Introduction to Probability and Stochastic Processes I 6/ARMA Model … · One exception is the...

Introduction to Probability andStochastic Processes I

Lecture 5

Henrik Vie Christensen

[email protected]

Department of Control Engineering

Institute of Electronic Systems

Aalborg University

Denmark

Slides originally by: Line Ørtoft Endelt

Introduction to Probability and Stochastic Processes I – p. 1/43

http://www.control.auc.dk/~vie/

LLTIVC Systems

1. Lumped: A dynamic system is called lumped if it can bemodeled by a set of ordinary differential or differenceequations.

2. Linear:

f(a1x1(t) + a2x2(t)) = a1f(x1(t)) + a2f(x2(t))

3. Time Invariant: If y(t) = f(x(t)), then

y(t − t0) = f [x(t − t0)]

4. Causal:y(t0) = f [x(t); t ≤ t0]


Deterministic LTIVC I

The input-output relation can be written as

N∑

m=0

amy[(m + n)Ts] =N

∑

m=0

bmx[(m + n)Ts]

It is assumed that x, y, ai’s and bi’s are real-valued.

And the input-output relation can also be written

y(n) = x(n) ∗ h(n) =∞

∑

m=−∞

h(n − m)x(m) =∞

∑

m=−∞

h(m)x(n − m)

where h is the impulse response, since the system iscausal, then h(k) = 0 for k < 0.


Deterministic LTIVC II

For a stable system

∞∑

k=0

|h(k)| < ∞

The transfer function is found as

H(f) = F{h(n)} =∞

∑

m=−∞

h(n) exp(−j2πnf), |f | <1

2

The impulse response is recovered by

h(n) = F−1{H(f)} =

∫1

2

−1

2

H(f) exp(j2πnf) df


Deterministic LTIVC III

Assume that F (x(n)) and F (y(n)) exists, then

YF (f) =∞

∑

n=−∞

[

∞∑

m=−∞

h(m)x(n − m)]

exp(−j2πnf)

If the system is stable, this can be rewritten to

YF (f) = XF (f)H(f)

The output y(n) is recovered as

y(n) = F−1{YF (f)} =

∫1

2

−1

2

XF (f)H(f) exp(j2πnf)df


Deterministic LTIVC IV

The Z-transform of a discrete sequence is defined as

XZ(z) =

∞∑

n=0

x(n)z−n

HZ(z) =

∞∑

n=0

h(n)z−n

The Z-transform is only applicable when the sequence isdefined on the nonnegative integers.

For a stable system, the fourier transform and theZ-transform are equal if z = exp(j2πf).

The Z-transform of a stable system will exists if |z| > 1.


Random Input I

If the input to the system is a random sequence X(n), theinput-output relation may be written as

Y (n) =

∞∑

m=−∞

h(m)X(n − m)

The mean of the output is found as

E{Y (n)} = µY (n) =∞

∑

m=−∞

h(m)E{X(n − m)}


Random Input II

The autocorrelation of the output is found as

RY Y (n1, n2) = E{Y (n1)Y (n2)}

=∞

∑

m1=−∞

∞∑

m2=−∞

h(m1)h(m2)RXX(n1 − m1, n2 − m2)

TheDistribution function of the output sequence are ingeneral very difficult to obtain, and cannot in general beexpressed in closed functional form.

One exception is the Gaussian process, since Y (n) isobtained as a linear combination of Gaussians, and henceGaussian. The mean-value and the Variance can be foundfrom the two equations above.


WSS input

If X(n) is WSS, then

µY = E{Y (n)} =∞

∑

m=−∞

h(m)µX = µX

∞∑

m=−∞

h(m) = µXH(0)

and

RY Y (n1, n2) =∞

∑

m1=−∞

∞∑

m2=−∞

h(m1)h(m2)

× RXX [(n2 − n1) − (m2 − m1)]

Hence the output of a LTIVC system is WSS when the inputis WSS. (It can be shown that SSS → SSS).


