1 \,|\,A) P(A)$ and the same for $P(R=1, B)$.

   • $P(A \,|\, R=1) =$
     
   • $P(B \,|\, R=1) =$
     

   In Bayesian terminology, the probabilities $P(A)$ and $P(B)$ are the prior probabilities of the rat's location. Without measuring the neuron, the prior captures the likelihood we'd think the rat would be in each room. The probabilities $P(A \,|\, R=1)$ and $P(B \,|\, R=1)$ are the posterior probabilities that give our new estimates of the probabilities. If we record the neuron's response in a 10 ms window and observe that it spiked, then we use the posterior probabilities as the likelihood we'd think the rat would be in each room.

   (In practice, it'd be overkill to use Bayes' Theorem for both $P(A \,|\, R=1)$ and $P(B \,|\, R=1)$ since we know (or assume) that the rat must be in one of the two rooms: $P(A \,|\, R=1) + P(B \,|\, R=1) =$
   . We can easily calculate one of these probabilities from the other. There is no harm in using Bayes' theorem for both if you like, though.)

4. You can likewise use Bayes' theorem in the case where you record the activity of the neuron in a 10 ms window and observe that the neuron did not fire a spike. In this case, we know that an outcome from the first column of the contingency table occurred, and Bayes' theorem is equivalent to rescaling that column by its total.

   Write Bayes' theorem for determining the probability that the rat is in room A given that you measured that the neuron did not spike.
   $P(A \,|\, R=0) =$
   

   Write Bayes' theorem for determining the probability that the rat is in room B given that you measured that the neuron did not spike.
   $P(B \,|\, R=0) =$
   

   (Again, it'd be overkill to use Bayes' theorem twice for both $P(A \,|\, R=0)$ and $P(B \,|\, R=0)$, but you could if you like.)

   Given that the neuron did not spike in a 10 ms window, what is the probability that the rat is in room A?
   $P(A \,|\, R=0) = $
   .

   What is the probability that the rat is in room B when you observe a lack of a spike?
   $P(B\,|\, R=0) = $
   .

1. Bayes' theorem is useful for decoding because it allows one to convert quantities like $P(R=1 \,|\,A)$ (the probability of spiking conditioned on being in room A) into quantities like $P(A \,|\, R=1)$ (the probability of being in room A conditioned on spiking). It is easy to estimate quantities like $P(R=1 \,|\,A)$; one just observes the likelihood of a spike when the rat is in room A. A quantity such as $P(A \,|\, R=1)$, on the other hand, allows us to read from the rat's mind (by looking at the neuron's activity) and estimate the rat's location.

   In the decoding, you never determine exactly the room in which room the rat is. Instead, you simply work with probability distributions of the rat position, giving you the likelihood that the rat is in each room. The decoding actually involves three probability distributions of the rat's position.

   The first is the prior distribution that you started with before recording the neuron. We wrote this probability distribution as $P(A)$ and $P(B)$ (which is $1-P(A)$).

   The second is the distribution of rat position conditioned on measuring a spike in one 10 ms window. We wrote this probability distribution as $P(A\,|\,R=1)$ and $P(B\,|\,R=1)$ (which is $1-P(A\,|\,R=1)$).

   The third probability distribution is the one conditioned on measuring the absence of a spike in one 10 ms window. We wrote this probability distribution as $P(A\,|\,R=0)$ and $P(B\,|\,R=0)$ (which is $1-P(A\,|\,R=0)$).

   To visualize the results of the decoding, sketch bar graphs of these probability distributions on the following figures.

   Feedback from applet
   Bar heights:
   Feedback from applet
   Bar heights:
   Feedback from applet
   Bar heights:

2. In this example, without recording from the neuron, you estimate that the probability that the rat is in room A to be $P(A) = $
   . But, if you record the neuron's activity in a 10 ms window and observe that the neuron spiked, you can conclude that the rat is roughly equally likely to be in room A or room B. that the rat is highly likely in room B. that the rat is highly likely in room A. about the same thing as before you measured the neuron.
   

   If, on the other hand, you observe that the neuron didn't spike in the 10 ms window, you can conclude that the rat is highly likely in room B. that the rat is roughly equally likely to be in room A or room B. about the same thing as before you measured the neuron. that the rat is highly likely in room A.
   So, measuring the fact that the neuron did not spike does not does
   give you much additional information about the rat's location. Is it ever possible to conclude from measuring this neuron once that the rat is highly likely to be in room B? no yes
   .

3. If you run this experiment once, what's the likelihood that you'll have a “successful” mind reading experiment, in the sense that you can determine, from measuring from the rat's neuron, which room the rat is highly likely to be in?
   
   $=$
   
   (In the first blank, enter the probability in symbols, i.e. P(something). In the second blank, enter the number for this example.)

   That likelihood isn't so crucial, as we can always repeat the experiment multiple times. (We don't want to have to address the issue of whether or not different experiments are independent, so we won't explore this option here.) The important question is how well we can determine the rat's location when we have such a “successful” experiment. In the cases where we believe we can successfully mind read, what is the likelihood the rat will actually be in the room that the neuron seems to indicate?
   
   $=$
   
   (Use the same convention for the two blanks as above.) This is the probability we are truly interested in, as it indicates how well we can actually decode the rat's location when it does seem that the neuron is telling us something.

