http://folk.uio.no/trondr/nve_course_3/hoelen1.R
Looks quite okay

Also looks ok.

c) Do a and b over again, but this time using the lognormal distribution. (Use the mean and standard deviations of the logarithmic values to adapt the lognormal distribution to the data).
Also looks ok, but not quite so good for low values (there are values that are extremely low according to the lognormal distribution). d) Repeat a-c for daily discharges also (found at http://folk.uio.no/trondr/nve_course_3/TrendDognHoelen.txt
). What conclusions change and why?
Extremely bad!
Better but not good
Yearly means are means of a large number of discharges, so the central limit theorem comes partially to the rescue. Not so for daily discharges.

Exercise 2: Is the expectancy of the yearly mean discharges at Hølen equal to
10m 3 /s?
http://folk.uio.no/trondr/nve_course_3/hoelen2.R a) Estimate the expectancy:
11.67041
b) Check the assumption expectancy=10m 3 /s using a t-test (takes the variance uncertainty into account). Use preferrably a significance level of 5% (confidence level of 95%). t = 5.4363, df = 84, p-value = 5.233e-07 alternative hypothesis: true mean is not equal to 10
95 percent confidence interval:
11.05937 12.28145 sample estimates: mean of x
11.67041
95% conf. int. from 11.06 to 12.28 does not contain 10, i.e. we can reject the assumption expectancy=10 with 95% confidence. P value=0.5ppm, so we could put more confidence on this also if we wanted.

c) Show the data together with the confidence interval.
Should we worry that most of the data is outside this interval?
No. The confidence interval indicates our ucnertainty concerning the overall expectancy (long-term mean), not out uncertainty concerning single outcomes.
Is it 95% probability that the real expectancy lies within this specific interval?
No (trick question). A 95% confidence interval is a method that before the data has a 95% probability of encompassing the true value, whatever that value is.
When you plug in the data, there is nothing more that can be assigned a probability. Frequentist methods do not have a concept about the probability concerning parameters. It’s the estimators as function of the data (when it has not yet arrived and thus has a probability) that has a probability. Once you plug data into estimators, you rather say you have a confidence regarding it

Exercise 2 contd.
d) Would we be justified in using the techniques in a-c for daily data also?
Exercise a could be performed and would give an unbiased estimate of the expectancy.
The analysis in exercise b-c depends on independence assumptions that are not at all valid for daily data!
e) Do the same analysis using the lognormal distribution also. Perform a bootstrap analysis in order to make a 95% confidence interval. What does this say about the assumption expectancy=10m 3 /s?
The estimated 95% confidence interval does not encompass 10. (It must be told that the way to get a 95% confidence interval from bootstrap samples is a little bit naive here, to keep things simple). The confidence interval is quite similar to the one we got from the t test (i.e. under normal assumptions). Conklusjon: expectancy=10m 3 /s doesn’t seem to be the case.

Extra exercise 2: Outcome of dice throws
On a cubic die, there is a 1/6th probability for each outcome from 1 to 6. When you have two dice, let’s assume the outcomes on the two dice are independent. a) What is the probability for a specific outcome on the two dice, for instance the probability of getting 5 on the first die and 2 on the second die?
Independent outcomes:
Pr(5 on the first die and 2 one the second die) =
Pr(5 on the first die)*Pr(2 on the second die) = 1/6*1/6 = 1/36 b) What is the probability of getting sum=2 one the two dice? Repeat for sum=3, sum=4, sum=5, sum=6 , sum=7, sum=8.
Sum=2. One way of getting this; 1-1. I.e. Pr(sum=2)=1/36
Sum=3. Two ways of getting this; 1-2, 2-1.
Thus P(sum=3)=Pr(1-2)+Pr(2-1)=2/36=1/18
Sum=4. Three ways of getting this; 1-3, 2-2, 3-1.
Thus Pr(sum=4)=1/36+1/36+1/36=3/36=1/12
Sum=5. Five ways of getting this; 1-4, 2-3, 3-2, 4-1.
Thus Pr(sum=5)=4/36=1/9
Sum=6. Five ways; 1-5, 2-4, 3-3, 4-2, 5-1. Pr(sum=6)=5/36
Sum=7. Six ways; 1-6, 2-5, 3-4, 4-3, 5-2, 6-1. Pr(sum=7)=6/36=1/6
Sum=8. Five ways; 2-6, 3-5, 4-4, 5-3, 6-2. Pr(sum=8)=5/36

