Now let's imagine we had multiple classifiers.
One for rain and one for whether the sprinkler is on or not.
The rain classifier deals with features like wetness of the grass and thunder,
whereas the sprinkler classifier deals with wetness of the grass, of course, but
also whether or not one sees a hose on the ground.
The probabilities for each of the classifiers work just like before.
The probability of rain, given w and t, is the probability of w given r, times the
probability of t given r, times p of r, divided by the probability of w and t.
Which is essentially the probably of whatever evidence one actually observes.
Assuming one observes both these features. Similarly, we get the equation for the
probability of a sprinkler, given an observation of h and w.
But, w is the same observation. If you are observing w for this
classifier, we should have the same value of w in this classifier.
So, we have to actually combine these networks into one network, and what we get
is called a Bayesian network. Now, it's not a simple classifier, there
are two things we could make conclusions about, sprinkler and rain,
And some of the evidence is shared, let's try to work out the probability For this
network, starting with the joint. The joint probability of H, W, T, S and R.
Using Bayes rule to factor R out, we get this first term, and we factor S out,
again, we'll get H, W, T given S and R, and the probability of S given R, and the
probability of R. This is simply a recursive application of
Bayes rule. We also know that S and R are independent.
Whether or not the sprinkler is on or whether or not it, it had rained last
night, we assume that, that these are independent.
In, in practice we might have some dependence.
That is one decides to put the sprinkler on only if it doesn't rain, but we ignore
that for the time being. We also assume as before that the, the
observations H, W and T are independent given S and R.
And this lets us factor this further as probability of H given S and R, times the
probability of W given S and R, times the probability of T given S and R and.
Because of independence of s and r, we get probability of s times probability of r.
Now, this part is tricky. H and R are also independent because the
probability of a hose lying on the ground doesn't depend on whether it's rained and
the probability of thunder occurring certainly doesn't depend on whether the
sprinkler has been put on. So these two terms, can be replaced by the
probability of H given S and the probability of T given R.
As a general rule in a vision network, and this is all you really need to understand
or remember. All you need to deal with are the
conditional probabilities of nodes given their parents.
So we have H given S, T given R and W given S and R, because it has two parents.
S and R don't have any parents, so they sit on their own as prior probabilities.
So this basic Bayesian that were in equation.
Gives us the joint probability. From this one can compute whatever one
actually requires simply using sequel by applying evidence joining and aggregating.
Let's do a couple of examples. For example, we will ignore the hose and
thunder and just deal with wetness, but now there are two possible causes for
wetness: sprinkler being on and the rain being on.
This is exactly what we started with, the confusing situation that classical logic
had trouble dealing with, and let's see how probabilistic logic using a Bayesian
network does better, perhaps. We have the conditional probability of w
given s and r, note that this is not the joint probability.
So for any particular combination of s and r, these two la- rows for different values
of w have to add up to one. So point nine and point one, point seven
and point three, point eight point two, etcetera.
We also have the priors of probability that it rains in general, and the
probability the sprinkler is on, in general, based on history.
Let's write our joint distribution now. Probability of r, s, and w, as probability
of s and r given w probability of w. And alternatively, as probability of w
given s and r probability of s probability of r.
Because we know that these two are independent.
Given that the evidence that we see on the ground as the grass is wet, we can now say
that the probability of r and s given w is the probabilty of W given R and S,
Times probability of S, times probability of R.
And now we condition it by the evidence that the grass is wet, but we are only
interested in the probability of R so we SUM out S.
So now we're going to apply the Summation operator in addition to the Selection
operator of'W equal to Yes. Sigma, as before, is the probability of the inverse
of the probability that W equal to" Yes. Essentially, we get that simply by
dividing byt P of W. In this equation.
Convince yourself that this is the case, and let's see how one can execute this in
SQL. Just like before, we select R, the sum of
all the products of the Ps from all the three tables, from all the three tables, W
equal to yes, R equal to R, and S equal to S in all the tables.
So I have, haven't written the full SQL over here, just have to make sure that all
the common variables are equal, and the evidence is applied.
And we finally group by R, Because we don't want anything to do with
S. The results should have one row for every
value of R. The result, it turns out, is the
following. Let's look at the first row.
We take the nine. From here.
Multiply it by the case where s is equal to yes.
Because here, s equal to yes. So we multiply by.3.
There's another situation where s equal to no where we also have rain equal to yes
and w equal to yes so that's.8 and that's multiplied by.7.
Both of these are cases where rain equal to yes.
So they're multiplied by.2 from this table and we get.166.
Similarly, in the case when rain equal to no.
S equal to yes and W equal to yes, of course..7 times.3.
Then we have.1 from this row, and S equal to no, so it's multiplied by.7, and of
course, both are the case where rain equal to no, so we multiply by.8.
We get.224. Normalizing this so that the sum is one,
we get.166, divide by.166 plus.224 is.42. Or in other words, the expectation or the
probability that the rain is, it happened last night given that the grass is wet is
42%. It's very important to note that earlier,
when we just had rain and wetness and no sprinkler to talk about, we got a
probability of 53% that it had rained if the grass was observed to be wet.
Now, without knowing anything about the sprinkler being on or not on, simply the
fact that there is a possible alternate cause for the grass being wet with an
explicit probability table that includes this.
We get a lower value for the probability that it had rained.
The fact is, that given the fact, that the grass is wet.
The probability that it had rained, depends on whether or not it had, the
sprinkler was on. Even if we don't observe S, the fact that
there's an alternate cause does change the probability from the situation where this
node didn't even occur. Let's see that more explicitly now by
seeing what happens if we do observe that the sprinkler is on.