- About The Coursera
- About Bayesian Statistics: Time Series Analysis Course
- Bayesian Statistics: Time Series Analysis Quiz Answers
- Practice Quiz : Objectives of the course
- Quiz: Stationarity, the ACF and the PACF
- Quiz: The AR(1) definitions and properties
- Week 02 : Properties of AR processes
- Week 03: Practice Quiz The Normal Dynamic Linear Model
- Quiz : NDLM, Part I: Review
- Week 04 : Quiz Seasonal Models and Superposition
- Quiz : NDLM, Part II
- Conclusion
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About Bayesian Statistics: Time Series Analysis Course
This course is designed for working data scientists and aspiring statisticians alike.
This is the fourth and final course in a sequence of four that will introduce the fundamentals of Bayesian statistics.
It expands upon the topics covered in the course Bayesian Statistics: From Concept to Data Analysis, Techniques and Models, as well as Mixture models.
Analysis of time series focuses on modeling the dependencies that exist between the components of a sequence of variables that are time-related.
In order for you to do well in this class, you should have some background knowledge in calculus-based probability, the principles of maximum likelihood estimation, and Bayesian inference.
You will learn how to construct models that are able to describe temporal dependencies, as well as how to perform Bayesian inference and forecasting for the models that you have constructed.
Using the open-source and freely available software R, you will put what you’ve learned into practise using sample databases.
Your instructor, Raquel Prado, will guide you from the fundamental ideas behind modeling temporally dependent data to the implementation of particular classes of models.
Course Apply Link – Bayesian Statistics: Time Series Analysis
Bayesian Statistics: Time Series Analysis Quiz Answers
Practice Quiz : Objectives of the course
Q1. In this course will focus on models that assume that (mark all the options that apply):
- The observations are realizations from spatial processes, where the random variables are spatially related
- The observations are realizations from time series processes, where the random variables are temporally related
- The observations are realizations from independent random variables
Q2. In this course we will focus on the following topics
- Some classes of models for non-stationary time series
- Models for univariate time series
- Models for multivariate time series
- Some classes of models for stationary time series
Q3. Some of the goals of time series analysis that we will illustrate in this course include:
- Online monitoring
- Analysis and inference
- Forecasting
- Clustering
Q4. In this course we will study models and methods for
- Equally spaced time series processes
- Discrete time processes
- Unequally spaced time series processes
- Continuous time processes
Q5. In this course you will learn about
- Nonparametric methods of estimation for time series analysis
- Normal dynamic linear models for non-stationary univariate time series
- Bayesian inference and forecasting for some classes of time series models
- Spatio-temporal models
- Non-linear dynamic models for non-stationary time series
- Autoregressive processes
Quiz: Stationarity, the ACF and the PACF
Q1. YtββYtβ1β=etββ0.8etβ1β
How is this process written using backshift operator notation ({B}B) ?
- (1βB)Ytβ=(1β0.8B)etβ
- None of the above
- BYtβ=(1β0.8B)etβ
- B(YtββYtβ1β)=0.8Betβ
Q2. Which of the following plots is the most likely to correspond to a realization of a stationary time series process?
- .
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- .
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- .
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Q3. If \{Y_t\}{Ytβ} is a strongly stationary time series process with finite first and second moments, the following statements are true:
- {Ytβ} is also weakly or second order stationary
- {Ytβ} is a Gaussian process
- The variance of Ytβ, Var(Y_t),Var(Ytβ),changes over time
- The expected value of Ytβ, E(Y_t),E(Ytβ),does not depend on t.t.
Q4. If \{Y_t\}{Ytβ} is weakly or second order stationary with finite first and second moments, the following statements are true:
- If \{Y_t\}{Ytβ} is also a Gaussian process then \{Y_t\}{Ytβ} is strongly stationary
- {Ytβ} is also strongly stationary
- None of the above
Q5. Wβhich of the following moving averages can be used to remove a period d=8d=8 from a time series?
