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STOCHASTIC HYDROLOGY Lecture -21 Course Instructor : Prof. P. P. MUJUMDAR Department of Civil Engg., IISc. INDIAN INSTITUTE OF SCIENCE
STOCHASTIC HYDROLOGY Lecture -21
Course Instructor : Prof. P. P. MUJUMDAR Department of Civil Engg., IISc.
INDIAN INSTITUTE OF SCIENCE
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Summary of the previous lecture
– Case study -3: Monthly streamflows at KRS reservoir
• Validation of the model – Case study -4: Monthly streamflow of a river
• Plots of Time series, Correlogram, Partial Autocorrelation function and Power spectrum
• Candidate ARMA models – Log Likelihood – Mean square error – Validation test (Residual mean)
CASE STUDIES - ARMA MODELS
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Case study – 5
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Sakleshpur Annual Rainfall Data (1901-2002)
Time (years)
Rai
nfal
l in
mm
Case study – 5 (Contd.)
5
Correlogram
( )( )1
02
0
1 n k
k t t ki
kk
X
c X X X XNcrc
c S
−
+=
= − −
=
=
∑
Case study – 5 (Contd.)
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PAC function Power spectrum x105
( ) 2 2
2 k kNI k α β⎡ ⎤= +⎣ ⎦
2k
kNπ
ω =
( )1
2 cos 2n
k t ktx f t
Nα π
=
= ∑ ( )1
2 sin 2n
k t ktx f t
Nβ π
=
= ∑
Pp * φp = ρp Auto Correlations
Partial Auto Correlation Auto Correlation function
Case study – 5 (Contd.)
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Model Likelihood AR(1) 9.078037 AR(2) 8.562427 AR(3) 8.646781 AR(4) 9.691461 AR(5) 9.821681 AR(6) 9.436822
ARMA(1,1) 8.341717 ARMA(1,2) 8.217627 ARMA(2,1) 7.715415 ARMA(2,2) 5.278434 ARMA(3,1) 6.316174 ARMA(3,2) 6.390390
Case study – 5 (Contd.)
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• ARMA(5,0) is selected with highest likelihood value
• The parameters for the selected model are as follows φ1 = 0.40499 φ2 = 0.15223 φ3 = -0.02427 φ4 = -0.2222 φ5 = 0.083435 Constant = -0.000664
Case study – 5 (Contd.)
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• Significance of residual mean
Model η(e) t0.95(N ) ARMA(5,0) 0.000005 1.6601
Significance of periodicities:
Case study – 5 (Contd.)
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Periodicity η F0.95(2, N-2 )
1st 0.000 3.085
2nd 0.00432 3.085
3rd 0.0168 3.085
4th 0.0698 3.085
5th 0.000006 3.085
6th 0.117 3.085
Case study – 5 (Contd.)
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• Whittle’s white noise test:
Model η F0.95(n1, N–n1) ARMA(5,0) 0.163 1.783
Case study – 5 (Contd.)
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Model MSE AR(1) 1.180837 AR(2) 1.169667 AR(3) 1.182210 AR(4) 1.168724 AR(5) 1.254929 AR(6) 1.289385
ARMA(1,1) 1.171668 ARMA(1,2) 1.156298 ARMA(2,1) 1.183397 ARMA(2,2) 1.256068 ARMA(3,1) 1.195626 ARMA(3,2) 27.466087
Case study – 5 (Contd.)
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• ARMA(1, 2) is selected with least MSE value for one step forecasting
• The parameters for the selected model are as follows φ1 = 0.35271 θ1 = 0.017124 θ2 = -0.216745 Constant = -0.009267
Case study – 5 (Contd.)
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• Significance of residual mean
Model η(e) t0.95(N ) ARMA(1, 2) -0.0026 1.6601
Significance of periodicities:
Case study – 5 (Contd.)
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Periodicity η F0.95(2, N-2 )
1st 0.000 3.085
2nd 0.0006 3.085
3rd 0.0493 3.085
4th 0.0687 3.085
5th 0.0003 3.085
6th 0.0719 3.085
Case study – 5 (Contd.)
