1
00:00:00,190 --> 00:00:04,880
In this lecture we will talk about
invertibility and stationarity conditions.

2
00:00:06,150 --> 00:00:10,130
Objectives is to articulate
invertibility condition for

3
00:00:10,130 --> 00:00:12,430
moving average processes with the order q.

4
00:00:13,570 --> 00:00:18,530
Discover stationarity condition for
auto-regressive processes with order p.

5
00:00:19,820 --> 00:00:23,525
And then we will relate at
moving average processes and

6
00:00:23,525 --> 00:00:26,000
auto-regressive processes,
through duality.

7
00:00:26,000 --> 00:00:29,506
So, let's start with MA(q) process.

8
00:00:29,506 --> 00:00:33,330
This MA(q) process is basically
Xt is a linear combination

9
00:00:33,330 --> 00:00:35,890
of innovations q lags back.

10
00:00:35,890 --> 00:00:41,290
And if you write B to use
backward shift operator,

11
00:00:41,290 --> 00:00:45,570
and write this in terms of
Xt is equal to beta B Zt and

12
00:00:45,570 --> 00:00:49,880
beta B is going to be
our polynomial operator.

13
00:00:49,880 --> 00:00:53,035
And what we are trained to do,
as we did in the last lecture for

14
00:00:53,035 --> 00:00:57,920
MA(1) process, we tried to find
inverse operator for beta B and

15
00:00:57,920 --> 00:01:02,640
write this as alpha zero + alpha one B and
so forth.

16
00:01:02,640 --> 00:01:06,780
The thing is finding this alpha zero,
alpha one, alpha two is not that easy, but

17
00:01:06,780 --> 00:01:12,210
at least you would like to be able to
find inverse of this polynomal operator.

18
00:01:12,210 --> 00:01:14,730
It throws up that we have a condition.

19
00:01:14,730 --> 00:01:17,280
And we're going to use this
condition as a black box.

20
00:01:17,280 --> 00:01:19,028
I will not give the proof.

21
00:01:19,028 --> 00:01:25,497
A proof for MA(1) process will
be in this optional lecture.

22
00:01:25,497 --> 00:01:28,663
MA(q) process is in invertal.

23
00:01:28,663 --> 00:01:33,470
In other words, we can write Zt in
terms of Xts as an infinite sum.

24
00:01:33,470 --> 00:01:39,130
If the roots of this polynomial,
which might be a real R complex,

25
00:01:39,130 --> 00:01:43,465
lies outside of a unit circle,
where we regard B as

26
00:01:43,465 --> 00:01:48,370
a complex number, not as an operator.

27
00:01:50,410 --> 00:01:56,857
So for example let's look at the MA(1)
process, this is our Xt, Zt + beta Zt- 1.

28
00:01:56,857 --> 00:02:02,053
Our operator, polynomial operator
is 1 + beta B, the only route,

29
00:02:02,053 --> 00:02:07,265
in this case there's only one real route,
is negative 1 over beta.

30
00:02:07,265 --> 00:02:10,202
Which has to have a magnitude
with greater than 1 so

31
00:02:10,202 --> 00:02:14,641
that it's outside of the unit circle
that would tell us that magnitude of, or

32
00:02:14,641 --> 00:02:18,630
absolute value in this case,
of beta is actually less than one.

33
00:02:18,630 --> 00:02:23,420
Then we can work Zt as
infinite sum of Xts.

34
00:02:23,420 --> 00:02:24,860
Let's look at MA(2) process.

35
00:02:24,860 --> 00:02:26,480
Now this is MA(2) process.

36
00:02:26,480 --> 00:02:33,010
And polynomial operator is 1 +
pi over 6B + 1 over 6B squared.

37
00:02:33,010 --> 00:02:38,943
If you would like to check if this
process is invertible, then all we

38
00:02:38,943 --> 00:02:45,503
have to do according to this invertibility
condition is to check whether or

39
00:02:45,503 --> 00:02:51,250
not all routes of this polynomial
lie outside of the unit circle.

40
00:02:52,820 --> 00:02:56,100
In terms of complex numbers
this is our quadratic equation.

41
00:02:56,100 --> 00:03:00,950
And it turns out that in this case we
still have real roots, which is 2 and

42
00:03:00,950 --> 00:03:06,170
3, both of which are lines outside of
the unit circle, they're on its axis.

43
00:03:06,170 --> 00:03:11,607
But both of them are on one side of it,
they're above 1.

44
00:03:11,607 --> 00:03:14,682
So we can try and
find inverse of the polynomial operator,

45
00:03:14,682 --> 00:03:16,829
usually you would not be able to find it.

46
00:03:16,829 --> 00:03:21,458
But in this case we can actually do it,
we write this 1 over our polynomial and

47
00:03:21,458 --> 00:03:25,370
factorize our polynomial,
and use partial fractions.

48
00:03:25,370 --> 00:03:29,170
And write this into two fractions, if you
actually find common denominator here,

49
00:03:29,170 --> 00:03:34,580
you will see you that you will get
back to this rational function.

50
00:03:34,580 --> 00:03:37,170
There B is regarded as a complex number.

51
00:03:37,170 --> 00:03:41,682
Now, we can expand each
fraction as a geometric series.

52
00:03:41,682 --> 00:03:47,250
There, we start at 3 and our R is going
to be half B, or going to be one third B,

53
00:03:47,250 --> 00:03:54,460
and if we combine these 2 geometric
series, we'll have a expression for

54
00:03:54,460 --> 00:03:59,410
inverse of this polynomial operator,
and this is going to be the inverse.

55
00:04:01,260 --> 00:04:05,970
In other words, Zt can be written
as this operator acting on an Xt,

56
00:04:05,970 --> 00:04:09,990
which will give us this pi k, and
pi ks are basically these coefficients.