- 1β/8ytβ4β+41β(ytβ3β+ytβ2β+ytβ1β+ytβ+yt+1β+yt+2β+yt+3β)+81βyt+4β
- 1/8βj=β88βytβk
- 1/2β(ytβ4β+ytβ3β+ytβ2β+ytβ1β+ytβ+yt+1β+yt+2β+yt+3β+yt+4β)
- 1/8β(ytβ4β+ytβ3β+ytβ2β+ytβ1β+ytβ+yt+1β+yt+2β+yt+3β+yt+4β)
Q6. Which of the following moving averages can be used to remove a period d=3d=3 from a time series?
- 1/2β(ytβ1β+ytβ+yt+1β)
- 1/3β(ytβ1β+ytβ+yt+1β)
- None of the above
Quiz: The AR(1) definitions and properties
Q1. Which of the following plots corresponds to the PACF for an AR(1) with \phi = 0.8Ο=0.8?
- None of above is correct
- .
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Q2.
- Which of the following AR(1) processes are stable and therefore stationary?
- Ytβ=0.9Ytβ1β+Ο΅tβ,Ο΅tββΌi.i.d.N(0,v)
- Ytβ=Ytβ1β+Ο΅tβ,Ο΅tββΌi.i.d.N(0,v)
- Ytβ=β2Ytβ1β+Ο΅tβ,Ο΅tββΌi.i.d.N(0,v)
- Ytβ=β0.8Ytβ1β+Ο΅tβ,Ο΅tββΌi.i.d.N(0,v)
Q3. Which of the statements below are true?
- The ACF coefficients of an AR(1) with AR coefficient \phi \in (-1,1)Οβ(β1,1) and \phi \neq 0Οξ β=0 are zero after lag 1
- The PACF coefficients of an AR(1) with AR coefficient \phi \in (-1,1)Οβ(β1,1) and \phi \neq 0Οξ β=0 are zero after lag 1
- The ACF of an AR(1) with coefficient \phi=0.5Ο=0.5 decays exponentially in an oscillatory manner
- The ACF of an AR(1) with AR coefficient \phi=0.8Ο=0.8 decays exponentially
Q4. Which of the following corresponds to the autocovariance function at lag h=2,h=2, \gamma(2)Ξ³(2), of the autoregressive process Ytβ=0.7Ytβ1β+Ο΅tβ,Ο΅tββΌi.i.d.N(0,v), with v=2.v=2.
- 3.9216
- 0.490.49
- 1.9216
Q5. What is the PACF coefficient at lag 1 for the AR(1) process
ytβ=β0.7ytβ1β+Ο΅tβ with \epsilon_t \stackrel{iid}\sim N(0,1)Ο΅tββΌiidN(0,1)?
- 0.70.7
- -0.7β0.7
- \approx 1.96β1.96
- 00
Q6. What is the autovariance function at lag 1, \gamma(1)Ξ³(1) of the AR(1) process
ytβ=0.6ytβ1β+Ο΅tβ with \epsilon_t \stackrel{i.i.d.}{\sim} N(0,v)Ο΅tββΌi.i.d.N(0,v) ? with variance v=2v=2.
- 1.5625
- 1.875
- 1
- 0.6
Q7. Consider an AR(1) process y_t = -0.5 y_{t-1} + \epsilon_t,ytβ=β0.5ytβ1β+Ο΅tβ, with \epsilon_t \stackrel{i.i.d.}{\sim} N(0,1)Ο΅tββΌi.i.d.N(0,1). Which of the following statements are true?
