STAT485/685代做、R程序设计调试、代写analysis留学生、代做R语言

日期：2019-10-10 10:05

STAT485/685 - Notes 4C STAT485/685 - Notes 4C STAT485/685 - Notes 4C

Simulations with AR(1) Processes

• To simulate 500 values from Yt = 0.6Yt−1 + et with et ∼ NID(0, 1), use the R

command:

arima.sim(list(ar=0.6), n=500)

• To simulate 600 values from Yt = −0.4Yt−1 + et with et ∼ NID(0, 1), use the R

command:

arima.sim(list(ar=-0.4), n=600)

Note: We usuall discard some values at the beginning of each simulated series for

further analysis.

Examples: Obtain and plot the last 300 values together with their acf for each

simulated series.

Solution:

• d1=arima.sim(list(ar=0.6), n=500)[201:500]

ts.plot(d1)

acf(d1)

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Time

d1

0 50 100 150 200 250 300

−4 0 2

Simulated AR(1) with phi=0.6

0 5 10 15 20

0.0 0.6

Lag

ACF

Series d1

• d2=arima.sim(list(ar=-0.4), n=600)[301:600]

ts.plot(d2)

acf(d2)

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Time

d2

0 50 100 150 200 250 300

−3 0 2

Simulated AR(1) with phi=−0.4

0 5 10 15 20

−0.5 0.5

Lag

ACF

Series d2

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AR(1) with lag operator L

Consider the stationary AR(1) process given by

Yt = φYt−1 + et, where |φ| < 1 and {et} ∼ W N(0, σ2e).

Using the lag operator L, this can be written as

Yt = φLYt + et, where Yt−1 = LYt.

This is equivalent to (1 − φL)Yt = et

. Since |φ| < 1, it is clear that the solution to

1 − φL = 0, L =1φ > 1.

Note: The equation 1 − φL = 0 associate with this AR(1) process is known as

its charateristic equation. (see P71 of the textbook). Therefore, |L| > 1 for

stationarity. This is known as the root is outside the unit circle.

Diagram:

Example: Consider the stationary AR(1) process given by Yt = 0.7Yt−1 + et

. Write

down this using the lag operator L and the corresponding charateristic equation.

Hence find the root of this equation and show that the root satisfies |L| > 1 or the

root is outside the unit circle.

Solution:

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Exercise: Consider the AR(1) process given by Yt = −0.3Yt−1 + et

. Write down this

using the lag operator L and the corresponding charateristic equation. Hence find

the root of this equation and determine whether this process is stationary.

Solution:

Remark: In AR(1) processes, the current Yt depends only on the immediate past

value Yt−1 or the conditional probability distribution of Yt given the history dependent

only on the immediate past Yt−1. That is,

P(Yt

|Yt−1, Yt−2, · · ·) = P(Yt

|Yt−1)

and is known as the Markov Property. Therefore, a stationary AR(1) is also known

as a Markov Process.

Now we look at the second order AR(2) process in detail.

4.5 AR(2) Processes - P71 of the textbook

Consider an AR(2) process generated by the recursive equation

Yt = φ1 Yt−1 + φ2 Yt−2 + et, {et} ∼ W N(0, σ2e).

Using the lag operator, this can be written as

(1 − φ1 L − φ2 L2) Yt = et.

The associate charateristic equation for this AR(2) process is 1 − φ1 L − φ2 L

2 = 0.

For convenience, let φ(L) = 1 − φ1 L − φ2 L

2

. Then we can write the above AR(2)

process as

φ(L) Yt = et,

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Example: Assuming {Yt} given in Yt = φ1 Yt−1 + φ2 Yt−2 + et

, {et} ∼ W N(0, σ2e) is

stationary, find

(i) E(Yt).

(ii) By evaluating the both sides of V ar(Yt) = V ar(φ1 Yt−1 + φ2 Yt−2 + et), find an

expression relating γ0, γ1 and σ

2

e

. Recall that under the stationary conditions,

Cov(Yt−k, et) = E(Yt−ket) = 0, k ≥ 1.

(ii) By evaluating the both sides of V ar(Yt − φ1 Yt−1 − φ2 Yt−2) = V ar(et), find an

expression relating γ0, γ1, γ2 and σ

2

e

.

Solution:

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Solution to this charateristic equation is useful in many applications. Therefore, we

look the methods of solving quadratic equations.

A Review of Solving Quadratic Equations

Consider the charateristic equation φ(L) = 1 − φ1 L − φ2 L

2 = 0 for

Yt = φ1 Yt−1 + φ2 Yt−2 + et

.

Since this charateristic equation is a quadratic, it is easy to solve either by factorisation

or from the fomula.

• Factorise: Let 1 − φ1 x − φ2 x

2 = (1 − δ1x)(1 − δ2x). Then the solutions are

• Formula: Let ax2 + bx + c be a quadratic equation. Then the solutions are.

Example: Consider the AR(2) process given by Yt = 1.3Yt−1 − 0.4Yt−2 + et

.

(i) Write down this AR(2) process using the lag operator L.

(ii) What is the corresponding charateristic polynomial φ(L)?

(iii) Solve the equation φ(L) and show that both roots are outside the unit circle.

Solution:

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Example: For each of the following AR(2) process:

(i) Write down this AR(2) process using the lag operator L.

(ii) What is the corresponding charateristic polynomial φ(L)?

(iii) Solve the equation φ(L) and show that both roots are outside the unit circle.

(a) Yt = −0.3Yt−1 + 0.1Yt−2 + et

.

(b) Yt = 1.5Yt−1 − 0.4Yt−2 + et

.

(c) Yt = 1.3Yt−1 + 0.5Yt−2 + et

.

(d) Yt = −Yt−1 + 6Yt−2 + et

.

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Solution:

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.

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Stationarity of AR(2) Processes

An Important Result:

A Stationary solution to the AR(2) process

Yt = φ1Yt−1 + φ2Yt−2 + et

exists if and only if the roots of the associtaed charateristic equation

1 − φ1L − φ2L

2 = 0

are outside the unit circle. That is, both roots larger than 1 in absolute value.

Example: Identify the stationary AR(2) processes in the previous example. Give

reasons.

Solution

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Note: It can be shown that the quadratic equation 1 − φ1L − φ2L

2 = 0 has roots

outside the unit cicle if the following three inequalities hold:

|φ2| < 1

φ2 + φ1 < 1

φ2 − φ1 < 1.

Example: For each of the AR(2) process (a), (b), (c) and (d) above

(i) Write down φ1 and φ2.

(ii) Write down all the three inequalities as given above.

(iii) Using (ii), determine the stationarity of each process.

(iv) Are the results in (iii) consistent with stationarity/nonstationarity from the

previous example?

Solution:

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90