Correlation and psd of Output I

Assume that X(n) is a real WSS input to a LTIVC system,the cross-correlation function between the input and output

RY X(k) = E{Y (n)X(n + k)}

= E{[

∞∑

m=−∞

h(m)X(n − m)]

X(n + k)}

=∞

∑

m=−∞

h(m)E{X(n − m)X(n + k)}

=∞

∑

m=−∞

h(m)RXX(k + m)

=∞

∑

n=−∞

h(−n)RXX(k − n) = h(−k) ∗ RXX(k)


Correlation and psd of Output II

Since RY X(τ) = RXY (−τ) and RXX is an even function

RXY (k) = h(k) ∗ RXX(k)

It can also be shown that

RY Y (k) = RY X(k) ∗ h(k)

and hence

RY Y (k) = RXX(k) ∗ h(k) ∗ h(−k)


Correlation and psd of Output III

The psd of Y (n) is the fourier transform of theautocorrelation function of Y (n)

SY Y (f) =∞

∑

n=−∞

RY Y (n) exp(−j2πnf), |f | <1

2

also

SY Y (f) = F{RY Y (k)}= F{RXX(k) ∗ h(k) ∗ h(−k)}= F{RXX(k)}F{h(k)}F{h(−k)}= SXX(f)H(f)H(−f)

= SXX(f)H(f)H∗(f)

= SXX(f)|H(f)|2Introduction to Probability and Stochastic Processes I – p. 12/43

Example

X(n) is a zero-mean white noise random sequence, withvariance σ2. Let X(n) be the input sequence to a LTIVCsystem, then

SXX(f) = F (RXX(n)) = σ2

since RXX(0) = σ2 and RXX(n) = 0 for n 6= 0, so

SY Y (f) = SXX(f)|H(f)|2 = σ2|H(f)|2

Also

RXY (k) = h(k) ∗ RXX(k) = h(k)σ2

So if the input and the output is known, the impulseresponse can be found.


Correlation and psd of Output IV

From the definition of the Z-transform

HZ [exp(j2πf)] = H(f)

Defining

S#XX(z) =

∞∑

n=0

z−nRXX(n)

S#XX [exp(j2πf)] = SXX(f)

It can be shown that

S#Y Y (z) = S#

XX(z)|H(z)|2

= S#XX(z)H(z)H(z−1)


Example a I

X(n): Stationary random sequence, is input to a LLTIVCsystem, with µX = 0, RXX(0) = 1 and RXX(k) = 0 for k 6= 0.

h(k) =

{

1 for k = 0, 1

0 for k > 1

The mean of the output Y (n):

µY (n) = µXH(0) = 0

The Transfer Function

H(f) =∞

∑

k=0

h(k) exp(−j2πkf)

= 1 + exp(−j2πf)


Example a II

The psd for X

SXX(f) = F{RXX(k)}

=∞

∑

k=0

RXX(k) exp(−j2πkf)

= 1, |f | <1

2

The psd of Y is found as

SY Y (f) = SXX(f)|H(f)|2

= 1 · |1 + exp(−j2πf)|2

= 2 + 2 cos 2πf, |f | <1

2Introduction to Probability and Stochastic Processes I – p. 16/43

Example a III

Using the Inverse Fourier Transform:

RY Y (k) = F−1{SY Y (f)}

=

∫1

2

−1

2

(2 + 2 cos 2πf) exp(j2πkf) df

=

∫1

2

−1

2

(2 + exp(−j2πf) + exp(j2πf)) exp(j2πkf) df

soRY Y (0) = 2

RY Y (±1) = 1

RY Y (k) = 0 |k| > 1


Example b I

X(n) is a zero-mean white noise random sequence, withvariance σ2. X(n) is input to a filter with transfer function

HZ(z) =a0 + a1z

−1 + a2z−2

1 + b1z−1

µY (n) = µXH(0) = 0

The psd of X(n) is

S#XX(z) =

∞∑

n=0

z−nRXX(n) = σ2


Example b II

The psd of Y (n) is

S#Y Y (z) = σ2

(a0 + a1z−1 + a2z

−2

1 + b1z−1

)(a0 + a1z + a2z2

1 + b1z

)

= σ2a20 + a2

1 + a22 + a1(a0 + a2)(z + z−1) + a0a2(z

2 + z−2)

1 + b21 + b1(z + z−1)

Switching to the Fourier domain, subst. z = exp(j2πf):

SY Y (f) = σ2a20 + a2

1 + a22 + 2a1(a0 + a2) cos 2πf + 2a0a2 cos 4πf

1 + b21 + 2b1 cos 2πf

|f | <1

2


Autoregressive Process I

An Autoregressive process is one represented by adifference equation of the form

X(n) =

p∑

i=1

φp,iX(n − i) + e(n)

Where X(n) is a real random sequence, φp,i areparameters, with φp,p 6= 0, and e(n) is a sequence of iidzero-mean white noise, i.e.