- The autocovariance process of this function decays exponentially as a function of the lag hh and it is always negative
- The autocovariance process of this function decays exponentially as a function of the lag hh and it is always positive
- The PACF coefficient at lag 1 \phi(1,1)Ο(1,1) is equal to -0.5β0.5
- The PACF coefficients for lags greater than 1 are zero
- The PACF coefficient at lag 1 \phi(1,1)Ο(1,1) is equal to 0.50.5
- The autocovariance process of this function decays exponentially as a function of the lag hh oscillating between negative and positive values
Week 02 : Properties of AR processes
Q1. Consider the following AR(2)AR(2) process,
Y_t = 0.5Y_{t-1} + 0.24Y_{t-2} + \epsilon_t, \quad \epsilon_t \sim \mathcal{N}(0, v).Ytβ=0.5Ytβ1β+0.24Ytβ2β+Ο΅tβ,Ο΅tββΌN(0,v).
Give the value of one of the reciprocal roots of this process.
Q2. Assume the reciprocal roots of an AR(2)AR(2) characteristic polynomial are 0.70.7 and -0.2.β0.2.
Which is the corresponding form of the autocorrelation function \rho(h)Ο(h) of this process?
- Ο(h)=(a+bh)0.3h,h>0, where $a$ and $b$ are some constants.
- Ο(h)=a(0.7)h+b(β0.3)h,h>0, where aa and bb are some constants.
- Ο(h)=(a+bh)0.7h,h>0, where aa and bb are some constants.
- Ο(h)=(a+bh)(0.3h+0.7h),h>0, where $a$ and $b$ are some constants.
Q3. Assume that an AR(2) process has a pair of complex reciprocal roots with modulus r = 0.95r=0.95 and period \lambda = 7.1.Ξ»=7.1.
Which following options corresponds to the correct form of its autocorrelation function, \rho(h)Ο(h) ?
- Ο(h)=a(0.95)hcos(7.1h+b), where aa and bb are some constants.
- Ο(h)=a(0.95)hcos(2Οh/7.1+b), where aa and bb are some constants.
- Ο(h)=a0.95h,h>0, where aa and bb are some constants.
- Ο(h)=(a+bh)0.95h, where aa and bb are some constants.
Q4. Given the following AR(2)AR(2) process,
Y_t = 0.5Y_{t-1} + 0.36Y_{t-2} + \epsilon_t, \quad \epsilon_t \sim \mathcal{N}(0, v).Ytβ=0.5Ytβ1β+0.36Ytβ2β+Ο΅tβ,Ο΅tββΌN(0,v).
The h=3h=3 steps-ahead forecast function f_t(3)ftβ(3) has the following form:
- ftβ(3)=c1tβ(1.1)3+c2tβ(β2.5)3 for c_{1t}c1tβ and c_{2t}c2tβ constants.
- ftβ(3)=(0.9)3(c1tβ+c2tβ3) for c_{1t}c1tβ and c_{2t}c2tβ constants
- ftβ(3)=c1tβ(3)0.9+c2tβ(3)β0.4 for c_{1t}c1tβ and c_{2t}c2tβ constants.
- ftβ(3)=c1tβ(0.9)3+c2tβ(β0.4)3 for c_{1t}c1tβ and c_{2t}c2tβ constants.
Week 03: Practice Quiz The Normal Dynamic Linear Model
Q1. Which of the models below is a Dynamic Normal Linear Model?
- Observation equation: y_t = a\theta^2_t + \epsilon_t, \quad \epsilon_t \sim \mathcal{N}(0, v), ytβ=aΞΈt2β+Ο΅tβ,Ο΅tββΌN(0,v),
- System equation: \theta_t = b\theta_{t-1} + c \frac{\theta_{t-1}}{1+ \theta^2_{t-1}} + \omega_t, \quad \omega_t \sim \mathcal{N}(0, w). ΞΈtβ=bΞΈtβ1β+c1+ΞΈtβ12βΞΈtβ1ββ+Οtβ,ΟtββΌN(0,w).