Pr(sum<=4) = Pr(sum=2 or sum=3 or sum=4) =
Pr(sum=2)+Pr(sum=3)+Pr(sum=4) = 1/36+2/36+3/36=6/36=1/6
Pr(sum<=4)=Pr(1-1)+Pr(1-2)+Pr(1-4)+Pr(2-1)+Pr(2-2)+Pr(3-1)=6/36=1/6

Extra exercise 2 (condt): Outcome of dice throws e) What is the probability of getting equal outcomes and sum<=4?
Pr(two equal dice and sum<=4) = Pr(1-1)+Pr(2-2)=2/36=1/18 f) What is the probability for either getting sum<=4 or two equal dice? You can use the answers from c, d and e or you can count the possibilities.
1). Pr(two equal dice or sum<=4) =
Pr(two equal)+Pr(sum<=4)-Pr(two equal and sum<=4) =
1/6+1/6-1/18 = 5/18
2). Pr(two equal or sum<=4)=
Pr(1-1)+Pr(1-2)+Pr(2-1)+Pr(2-2)+Pr(1-3)+Pr(3-1)+Pr(3-3)+Pr(4-4)+Pr(5-5)+
Pr(6-6) = 10/36=5/18

Extra exercise 2 (condt): Outcome of dice throws g) Both from the rule for conditional probability and from the list of ourcomes where sum<=4, what is the probability of equal dice given that sum<=4
1). Pr(two equal | sum<=4)=
Pr(two equal and sum<=4)/Pr(sum<=4)=
(1/18) / (1/6) = 1/3
2). List of outcomes where sum<=4: 1-1, 1-2, 1-3, 2-1, 2-2, 3-1
Six different (equally probable) outcomes where sum<=4. (see exercise c). Two outcomes with two equal and sum<=4: 1-1 and 2-2.
Therefore Pr(two equal | sum<=4)=2/6=1/3 h) Calculate the probability from sum<=4 given two equal dice, both from a list of possible outcomes and from Bayes formula.
1). List of outcomes with two equal: 1-1, 2-2, 3-3, 4-4, 5-5, 6-6. Six outcomes with equal probability.
Two of these has sum<=4. Ergo
Pr(sum<=4 | two equal)=1/3
2). Bayes formel:
Pr( sum

4 | two equal)

Pr(two equal | sum

4)Pr(sum
Pr(two equal)

4)

( 1 / 3 ) * ( 1 / 6 )
1 / 6

1 / 3

PS: Assume stationarity, that is that all probabilities do not depend of which day it is, when the circumstances are the same.
a) What is the probability that it rains any given day? (I.e. what is the marginal probability for rain?) Tip....
Pr(rain today)=
Pr(rain today and rain yesterday)+Pr(rain today but not yesterday)=
Pr(rain today | rain yesterday)*Pr(rain yesterday) +
Pr(rain today | no rain yesterday)*Pr(no rain yesterday) =
33.9%*Pr(rain yesterday)+12.9%*(1-Pr(rain yesterday)=
(stationarity)
33.9%*Pr(rain today)+12.9%*(1-Pr(rain today)

Pr(rain today)*(100%-33.9%+12.9%)=12.9% )

Pr(rain today)=12.9%/79%=16.3%

PS: Assume stationarity, that is that all probabilities do not depend of which day it is, when the circumstances are the same.
(Tip: Bayes formula)
Pr(rain yesterday | rain today )

Pr(rain to day | rain yesterday) * Pr(rain yesterday)

Pr(rain to day)
Pr(rain to day | rain yesterday) * Pr(rain to day)