E{e(n)} = 0

E{(e(n))2} = σ2N

fe(n)(λ) =1√

2πσN

exp{

− λ2

2σ2N

}


Autoregressive Process II

Autoregressive models are also called state models,recursive digital filters, and all-pole models.The AR-model can be transformed to a state model

X(n) = ΦX(n − 1) + E(n)

Often the AR-model is written as

X(n) =

p∑

i=1

hiX(n − i) + e(n)

The transfer function of the AR-model is

H(f) =1

1 −∑p

i=1 φp,i exp(−j2πfi), |f | <

1

2


First-order Autoregressive Model I

First-order Autoregressive model:

X(n) = φ1,1X(n − 1) + e(n)

The mean of X(n) is found as

µX = E{X(n)} = E{φ1,1X(n − 1) + e(n)}= φ1,1µX + 0

hence for φ1,1 6= 1

µX = 0


First-order Autoregressive Model II

0 20 40 60 80 100−3

−2

−1

0

1

2

3

Sample function of a first-order AR-process:X(n) = 0.48X(n − 1) + e(n).


First-order Autoregressive Model III

0 20 40 60 80 100−10

−5

0

5

Sample function of a first-order AR-process:X(n) = 0.97X(n − 1) + e(n).


First-order Autoregressive Model IV

The variance of the AR-model is

σ2X = E{X(n)2}

= E{φ21,1X(n − 1)2 + e(n)2 + 2φ1,1X(n − 1)e(n)}

= φ21,1σ

2X + σ2

N

hence

σ2X =

σ2N

1 − φ21,1

Since σ2X is finite and nonnegative, then

−1 < φ1,1 < 1


First-order Autoregressive Model V

The autocorrelation function is given by

RXX(m) = E{X(n)X(n − m)}, m ≥ 1

= E{[φ1,1X(n − 1) + e(n)][X(n − m)]

= φ1,1RXX(m − 1)

Hence

RXX(m) = φm1,1RXX(0) = φm

1,1σ2X

and the autocorrelation coefficient of the process is

rXX(m) =RXX(m)

RXX(0)= φm

1,1


First-order Autoregressive Model VI

0 2 4 6 8 10 120

0.2

0.4

0.6

0.8

1

Corrolation coefficient of the first-order AR-process.Introduction to Probability and Stochastic Processes I – p. 27/43

First-order Autoregressive Model VII

The psd can be found using

SXX(f) = |H(f)|2See(f)

Since

See(f) = σ2N |f | <

1

2

H(f) =1

1 − φ1,1 exp(−j2πf)|f | <

1

2

|H(f)|2 =1

1 − 2φ1,1 cos(2πf) + φ21,1

|f | <1

2


First-order Autoregressive Model VIII

Therefore

SXX(f) =σ2

N

1 − 2φ1,1 cos(2πf) + φ21,1

|f | <1

2

SXX(f) =σ2

X(1 − φ21,1)

1 − 2φ1,1 cos(2πf) + φ21,1

|f | <1

2

If z−1 is defined to be the backshift operator, which can bewritten as:

z−1[X(n)] = X(n − 1), z−1[e(n)] = e(n − 1)

z−k[X(n)] = X(n − k), z−k[e(n)] = e(n − k)


First-order Autoregressive Model IX

Then the first-order autoregressive model can be written as

X(n) = φ1,1z−1[X(n)] + e(n)

Hence

X(n) =e(n)

1 − φ1,1z−1

and, then

X(n) =[

∞∑

i=0

φi1,1z

−i]

e(n) =∞

∑

i=0

φi1,1e(n − i)

A weighted infinite sum of white noise.


Second-order Autoregressive Model IThe second-order autoregressive process is given by

X(n) = φ2,1X(n − 1) + φ2,2X(n − 2) + e(n)

0 20 40 60 80 100−3

−2

−1

0

1

2

3


Second-order Autoregressive Model II

The mean value of the process is

µX = φ2,1µX + φ2,2µX

hence if φ2,1 + φ2,2 6= 1, then

µX = 0


Second-order Autoregressive Model III

The variance

σ2X = E{X(n)X(n)}

= E{φ2,1X(n)X(n − 1) + φ2,2X(n)X(n − 2) + X(n)e(n)}= φ2,1RXX(1) + φ2,2RXX(2) + σ2