- Observation equation: y_t = \mu_t + \epsilon_t, \quad \epsilon_t \sim \mathcal{N}(0, v),ytβ=ΞΌtβ+Ο΅tβ,Ο΅tββΌN(0,v),
- System equation: \mu_t = \mu_{t-1} + \omega_t, \quad \omega_t \sim \mathcal{N}(0, w).ΞΌtβ=ΞΌtβ1β+Οtβ,ΟtββΌN(0,w).
- Observation equation: y_t = \theta_t + \epsilon_t, \quad \epsilon_t \sim \mathcal{N}(0, v), ytβ=ΞΈtβ+Ο΅tβ,Ο΅tββΌN(0,v),
- System equation: \theta_t = b\theta_{t-1} + c \frac{\theta_{t-1}}{1+ \theta^2_{t-1}} + \omega_t, \quad \omega_t \sim \mathcal{N}(0, w). ΞΈtβ=bΞΈtβ1β+c1+ΞΈtβ12βΞΈtβ1ββ+Οtβ,ΟtββΌN(0,w).
Q2. Consider the Normal Dynamic Linear Model \mathcal{M}: \left\{\bm{F}_t, \bm{G}_t, \cdot, \cdot\right\}, M:{Ftβ,Gtβ,β ,β }, for t = 1, \dots, T.t=1,β¦,T. Letβs assume \bm{F}_tFtβ is K \times 1KΓ1 vector. What is the dimension of \bm{G}_t?Gtβ?
- T \times 1TΓ1
- T \times TTΓT
- K \times KKΓK
- K \times 1KΓ1
Q3. Consider the third order polynomial Normal Dynamic Linear Model \mathcal{M}: \{\bm{F}, \bm{G}, \cdot, \cdot\}, M:{F,G,β ,β }, where \bm{F} = (1 \quad 0 \quad 0)βF=(100)β² and \bm{G} = \bm{J}_3(1),G=J3β(1), where \bm{J}J is Jordan block given by
J_3(1) = \left(
100110011
\right) J3β(1)=ββββ100β110β011ββ βββ
Given the posterior mean E (\bm{\theta}_t | D_t) = (m_t, b_t, g_t)β,E(ΞΈtββ£Dtβ)=(mtβ,btβ,gtβ)β², which of the following options is the one corresponding to the forecast function f_t(h) \quad (h \geq 0)ftβ(h)(hβ₯0) of the model?
- ftβ(h)=mtβ+hbtβ+h(hβ1)gtβ/2
- βftβ(h)=mtβ+hbtβ
- ftβ(h)=mtβ+hbtβ+h(h+1)gtβ
- ftβ(h)=mtβ+hbtβ+h2gtβ
Q4. Consider a forecast function of the following form:
ftβ(h)=at,0β+at,1βxt+hβ+at,3β+at,4βh, for h \geq 0.hβ₯0.
Which is the possible corresponding DLM \left\{ \bm{F}, \bm{G}, \cdot, \cdot \right\}?{F,G,β ,β }?
- F=(1xtβ10)β²
\bm{G} = \bm{I}_4,G=I4β,
where \bm{I}_4I4β is the identity matrix of dimension 4 \times 44Γ4
- F=(1010)β²
\bm{G} = \textrm{block diag} [\bm{I}_2, \bm{J}_2(1)],G=block diag[I2β,J2β(1)],
where \bm{I}_2I2β is the identity matrix of dimension 2 \times 22Γ2, and \bm{J}_2(1)J2β(1) is the Jordan block matrix given by
J_2(1) = \left(
1011 \right).J2β(1)=(10β11β).
- F=(1xtβ10)β²
\bm{G} = \textrm{block diag} [\bm{I}_2, \bm{J}_2(1)],G=block diag[I2β,J2β(1)],
where \bm{I}_2I2β is the identity matrix, and \bm{J}_2(1)J2β(1) is the Jordan block matrix given by
J_2(1) = \left(
1011 \right).J2β(1)=(10β11β).
- None is correct.