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• Whittle’s white noise test:
Model η F0.95(n1, N–n1) ARMA(1, 2) 0.3605 1.783
SUMMARY OF CASE STUDIES
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Summary of Case studies Case study-1:Time series plot
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Daily rainfall data of Bangalore city Monthly rainfall data of Bangalore city
0 5 10 15 20 25 30 35500
600
700
800
900
1000
1100
1200
1300
1400
Yearly rainfall data of Bangalore city
Summary of Case studies Case study-1: Correlogram
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0 100 200 300 400 500 600-0.1
-0.05
0
0.05
0.1
0.15
0.2
Lag
Sam
ple A
utoc
orre
lation
Sample Autocorrelation Function
Daily rainfall data of Bangalore city Monthly rainfall data of Bangalore city
Yearly rainfall data of Bangalore city
Summary of Case studies Case study-1: Power spectrum
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Daily rainfall data of Bangalore city Monthly rainfall data of Bangalore city
Yearly rainfall data of Bangalore city
Time (months)
Flow
in c
umec
Summary of Case studies Time series plot
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4. Monthly stream flow data of a river
Time
Flow
in c
umec
3. Monthly stream flow data for Cauvery
5. Sakleshpur Annual Rainfall Data
Summary of Case studies Correlogram
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4. Monthly stream flow data of a river 3. Monthly stream flow data for Cauvery
5. Sakleshpur Annual Rainfall Data
Summary of Case studies Power spectrum
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4. Monthly stream flow data of a river 3. Monthly stream flow data for Cauvery
-2E+08
0
200000000
400000000
600000000
800000000
1E+09
1.2E+09
0 1 2 3 4 5 6 7
5. Sakleshpur Annual Rainfall Data
Summary of Case studies ARMA Models
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3. Monthly stream flow data for Cauvery ARMA(4, 0) – For data generation ARMA(1, 0) – For one step forecasting 4. Monthly stream flow data of a river ARMA (8, 0) – For both data generation & one step forecasting 5. Sakleshpur Annual Rainfall Data ARMA(5, 0) – For data generation ARMA(1, 2) – For one step forecasting
MARKOV CHAINS
• A Markov chain is a stochastic process with the
property that value of process Xt at time t depends on its value at time t-1 and not on the sequence of other values (Xt-2 , Xt-3,……. X0) that the process passed through in arriving at Xt-1.
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Markov Chains
[ ] [ ]1 2 0 1, ,.....t t t t tP X X X X P X X− − −=
Single step Markov chain
• This conditional probability gives the probability at time t will be in state ‘j’, given that the process was in state ‘i’ at time t-1.
• The conditional probability is independent of the states occupied prior to t-1.
• For example, if Xt-1 is a dry day, we would be interested in the probability that Xt is a dry day or a wet day.
• This probability is commonly called as transition probability
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Markov Chains
1t j t iP X a X a−⎡ ⎤= =⎣ ⎦
• Usually written as indicating the probability of a step from ai to aj at time ‘t’.
• If Pij is independent of time, then the Markov chain is said to be homogeneous. i.e., v t and τ the transition probabilities remain same across time
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Markov Chains
tijP
1t
t j t i ijP X a X a P−⎡ ⎤= = =⎣ ⎦
t tij ijP P τ+=
Transition Probability Matrix(TPM):
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Markov Chains
1 2 3 . . m
11 12 13 1
21 22 23 2
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1 2
. .
. .
.
.
m
m
m m mm
P P P PP P P PP
P
P P P
⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥
= ⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦
1
2
3
.
m
.
m x m
t+1 t
• Elements in any row of TPM sum to unity • TPM can be estimated from observed data by
enumerating the number of times the observed data went from state ‘i’ to ‘j’
• Pj (n) is the probability of being in state ‘j’ in time
step ‘n’.
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Markov Chains
V i 1
1m
ijjP
=
=∑
• pj(0) is the probability of being in state ‘j’ in period
t = 0.
