Hi.
In this problem, we're going to use the set of probability
axioms to derive the probability of the difference
of two events.
Now, before we get started, there's one thing you might
notice that, the equation we're trying to prove is
actually quite complicated.
And I don't like it either, so the first thing we're going to
do will be to find a simpler notation for the events that
we're interested in.
So we start with two events, A and B, and there might be some
intersection between the two events.
We'll label the set of points or samples in A that are not
in B, as a set C. So C will be A intersection B complement.
Similarly, for all points that are in B but not in A, this
area, we'll call it D.
And D will be the set A complement intersection B. And
finally, for points that are in the intersection of A and
B, we'll call it E. So E is A intersection B. And for the
rest of our problem, we're going to be using the notation
C, D, and E instead of whatever's down below.
If we use this notation, we can rewrite our objective as
the following.
We want to show that the probability of C union D is
equal to the probability of the event A plus the
probability of B minus twice the probability of E. And that
will be our goal for the problem.
Now, let's take a minute to review what the axioms are,
what the probability axioms are.
The first one says non-negativity.
We take any event A, then the probability of A
must be at least 0.
The second normalization says the probability of the entire
space, the entire sample space omega, must be equal to 1.
And finally, the additivity axiom, which will be the axiom
that we're going to use for this problem says, if there
are two events, A and B that are disjoint--
which means they don't have anything in common, therefore.
the intersection is the empty set.
Then the probability of their union will be equal to the
probably A plus the probability of B. For the rest
of the problem, I will refer to this axiom as add.
So whenever we invoke this axiom, I'll write
"add" on the board.
Let's get started.
First, we'll invoke the additivity axioms to argue
that the probability of C union D is simply the sum of
probability of C plus probability of
D. Why is this true?
We can apply this axiom, because the set C here and the
set D here, they're completely disjoint from each other.
And in a similar way, we'll also notice the following.
We see that A is equal to the union of the set C and E.
And also, C and E, they're disjoint with each other,
because C and E by definition don't share any points.
And therefore, we have probably A is equal to
probability of C plus the probability of E. Now, in a
similar way, the probability of event B can also be written
as a probability of D plus the probability of E, because
event B is the union of D and E.
And D and E are disjoint from each other.
So we again invoke the additivity axiom.
Now, this should be enough to prove our final claim.
We have the probability of C union D. By the very first
line, we see this is simply probability of C plus the
probability of D.
Now, I'm going to insert two terms here to make the
connection with a second part of the equation more obvious.
That is, I will write probability C plus probability
E plus probability D plus probability of E. Now, I've
just added two terms here--
probability E. So to make the equality valid or subtract it
out two times, the probability of E.
Hence this equality is valid.
So if we look at this equation, we see that there
are two parts here that we've already seen
before right here.
The very first parenthesis is equal to the probability of A.
And the value of the second parenthesis is equal to the
probability of B. We just derived these here.
And finally, we have the minus 2 probability of E. This line
plus this line gives us the final equation.
And that will be the answer for the problem.