N

Since RXX(k) = σ2XrXX(k), then

σ2X =

σ2N

1 − φ2,1rXX(1) − φ2,2rXX(2)

Since σ2X is finite and positive, then

φ2,1rXX(1) + φ2,2rXX(2) < 1


Second-order Autoregressive Model IVRXX(m) for m ≥ 1:

RXX(m) = E{X(n − m)X(n)}= E{φ2,1X(n − m)X(n − 1) + φ2,2X(n − m)X(n − 2)

+ X(n − m)e(n)}= φ2,1RXX(m − 1) + φ2,2RXX(m − 2)

This is a second-order linear homogeneous differenceequation, having the solution

RXX(m) = A1λm1 + A2λ

m2 if λ1 6= λ2

= B1λm + B2mλm if λ1 = λ2

λ1 and λ2 are roots of the characteristic equation obtainedby assuming RXX(m) = λm for m ≥ 1, which gives . . .


Second-order Autoregressive Model V

λ2 = φ2,1λ + φ2,2

hence

λ =φ2,1 ±

√

φ22,1 + 4φ2,2

2

If the model must satisfy the initial condition, RXX(0) = σ2X ,

and we require stationarity, then

RXX(1) = φ2,1RXX(0) + φ2,2RXX(−1)

RXX(1) =φ2,1

1 − φ2,2σ2

X


Second-order Autoregressive Model VI

If λ1 6= λ2, then

RXX(0) = σ2X = A1 + A2

RXX(1) = A1λ1 + A2λ2 =φ2,1

1 − φ2,2σ2

X

Now A1 and A2 can be found, hence RXX is known.

With ai = Ai/σ2X for i = 1, 2

rXX(m) =RXX(m)

σ2X

= a1λm1 + a2λ

m2


Second-order Autoregressive Model VII

So

rXX(1) =φ2,1

1 − φ2,2

rXX(2) =φ2

2,1

1 − φ2,2+ φ2,2

Substituting this in the expression for the variance gives

σ2X =

σ2N (1 − φ2,2)

(1 + φ2,2)(1 − φ2,1 − φ2,2)(1 + φ2,1 − φ2,2)

This will be finite if

φ2,2 6= −1, φ2,1 + φ2,2 6= 1, φ2,2 − φ2,1 6= 1


Second-order Autoregressive Model VIII

And positive if

−1 < φ2,2 < 1, φ2,1 + φ2,2 < 1, −φ2,1 + φ2,2 < 1

The psd is given by

SXX(f) = |H(f)|2σ2N , |f | <

1

2

Where

H(f) =1

1 − φ2,1 exp(−j2πf) − φ2,2 exp(−j4πf), |f | <

1

2


Second-order Autoregressive Model IX

Giving

SXX(f) =σ2

N

|1 − φ2,1 exp(−j2πf) − φ2,2 exp(−j4πf)|2 , |f | <1

2

This could also be found as the Fourier transform ofRXX(m).


General Autoregressive Model I

The general autoregressive model:

X(n) =

p∑

i=1

φp,iX(n − i) + e(n)

or using the backshift operator

X(n) =

p∑

i=1

φp,iz−iX(n) + e(n)

By using the same technique as before, the mean

µX = 0


General Autoregressive Model II

The variance

σ2X = E{X(n)X(n)} = E

{

X(n)

p∑

i=1

φp,iX(n − i) + X(n)e(n)}

=

p∑

i=1

φp,iRXX(i) + σ2N

The autocorrelation coefficient is given by

rXX(k) =RXX(k)

σ2X

=E{X(n − k)X(n)}

σ2X

=

p∑

i=1

φp,irXX(k − i), for k ≥ 1


General Autoregressive Model III

rXX(k) =

p∑

i=1

φp,irXX(k − i), for k ≥ 1

This is a pth order difference equation, which can beexpressed in matrix form

rXX = RΦ

where R is the correlation coefficient matrix, rXX is thecorrelation coefficient vector, and Φ is the autoregressivecoefficient vector.The matrix equation is called the Yule-Walker equation.Since R is invertible

Φ = R−1

rXX


General Autoregressive Model IV

The psd can be shown to be

SXX(f) = See(f)|H(f)|2

=σ2

N∣

∣1 −∑p

i=1 φp,i exp(−j2πfi)∣

∣

2 , |f | <1

2


Introduction to Probability and Stochastic Processes I 6/ARMA Model … · One exception is the...

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