   What is likelihood that we got misinformation from the neuron, i.e., that the rat was actually in one room when the neuron seemed to indicate the rat was in the other room?
   
   $=$
   
   (Use the same convention for the two blanks as above.) This probability is equally important as the previous one, but it's just a different perspective on the same information.

1. When measuring the neuron that we've described, you were able to do a remarkable job at decoding the rat's location when you did measure a spike from the neuron. Let's make the mind reading a little more difficult by introducing some error into the measurement of the spikes. One approach to measure a neuron's spikes is to insert an electrode into the brain and try to get the electrode close to one neuron to isolate its signal. Nonetheless, the signal the electrode picks up is noisy and includes effects from other neurons' spikes. It's a challenge to isolate the neuron's spikes from that signal, and algorithms to determine the spikes might miss some of the neuron's spikes or misidentify other neurons' spikes as coming from the neuron being recorded.

   Let's imagine that, due to being conservative about classifying the signal as a spike, one misses half of the spikes of the neuron. We'll assume there is no relationship between the missed spikes and the location of the rat. Therefore, as far as the measurement is concerned, the neuron spikes only in 5 percent of the ms windows when the rat is in room A, and it spikes in only 1 percent of the ms windows when the rat is in room B. In other words, given that the event $R=1$ corresponds to measuring a spike, the conditional probabilities have dropped to
   $P(R=1 \,|\, A) = $
   ,
   $P(R=1 \,|\, B) = $
   .

   With this measurement error, the overall spiking probability becomes $P(R=1)= $
   .

   How does this measurement error, and consequent reduced spiking probability, affect the ability to decode the rat's position from the measurement of a spike? With the error, the probability distribution of the rat's position conditioned on measuring a spike becomes
   $P(A \,|\, R=1) = $
   ,
   $P(B \,|\, R=1) = $
   .

   Does the effect of the error on your ability to decode surprise you? yes no
   Mathematically, any mystery can be explained by Bayes' theorem for $P(A \,|\, R=1)$. The factor $P(R=1 \,|\, A)$ in the numerator was cut in half doubled stayed the same
   , and the denominator $P(R=1)$ stayed the same was cut in half doubled
   , so that the ratio stayed the same was cut in half doubled
   .

   Surely, this conservative approach where one misses half the spikes must affect the decoding in some way. Indeed, the difference is that, due to the decrease increase
   in $P(R=1)$, the probability of actually measuring a spike (or having a “successful” experiment) stayed the same was cut in half doubled
   . If one becomes too conservative and drops $P(R=1)$ a whole lot, then it might take an unreasonably long time to successfully decode the rat's location. But, even so, at least the quality of the decoding would be enhanced degraded unchanged
   .

2. What happens if, on the other hand, one is too liberal in attempting to detect the neuron's spikes and ends up including additional spikes from other neurons. We wouldn't expect these extraneous spikes to be related to the rat's location in the same way as the target neuron's spikes, so let's imagine that these extra spikes increase the probability of measuring $R=1$ by 0.05 regardless of whether the rat is in room A or room B. The measurement error due to the liberal spike detection algorithm has increased the conditional probabilities of measuring a spike to
   $P(R=1 \,|\, A) = $
   ,
   $P(R=1 \,|\, B) = $
   .

   With this measurement error, the overall spiking probability becomes $P(R=1)= $
   .

   How does this measurement error, and consequent increased spiking probability, affect the ability to decode the rat's position from the measurement of a spike? With the error, the probability distribution of the rat's position conditioned on measuring a spike becomes
   $P(A \,|\, R=1) = $
   ,
   $P(B \,|\, R=1) = $
   .

   With the measurement error, the quality of the decoding was unchanged enhanced degraded
   . This result can be seen through Bayes' theorem for $P(A \,|\, R=1)$. Although both $P(R=1 \,|\, A)$ in the numerator and $P(R=1)$ in the denominator increased by the same amount
   , $P(R=1 \,|\, A)$ increased by a smaller larger
   fraction. Thus the ratio of $P(R=1 \,|\, A)/P(R=1)$ decreased increased
   after the introduction of the measurement error due to the liberal spike detection algorithm.

3. In general, decoding the rat's position from the spike of one neuron will work the best when $P(A \,|\, R=1)$ is neither small nor larger very large very small or very large very small
   , as that indicates we've determined that the rat is highly likely to be in one room or the other.

   Given that we are viewing $P(A)$ and $P(B)$ as fixed (we aren't changing the rat's behavior), Bayes' theorem indicates that the decoding will work best when the ratio $P(R=1 \,|\, A)/P(R=1)$ is very large very small or very large very small neither small nor larger
   .

   It shouldn't be surprising that this ratio indicates good decoding when the neuron is much more likely to fire in one room than in the other, i.e., when the neuron's response is strongly selective to the rat's location. (You can try calculating the ratio for a few example neurons to convince yourself it is true. For example, compare the non-selective case where $P(R=1 \,|\, A)=0.2$ and $P(R=1 \,|\, B)=0.1$ to the selective case where $P(R=1 \,|\, A)=0.2$ and $P(R=1 \,|\, B)=0.01$.)