So in this lecture, we'll be discussing an important topic in Arima theory, which is converting between
different models.
So throughout this course, you've learned about various different models, such as a. m. a va voom
and so forth.
We've discussed these models as if they are different, but in fact, they are the same to some degree.
In this lecture, you learn how to convert a model from one form to another.
For example, how to convert an R to an May answer to a VA.
Now, we could have started this discussion in the arena section, but I think it gives you a better
overall picture when you can incorporate vector models as well.
So let's start with a scale or time series, suppose that we have an air one, this is equal to some
byas term plus five times the previous value of Y plus the error at time t, but note that this equation
is a recurrence.
Therefore, just as we can express Y of T in terms of T minus one, we can express C, T minus one in
terms of T minus two.
So we can plug in the expression for Y of T minus one in terms of Y of T minus two.
The next step is to simplify this expression, so now you can see that we have T minus two now has a
squared coefficient and two different error terms appear.
Of course, we can just do this again and replace Y of T minus two in terms of Y of T minus three.
After we expand this, we see that we get another term, another error term, and the coefficient in
front of Y of T minus three is now five one to the power of three.
The next step is to consider what happens if we repeat this pattern, an infinite number of times.
So when we repeat this pattern, an infinite number of times, we end up with three terms, the first
term involves be multiplied by one over every integer power from zero to infinity.
Note that since B appears in all of these terms, it can be factored out.
The next term I want to look at is the last term, which involves the sum of all past errors.
Now, in a typical model, this would only sum up.
Q But in this case?
Q is infinite.
However, this doesn't change the fact that we still have a weighted sum of past errors.
The final term we have involves the limit of one to the power K times Y of T minus K as K approaches
infinity.
I mentioned this one last because it's kind of a pesky term that we want to go away, but it only goes
away if we make further assumptions.
For example, if the value of the Time series is just zero beyond a certain point in the past, then
clearly this whole thing would be zero.
Or if the magnitude of PHY is less than one, this term would also approach zero.
However, these details are not really too much of a concern, we can just pretend it's zero or constant
without too much trouble.
The main point to realize is that this is an M.
Infiniti model.
We have a bias term and we have a linear combination of past era terms.
Therefore, in error, one is equivalent to an M infinity.
Now, just in case you're curious, in the final line here, I have assumed that the middle term is
zero and we use the summation formula from calculus to simplify the summation.
And clearly, this is just a number.
So it's considered a bias term.
The next model I want to consider is an M1, so let's begin by writing down the formula for M1.
You'll notice that I've written the parameters with these prime symbols.
This is because the trick here is to write them a one in a different form.
Specifically, we're going to bring all the terms to the other side in order to express Epsilon T in
terms of everything else.
Now we have a normal SI and a normal theta one which are just negative C prime and negative theta one
prime.
So the trick is to use this recurrence instead.
For the next step, we can replace Epsilon at T minus one with an expression in terms of Epsilon of
T minus two, and we can do this because we just found a recurrence that allows us to do this.
And this should seem familiar because it's exactly the same kind of substitution we just used after
expanding this.
We now have to y terms to see terms and one error term a time T minus two.
The next step, of course, is to replace the Epsilon at T minus two term with an expression in terms
of Epsilon at T minus three.
After expanding this expression, we get three wide terms, three terms and the error term at time T
minus three.
OK, so clearly this is the same kind of pattern we had before.
So, again, if we extend this an infinite number of times, we get an expression with three terms,
the first term is an infinite weighted sum of was.
The second term is just the constant, since theta one is constant and see is also constant.
Note that I've used the infinite summation formula.
Once again, the final term is the limit, which again for simplicity will assume a zero.
Now, remember that our Time series is Y of T, not Epsilon of T.
So in order to get back, we have T we can simply pull it out of the summation.
Now the summation starts at K equals one instead of equals zero.
The next step is to move everything to the other side into a form we can more easily recognize, as
you can see, we have a constant term, plus an infinite sum of lag terms, plus the usual error.
Thus, this is in r infinity.
OK, so we've just shown the equivalence between ARMM, the next step is to think of vector auto regressions.
In fact, for part of this, there's really no more work to do.
We can just use the same kind of mathematical substitutions to show that VAR one is the same as Vermes
Infinity.
Likewise, we can show that PVM one is the same as VAR Infinity.
The only difference is that instead of multiplying scalars by scalars, you'll be multiplying matrices
by vectors, but intuitively you should be able to see that the same math applies.
The interesting part, I think, is to show that an AARP can be expressed as a VA, want to see how
this can be done?
It's easiest to start with an actual process.
In this case, we have to coefficients five one in five to note that for the following slides, you'll
notice that these formulas look a little different.
This is because the program I normally use to make equations does not work with matrices.
So in order to have matrices, I had to use a different program, which is what you're seeing here.
In fact, if you read a lot of papers, this will probably look more similar to what you're used to.
So suppose we create a new vector, which is just a vector containing both YFC and YFC minus one.
We can then write the VA one model in terms of this whole vector instead of just Y of T.
So as an exercise, I recommend going backwards and expanding this matrix equation on paper for the
top row when you multiply this out.
You should get back to the original R two for the bottom row when you multiply it out.
You should get back to the trivial equation.
Y of T minus one equals y t minus one and of course recognize that this new vector equation is of our
one.
One way to see this, if it's not already clear, is the use substitution.
So suppose that we let the Vector ZT be equal to a vector containing YFC and Y of T minus one.
We can let the prime be A vector containing B and zero.
We can let capital thig be a matrix containing five one five to one and zero, and we can let eight
to be a vector containing Epsilon C and zero.
So when we do this, the equation becomes precisely the definition of a var one.
The trick is seeing that Z of T minus one is a vector containing Y of T minus one and Y of T minus two.
So let's now try this strategy again for three, so this time we'll create a vector with three components
containing YFC, YFC minus one and YFC minus two.
Again, we do the same thing where we express the top row as the original Air Three and the next two
rows as dummy equations.
So if you multiply out the bottom two rows, you should get the trivial equations.
YFC minus one equals YFC minus one and YFC minus two equals YFC minus two.
And again, note that this vector model is of our one.
So an even better trick is to show that we can do the same thing, not just with an AARP, but with
a VIP.
So let's start with a VA three.
This is just like what we had before, except now with matrices and vectors.
So the first step, again, is to concatenate Y of T, Y of C minus one and Y of T minus two into a
bigger vector.
So this new vector would be three times the size as each individual y.
The next step follows the same pattern where the top row becomes the original VARE three and the next
two rows are dummy equations.
But note that now instead of having ones, we have eyes, as you recall, any matrix times eye, which
is the identity matrix, just gives you back the same original matrix.
So what we have shown is any VPE can be expressed as of our one, except that the vector has become
P times larger.
So what have we just shown?
We've shown that both AARP and VA can be expressed as VA one and of course, as we saw earlier, this
can be expressed as VA infinity.
Now, one thing we didn't go through is to show that McHugh and the VA make you can be expressed as
VA one.
So as an exercise, you can show that this is true by applying the same tricks we learned earlier in
this lecture.
But we also know that PVM one can be expressed as VA infinity.
So we've sort of completed the cycle between VA to FEMA and back again.
Now, again, this lecture is optional, so if you don't care about this sort of thing, that's OK.
But to be clear, this is a typical part of any college level time series course.
So it's something worth knowing.
Just keep in mind that this leads to a lot more math that we could do, which eventually leads to practical
applications.
At this time, we won't look into these things since they require a lot more linear algebra.
But this at least exposes you to more possibilities.