Quiz : NDLM, Part I: Review
Q1. Assume we have following DLM representation:
\mathcal{M}: \qquad \left\{
(10)
,
(β0.7501β0.75)
, \cdot, \cdot \right\} M:{(10β),(β0.750β1β0.75β),β ,β } and
E(\bm{\theta}_t | \mathcal{D}_t) = (1, 0.5)βE(ΞΈtββ£Dtβ)=(1,0.5)β². What is the best description of the forecast function, f_t(h)ftβ(h) for h \geq 0hβ₯0 of this model?
- .
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Q2. Which of the options below correspond the DLMs \{\bm{F}, \bm{G}, \cdot, \cdot \}{F,G,β ,β } with forecast function
f_t(h) = k_{t1}\lambda^h_1 β k_{t2}\lambda_2^h, ftβ(h)=kt1βΞ»1hββkt2βΞ»2hβ,
where \lambda_1\lambda_2 \neq 0Ξ»1βΞ»2βξ β=0 and \lambda_1 \neq \lambda_2Ξ»1βξ β=Ξ»2β?
- F=(1,0)β²,
\bm{G} =
(Ξ»111Ξ»2).G=(Ξ»1β1β1Ξ»2ββ).
- F=(1,β1)β²,
\bm{G} =
(Ξ»100Ξ»2).G=(Ξ»1β0β0Ξ»2ββ).
- F=(1,β1)β²,
\bm{G} =
(Ξ»111Ξ»2).G=(Ξ»1β1β1Ξ»2ββ).
- F=(1,0)β²,
\bm{G} =
(Ξ»100Ξ»2).G=(Ξ»1β0β0Ξ»2ββ).
Q3. Consider a DLM with a forecast function of the form f_t(h) = k_{t,1} \lambda^h + k_{t,2} x_{t+h}.ftβ(h)=kt,1βΞ»h+kt,2βxt+hβ. Which of the following representations corresponds to this DLM?
- {Ftβ,Gtβ,β ,β } with
\bm{F}_t = (1,x_t)βFtβ=(1,xtβ)β² and \bm{G}_t= \left(
Ξ»001 \right).Gtβ=(Ξ»0β01β).
- {Ftβ,Gtβ,β ,β } with
\bm{F}_t = (1,0)βFtβ=(1,0)β² and \bm{G}_t= \left(
Ξ»00xt \right).Gtβ=(Ξ»0β0xtββ).
- {Ftβ,Gtβ,β ,β } with
\bm{F}_t = (1,x_t)βFtβ=(1,xtβ)β² and \bm{G}_t= \left(
1001 \right).Gtβ=(10β01β).
Q4. Which of the following statements are true?
- In order to obtain the filtering equations one must first obtain the smoothing equations
- In a DLM of the form \{ \bm{F}_t, \bm{G}_t, v_t, \bm{W}_t\}{Ftβ,Gtβ,vtβ,Wtβ} with \bm{F}_t, \bm{G}_t, v_t, \bm{W}_tFtβ,Gtβ,vtβ,Wtβ known for all t,t, the distribution of (\mathbf{\theta}_t| \mathcal{D}_t)(ΞΈtββ£Dtβ) is normal if the distribution of (\mathbf{\theta}_0 | \mathcal{D}_0)(ΞΈ0ββ£D0β) is normal
- The filtering equations allow us to obtain the moments of the distribution of (\mathbf{\theta}_t| \mathcal{D}_T)(ΞΈtββ£DTβ) for t=1:Tt=1:T and T >t.T>t.
- The smoothing equations allow us to obtain the moments of the distribution of (\mathbf{\theta}_t| \mathcal{D}_T)(ΞΈtββ£DTβ) for t=1:Tt=1:T and T >t.T>t.
