Math Insight

Your goal is be able to perform an experiment where we measure whether or not the neuron fired a spike and see what information you can decode (or mind read) about the location of the rat. As a first step, enumerate all the possible outcomes for the experiment and calculate their probabilities given the initial data you collected.

We assume two possible events for the spike measured (the event $R=0$ or the event $R=1$) and two possible events for the rat location (the event A or B, as we will assume from now on that the rat is either in room A or room B). Hence, there will be four possible outcomes of the experiment:

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 $R=0$).

For starters, let's determine the probability of the first outcome, that the rat is in room A and the neuron spikes.

We already estimated the probability that the rat is in room A, $P(A)=$
, and the conditional probability of a spike given A, $P(R=1\,|\,A)=$
.

To estimate the probability that the rat is in room A and the neuron spikes, we need to multiply the probability that the rat is in room A, $P(A)$, by the probability that the neuron spikes conditioned on the rat being in room A. In equations, we write this as $$P(R=1, A) = P(A) P(R=1 \,|\, A).$$ (The comma means “and”, so $P(A,B)$ means the probability of event $A$ and event $B$. The order doesn't matter, so $P(R=1, A) = P(A, R=1)$)

Calculate the probability of the first outcome. $P(R=1,A) = $

The bottom right corner is the overall total, i.e., the probability that any of the four outcomes occurred: $P(R=1, A) + P(R=0, A) + P(R=1, B) + P(R=0,B) $. The value for this corner makes sense because we allow a small probability that one of the four outcomes did not occur. one of the four outcomes must occur.

Now, if you record the neuron for one 10 ms window (without observing the rat), we want to calculate the probability that you will record a spike, i.e., the probability of the event $R=1$. We denote this probability by $P(R=1)$.

If you record a spike, which outcome(s) from the contingency table must have occurred? One of the two outcomes in the second column. The second outcome from the second column. The first outcome from the first column. The first outcome from the second column. The second outcome from the first column. One of the two outcomes in the first column. One of the two outcomes in the first row. One of the two outcomes in the second row.

Since the probability that a neuron spiked is the sum of the probabilities of those outcomes, the probability you will record a spike is $P(R=1)=$

On the other hand, the probability you don't record a spike is $P(R=0) = $
.

We can finally get to answer the question we started out to answer: if we record a spike from the neuron in a 10 ms window (without observing the rat's location), what's the probability it is in room A? (We can then get the probability it is in room B for free since these probabilities must add up to one.)

Here's a trivial question. If we record that the fact that the neuron spiked, what's the probability that $R=1$?
. 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), what's the 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) = $
.

What you calculated in that rescaled table were the conditional probabilities: the probability $P(A \,|\, R=1)$ of being in room $A$ conditioned on the presence of a spike and the probability $P(B \,|\, R=1)$ of being in room $B$ conditioned on the presence of a spike.

By rescaling the table column, you calculated $P(A \,|\, R=1)$ as the ratio of two probabilities: (i) the probability that the rat was in room A and the neuron spiked and (ii) the probability that the neuron spiked. In equations: \begin{align} P(A \,|\, R=1) = \frac{P(R=1,A)}{P(R=1)}. \label{eq:pAR1} \end{align} 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.

Let's summarize the results.

• $P(A \,|\, R=1) = $
  
  This is the likelihood (i.e., probability) that the rat was in room A given that you recorded the neuron's response in a 10 ms window and observed that it spiked.
• $P(B \,|\, R=1 ) = $
  
  If you observe a spike, this quantity is the probability the rat was in room B.

Equation \eqref{eq:pAR1} gives an expression for $P(A \,|\, R=1)$. The numerator of this expression is $P(R=1, A)$. Above, we calculated this value as $P(R=1, A) = P(R=1 \,|\,A) P(A)$. Recall that we started with the values of $P(R=1 \,|\,A)$ and $P(A)$, as they are the probability of a spike given that the rat was in room A and the probability of the rat being in room A, respectively.

Rewrite equation \eqref{eq:pAR1} using the numerator $P(R=1 \,|\,A) P(A)$.
$P(A \,|\, R=1) =$

This expression for $P(A \,|\, R=1)$ in terms of $P(R=1 \,|\,A)$ is called Bayes' Theorem. If you know the probability $P(A)$ that the rat is in room A and the probability $P(R=1)$ that the neuron fires a spike, then you can use Bayes' Theorem to convert $P(R=1 \,|\,A)$ (the probability of spiking conditioned on being in room A) into $P(A \,|\, R=1)$ (the probability of being in room A conditioned on spiking).

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

However, since we assume that rat must be in either room A or room B, we know that $P(A \,|\, R=0) + P(B \,|\, R=0) = $
. Hence, 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) = $
.

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.

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
.

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.

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
.

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.