57
00:04:11,580 --> 00:04:18,450
So, as we can see, MA(2) process can
be inverted into AR(infinity) process.

58
00:04:20,430 --> 00:04:23,620
If invertible, the condition is satisfied.

59
00:04:23,620 --> 00:04:28,750
Now remember MA(q) processes
are always stationary.

60
00:04:28,750 --> 00:04:31,520
But it is not the case for
AR(p) processes.

61
00:04:31,520 --> 00:04:34,568
It turns out that there's
a counterpart for

62
00:04:34,568 --> 00:04:40,050
AR(p) processes and
similar condition which hold it for

63
00:04:40,050 --> 00:04:45,070
immutability for MA(q) processes holds for
stationarity of AR(p) processes.

64
00:04:45,070 --> 00:04:46,520
So what is the condition?

65
00:04:46,520 --> 00:04:50,610
We have AR(p)process where Xt
is regressed from the previous

66
00:04:50,610 --> 00:04:52,980
p lags with some random noise.

67
00:04:52,980 --> 00:04:57,030
We write this using backwards shift
operator, our operator polynomial is going

68
00:04:57,030 --> 00:05:03,740
to look like 1- phi 1B- phi 2B square-
phi pB to the P, because order is P.

69
00:05:03,740 --> 00:05:08,270
And all we have to check, in this case,
is again, to make sure the polynomial,

70
00:05:08,270 --> 00:05:13,230
this polynomial, has all of its complex
rules outside of the unit circle.

71
00:05:14,440 --> 00:05:17,120
So in case of AR (1) process.

72
00:05:17,120 --> 00:05:22,850
Our polynomial operator is 1- phi 1B,
which has only one real root].

73
00:05:22,850 --> 00:05:25,182
Which is 1 over phi 1.

74
00:05:25,182 --> 00:05:28,880
You have to make sure that this
slides outside of the unit circle.

75
00:05:28,880 --> 00:05:30,300
In other words, the magnitude.

76
00:05:30,300 --> 00:05:34,020
In this case, the absolute
value has to be greater than 1.

77
00:05:34,020 --> 00:05:36,403
Which means that phi1 must be less than 1.

78
00:05:36,403 --> 00:05:40,692
In other words, if phi1 has
absolute value less than 1 then

79
00:05:40,692 --> 00:05:44,140
AR(1) process is going to be stationary.

80
00:05:44,140 --> 00:05:46,060
And we can actually invert it.

81
00:05:46,060 --> 00:05:51,610
And when I say invert, we can find
inverse of this operator acting on Zt.

82
00:05:51,610 --> 00:05:54,930
It's going to be 1 + phi
1B + phi 1B squared.

83
00:05:54,930 --> 00:05:59,180
This is basically geometric expansion,
and it's acting on Zt, and

84
00:05:59,180 --> 00:06:04,100
we write Xt as infinite sum of Zts, right?

85
00:06:04,100 --> 00:06:09,730
So in other words, we actually express
the AR(P) process as MA(infinity) process.

86
00:06:09,730 --> 00:06:12,900
We'll come back to that, but
let's just take another look at phi 1.

87
00:06:12,900 --> 00:06:17,070
If you actually take the variance of Xt,
variance of infinite sum,

88
00:06:17,070 --> 00:06:22,040
assuming that infinite sum is
convergent in some sense which is again

89
00:06:22,040 --> 00:06:27,580
the optional video,
it's in mean square sense.

90
00:06:27,580 --> 00:06:32,295
Then, we can basically distribute
the variance because all of the Zs

91
00:06:32,295 --> 00:06:36,279
are uncorrelated, and
variance of Zs are sigma squared,

92
00:06:36,279 --> 00:06:41,640
become pull off sigma squared,
we'll get this infinite sum of numbers.

93
00:06:41,640 --> 00:06:44,210
But this is a geometric series, right?

94
00:06:44,210 --> 00:06:46,600
This is our usual geometric series.

95
00:06:46,600 --> 00:06:50,790
R is phi 1 squared, so
it has to have magnitude less than 1, so

96
00:06:50,790 --> 00:06:56,490
that this infinite sum is convergent,
so that the variance do exists.

97
00:06:56,490 --> 00:06:59,019
In other words,
you want phi 1 to be less than 1.

98
00:06:59,019 --> 00:07:00,880
So you just found that,

99
00:07:00,880 --> 00:07:06,279
if I want my AR(1) process to be
stationary in two different ways,

100
00:07:06,279 --> 00:07:11,720
we actually proved that absolute
value of phi 1 must be less than 1.

101
00:07:11,720 --> 00:07:15,600
So, AR(p) processes can be
written as MA(infinity) process,

102
00:07:15,600 --> 00:07:20,620
if we have this stationarity
condition that that holds.

103
00:07:22,010 --> 00:07:25,610
So that gives us duality, so
we have a duality between AR,

104
00:07:25,610 --> 00:07:30,250
other aggressive processes, and
MA processes, moving other processes.

105
00:07:30,250 --> 00:07:33,830
Under invertibility condition
MA(q) processes can be written

106
00:07:33,830 --> 00:07:35,480
as AR(infinity) process.

107
00:07:35,480 --> 00:07:37,573
Under stationarity condition,

108
00:07:37,573 --> 00:07:42,340
AR(p) process can be written
as MA(infinity) process.

109
00:07:42,340 --> 00:07:43,170
So what have we learned?

110
00:07:45,250 --> 00:07:49,631
We have learned invertibility
conditions for MA(q) processes.

111
00:07:49,631 --> 00:07:54,448
We have learned stationarity condition for
AR(p) processes and we have learned

112
00:07:54,448 --> 00:07:58,998
duality between moving average processes
and other regressive processes.