- In a DLM of the form \{ \bm{F}_t, \bm{G}_t, v_t, \bm{W}_t\}{Ftβ,Gtβ,vtβ,Wtβ} with \bm{F}_t, \bm{G}_t, v_t, \bm{W}_tFtβ,Gtβ,vtβ,Wtβ known for all t,t, the distribution of (\mathbf{\theta}_t| \mathcal{D}_t)(ΞΈtββ£Dtβ) is not normal even if the distribution of (\mathbf{\theta}_0 | \mathcal{D}_0)(ΞΈ0ββ£D0β) is normal
- In order to obtain the DLM smoothing equations one must first obtain the filtering equations
Week 04 : Quiz Seasonal Models and Superposition
Q1. Consider a full seasonal Fourier DLM with fundamental period p=3.p=3. Which of the choices below corresponds to the specification of \bm{F}F and \bm{G}G for such model?
- F=(10)β²
\bm{G} = \left(
β12β3β23β2β12 \right) G=(β21ββ23βββ23βββ21ββ)
- F=(10)β²
\bm{G} = \left(
1011 \right) G=(10β11β)
- F=(101)β²
\bm{G} = \left(
0β1010000β1
\right) G=ββββ0β10β100β00β1ββ βββ
- F=(10)β²
\bm{G} = \left(
12β3β23β212 \right) G=(21ββ23βββ23ββ21ββ)
Q2. Assume monthly data have an annual cycle and so the fundamental period is p=12.p=12. Further assume that we want to fit a model with a linear trend and seasonal component to this dataset. For the seasonal component, assume we only consider the fourth harmonic, i.e., we only consider the Fourier component for the frequency \omega= 2\pi 4/12= 2 \pi/3.Ο=2Ο4/12=2Ο/3. What is the forecast function f_t(h), h \geq 0,ftβ(h),hβ₯0, for a DLM with this linear trend and a seasonal component that considers only the fourth harmonic?
- ftβ(h)=at,0β+at,1βh
- ftβ(h)=at,0β+at,1βh+at,3βcos(32Οhβ)+at,4βsin(32Οhβ)
- ftβ(h)=at,1βcos(32Οhβ)+at,2βsin(32Οhβ)
- ftβ(h)=at,0β+at,1βh+at,3βcos(32Οhβ)+at,4βsin(32Οhβ)+at,5β(β1)h
Q4. A DLM \{ \bm{F}_t, \bm{G}_t, \cdot, \cdot \}{Ftβ,Gtβ,β ,β } has the following forecast function:
f_t(h) = a_{t,1} x_{t+h} + (-1)^{h} a_{t,2}.ftβ(h)=at,1βxt+hβ+(β1)hat,2β.
What are the corresponding \bm{F}_tFtβ and \bm{G}_tGtβ matrices?
- Ftβ=(1,xtβ)β² and \bm{G}_t =\left(
1001 \right)Gtβ=(10β01β)
- Ftβ=(1,xtβ)β² and \bm{G}_t =\left(
100β1 \right)Gtβ=(10β0β1β)
- Ftβ=(1,0)β² and \bm{G}_t =\left(
10xtβ1 \right)Gtβ=(10βxtββ1β)
Quiz : NDLM, Part II
Q1. Assume that we want a model with the following $2$ components:
- Linear trend: \{\bm{F}_1, \bm{G}_1, \cdot, \cdot\}{F1β,G1β,β ,β } with \bm{F}_1 = (1,0,0)β,F1β=(1,0,0)β², \bm{G}_1 = \bm{J}_3(1) G1β=J3β(1) and state vector \bm{\theta}_{1t} = (1, -0.5, 0.1)β.ΞΈ1tβ=(1,β0.5,0.1)β².
- Seasonal component: \{\bm{F}_2, \bm{G}_2, \cdot, \cdot\}{F2β,G2β,β ,β } with \bm{F}_2 = (1, 0)βF2β=(1,0)β², \bm{G}_2 = \bm{J}_2(\lambda, \omega)G2β=J2β(Ξ»,Ο), where \omega = \frac{\pi}{2}Ο=2Οβ and \lambda = 0.9Ξ»=0.9 and state vector \bm{\theta}_{2t} = (1, 0.45)β.ΞΈ2tβ=(1,0.45)β².