• If p(0) is given and TPM is given
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Markov Chains
( ) ( ) ( ) ( )0 0 0 01 2 1
. . m mp p p p
×⎡ ⎤= ⎣ ⎦
…. Probability vector at time 0
( ) ( ) ( ) ( )1 2 1
. .n n n nm m
p p p p×
⎡ ⎤= ⎣ ⎦…. Probability
vector at time ‘n’
( ) ( )1 0p p P= ×
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Markov Chains
( ) ( ) ( ) ( )
11 12 13 1
21 22 23 21 0 0 0
1 2 31
1 2
. .
. .. .
.
m
m
m
m m mm
P P P PP P P P
p p p p P
P P P
⎡ ⎤⎢ ⎥⎢ ⎥
⎡ ⎤ ⎢ ⎥= ⎣ ⎦ ⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦
( ) ( ) ( )0 0 01 11 2 21 1.... m mp P p P p P= + + + …. Probability of
going to state 1
( ) ( ) ( )0 0 01 12 2 21 2.... m mp P p P p P= + + + …. Probability of
going to state 2 And so on…
Therefore
In general,
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Markov Chains
( ) ( ) ( ) ( )1 1 1 11 2 1
. . m mp p p p
×⎡ ⎤= ⎣ ⎦
( ) ( )
( )
( )
2 1
0
0 2
p p P
p P P
p P
= ×
= × ×
= ×
( ) ( )0n np p P= ×
• As the process advances in time, pj(n) becomes less
dependent on p(0) • The probability of being in state ‘j’ after a large
number of time steps becomes independent of the initial state of the process.
• The process reaches a steady state at large n
• As the process reaches steady state, the probability vector remains constant
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Markov Chains
np p P= ×
Example – 1
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Consider the TPM for a 2-state first order homogeneous Markov chain as State 1 is a non-rainy day and state 2 is a rainy day Obtain the 1. probability of day 1 is non-rainfall day / day 0 is rainfall day 2. probability of day 2 is rainfall day / day 0 is non-rainfall day 3. probability of day 100 is rainfall day / day 0 is non-rainfall
day
0.7 0.30.4 0.6
TPM ⎡ ⎤= ⎢ ⎥⎣ ⎦
Example – 1 (contd.)
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1. probability of day 1 is non-rainfall day / day 0 is rainfall day
The probability is 0.4
2. probability of day 2 is rainfall day / day 0 is non-rainfall day
0.7 0.30.4 0.6
TPM ⎡ ⎤= ⎢ ⎥
⎣ ⎦
No rain
rain
No rain rain
( ) ( )2 0 2p p P= ×
Example – 1 (contd.)
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The probability is 0.39
3. probability of day 100 is rainfall day / day 0 is non-rainfall day
( ) [ ]
[ ]
2 0.7 0.30.7 0.3
0.4 0.6
0.61 0.39
p ⎡ ⎤= ⎢ ⎥
⎣ ⎦
=
( ) ( )0n np p P= ×
Example – 1 (contd.)
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2
4 2 2
8 4 4
16 8 8
0.7 0.3 0.7 0.3 0.61 0.390.4 0.6 0.4 0.6 0.52 0.48
0.5749 0.42510.5668 0.4332
0.5715 0.42850.5714 0.4286
0.5714 0.42860.5714 0.4286
P P P
P P P
P P P
P P P
= ×
⎡ ⎤ ⎡ ⎤ ⎡ ⎤= =⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦ ⎣ ⎦
⎡ ⎤= × = ⎢ ⎥
⎣ ⎦
⎡ ⎤= × = ⎢ ⎥
⎣ ⎦
⎡ ⎤= × = ⎢ ⎥
⎣ ⎦
Example – 1 (contd.)
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Steady state probability
[ ]
[ ]
0.5714 0.42860.5714 0.4286
0.5714 0.4286
0.5714 0.4286
np p P= ×
⎡ ⎤= ⎢ ⎥
⎣ ⎦
=
For steady state,
[ ]0.5714 0.4286p =
