And that's it, in terms of independencies.

And so, this, in fact, is an I-map for the distribution because in this case, I

of G is a subset of I of P, which is this set over here.

But it doesn't capture all of the independencies, it only captures one out

of the two. So, this is the distribution that does not have a perfect map of a

Bayesian network. Let's come up.

Let's provide another example of an imperfect map.

And, in fact, this is a this is a distribution that doesn't have an

imperfect map as either a Bayesian network or, in fact, as a Markov network.

and this is the famous example of the XOR, which is, as we'll see, a

counterexample to many, many things. So here, we have two random variables X1

and X2, which we're assuming are binary valued and say,

each of them is, takes the value of 0,1 with the probability of 50, 50.

Why, on the other hand, is the exclusive or of X1 and X2.

Which means that Y is equal to one, if and only, if exactly one of X1 or X2 is

equal to one, okay?

So, let's look at what this probability distribution P looks like.

Here it is. we have that there's four possible

configurations that have nonzero probability.

and each of them has equal probability of 0.25.

So, X1, X2 can be either zero or one with probability 50,

50 and here's the value of Y, over here. But now, let's think about other so let's

think about what independent statements are true for this distribution.

So, most obviously looking at this graph, we see that X1 is marginally independent

of X2. But if you close your eyes on this part

of, this, of the image, and just look at the right-hand side, you'll see that

really X1, X2, and Y are all symmetrical in terms of their structure.

And so, it's not difficult to verify that we also have the X1, in fact, is

independent of Y, and that X2 is independent of Y.

And all three of these are pairwise independencies hold in this distribution.

And so, this is not, in fact, the graph on the left is not a perfect map for this

distribution because their independencies that hold in P are not visible in the

graph. And, in fact, you can organize the nodes

in this graph in any which way and, but you cannot get all through these

independencies captured at the same time because the only way to do that, would be

to have X1, X2, and Y be separate variables and, of course, that wouldn't

be an I-map for the distribution. Now, we've talked about Bayes nets as a

perfect map. What about Markov networks as a perfect

map? The definition here is the same, except

that we've replaced G by H, so a Markov network H is a perfect map if the

independencies included by H are in, exactly correspond to the independencies

in P, and can we capture all possible distributions in terms of a Markov

network, is a perfect map. So, I'm sure you're all expecting at this

point for the answer to be no and sure enough, that's true.

so here, is an example of a distribution that has a perfect map, in this case, is

a Bayesian network, but not as a Markov network.

So, this is the famous V structure example, in this case the, difficulty

intelligence [UNKNOWN] structure. And let's think about what, in the, what we

would need to encode as, in terms of edges or, for being an I-map of this

distribution. So clearly, we need to introduce an edge

between B and G and between I and D because it's certainly not the case that

we can make D and G independent given I or vice versa.

And so, this would be the obvious I-map for this distribution.

But this example, if we choose this as a, as a candidate I-map, it would imply,

among other things, that B is independent of I given d.

And that, of course, is exactly wrong because we know that when we condition on

G, D is indepedenet, D D and I become dependent.

And so, the only I-map for this distribution is one that has all three

edges and that loses the independence that we had over here which is D is

marginally independent [UNKNOWN]. So once again, there's no perfect map for

this distribution as a Markov network. The final question that we might ask

ourselves is the extent to which a representation of a distribution is, in

fact, unique? And so, say, that we could represent the

distribution using some graphs, say, as a perfect map,

is that a unique representation? So, to understand that, let's look first

at the simplest example, the one where we have two variables X and Y. And here, we

have one graph that captures in this case, no independence assumptions so I of

G is equal to the empty set, H1.

And here is G2, it looks the same except that the edges are inverted, the one edge

is inverted, and once again, this is a different graph that has adopted the same

value set of independencies. So, we can see that here, we have two

distinct graphs that have the exact same of independence assumptions.

And because of that, they can represent the exact,