Which graph is the description of forecast function f_t(h)ftβ(h) with h \geq 0?hβ₯0?
- .
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- .
![Bayesian Statistics: Time Series Analysis Coursera Quiz Answers 2022 | All Weeks Assessment Answers [π―Correct Answer] 13 myM6qWcVRpajOqlnFfaW2Q f326a46501b645179e3b8e3816a72ef1 wrong 1](https://d3c33hcgiwev3.cloudfront.net/imageAssetProxy.v1/myM6qWcVRpajOqlnFfaW2Q_f326a46501b645179e3b8e3816a72ef1_wrong_1.png?expiry=1654992000000&hmac=tH-hNRfwPtILGEKSddjBC0fBXirPP7QC8yMVrXENsQs)
- .
![Bayesian Statistics: Time Series Analysis Coursera Quiz Answers 2022 | All Weeks Assessment Answers [π―Correct Answer] 14 7 gzBdK1RDq4MwXStcQ6ow 1703789779d64599a7dbd20de14fc0f1 wrong 2](https://d3c33hcgiwev3.cloudfront.net/imageAssetProxy.v1/7_gzBdK1RDq4MwXStcQ6ow_1703789779d64599a7dbd20de14fc0f1_wrong_2.png?expiry=1654992000000&hmac=S52AbmTFCxOI5yiJPMqHyCCsS_diyrXUHiC865db6BU)
- .
![Bayesian Statistics: Time Series Analysis Coursera Quiz Answers 2022 | All Weeks Assessment Answers [π―Correct Answer] 15 U5DME34SSGiQzBN EqhoBw 67730175c5324b91886eb4c458daddf1 correct](https://d3c33hcgiwev3.cloudfront.net/imageAssetProxy.v1/U5DME34SSGiQzBN-EqhoBw_67730175c5324b91886eb4c458daddf1_correct.png?expiry=1654992000000&hmac=KBaFK2-gzzdLCEfWB9JiqEKbG_aqQfYTctbb6R6ij6E)
Q2. Which is the right Fourier DLM model \{\bm{F}, \bm{G}, \cdot, \cdot \}{F,G,β ,β } with period p = 6?p=6?
1 point
- F=(1,0,1,0,0)β²
\bm{G} =
ββββββββββββββββ12β3β20003β21200000β12β3β20003β2β1200000β1ββ ββββββββββββββG=βββββββββ21ββ23ββ000β23ββ21β000β00β21ββ23ββ0β0023βββ21β0β0000β1ββ ββββββββ
- F=(1,0,1,0,1)β²
\bm{G} =
ββββββββββββββββ12β3β20003β21200000β12β3β20003β2β1200000β1ββ ββββββββββββββG=βββββββββ21ββ23ββ000β23ββ21β000β00β21ββ23ββ0β0023βββ21β0β0000β1ββ ββββββββ
- F=(1,0,1,0,0)β²
\bm{G} =
ββββββββββββββββ3β21200012β3β2000003β2β12000β12β3β200000β1ββ ββββββββββββββG=βββββββββ23ββ21β000β21ββ23ββ000β0023βββ21β0β00β21ββ23ββ0β0000β1ββ ββββββββ
- F=(1,0,1,0,1)β²
\bm{G} =
ββββββββββββββββ3β21200012β3β2000003β2β12000β12β3β200000β1ββ ββββββββββββββG=βββββββββ23ββ21β000β21ββ23ββ000β0023βββ21β0β00β21ββ23ββ0β0000β1ββ ββββββββ
Q3. Consider a full seasonal Fourier DLM with fundamental period p=10.p=10. What is the dimension of the state vector \bm{\theta}_tΞΈtβ at each time tt?
- None of the above
- 10
- 9
- 11
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