Pr(rain to day)
Pr(rain to day | rain yesterday)

33.9%
(stationarity)

The Hobbit Shire council has decided to expand the hobbit settlements westwards.
Unfortunately, these lands are infested with dragons!
Of the 10kmx10km areas studied so far, 70% av were dragon-infisted.
A standard protocol for examining and area was made. A standardized test area of lesser size is examined, where a field hobbit biologist goes through the area in detail. The
Shire Biology Institute has determined that there is a 50% chance of detecting a dragon in an area with this procedure, if there is a dragon in that area.
If there is no dragon in the area, there is of course no probability of detecting one either!
Hobbit
Dragon
No dragons
?
?

Extra exercise 4 contd.
What is the (marginal) probability of detecting a dragon, when you don’t know whether the area is infected or not?
(Hint: The law of total probability) eller
The law of total probability:
Pr(detecting a dragon)=Pr(detecting a dragon | dragon in the area)*Pr(dragon in the area) +
Pr(detecting a dragon | no dragon in the area)*Pr(no dragon in the area) =
50%*70%+0%*30% = 35%

Extra exercise 4 contd.
Show using Bayes formula that the probability of there being a dragon in the area if you actually detected one, is 100%.
Dragon in the area
Dragon detected
No dragon
Dragon detected and in the area



Find the probability of there being a dragon in the area given that you did not detect one. Could you have expected the probability to decrease even without knowing the detection rate?
Dragon in the area
Dragon detected
No dragon
Dragon in the area
No dragon
P(dragon in the area | dragon not detected)

P(dragon not detected | dragon in the area) * P(dragon in the area)

P(dragon not detected)
50 % * 70 % /( 1

35 %)

35 % / 65 %

7 / 13

53 .
85 %
Last question: Yes. See graphical representation or use evidence rules of course 2.

Will look at the danger for the discharge to go above a fixed threshold. The station Gryta has discharge>1.5m
3 /s 27(=y) times in the course of t=44 years. Assume such events to be independent in time, i.e. a Poisson process. Instead of rate, we will use the expected repeat period, T=1/

(where
 is the rate). When one has data over a time period, t, the probability for y events will then be (Poisson distribution):
L ( T )

P ( y events in time t | T )

( t / T ) y y !
e
 t / T
Show that the ML estimate is T
ˆ  t y l ( T )
 log( L ( T ))

 l ( t )

T
  y
T y log(

T t
2
 l

(
T
ˆ t ) t )


0 y log( T )
 log(
 y T
 t

T
ˆ y

!
) t y
 t / T

2 l ( t )

T
2
 y
T
2

2 t
T
3
I(T)

-E

2 l ( t )

T
2
 
Ey
T
2

2
T
3 t
Ey
 t / T

I(T)
 t
T 3
At the same time, calculate the information matrix, I(T).

T
ˆ  t
The likelihood is maximal at . a) What is the ML-estimate for T in this particular case?
44years/27=1.63years.
b) 95% confidence interval. Sd(T)=1/

I(T)=T 3/2 /
 t=t/y 3/2 =0.31years.
1.63years+/-1.96*0.31years = (1.01years,2.24years).

Want to test whether thresholded precipitation (rain or no rain) is time dependent or not. The alternative model is a Markov chain. The rain state (0 or 1) is called x i
, where i denotes the day index. Since we need to condition on the previous state, we set aside the first day, x except this, is denoted n.
0
. The number of days
Zero hypothesis: The rain status (0 or 1) is independent from day to day
(Bernoilli process). We can thus specify the model with one parameter,
p=probability for rain a given day. The probability for a series of data is then:
P ( x then
1
,  , x
ˆ  k / n n
)
 p k
( 1
 p ) n
 k where k=number of days with rain. This will thus be the likelihood-function of this model. The ML estimate of p is p
RR
Alternative hypothesis: The rain status (0 or 1) is a Markov chain. There is a transition probability from no rain to rain p rain given rain the previous day also, p
RR
NR and a transition probability for
. (The two other probabilities are given as the negation of the two that have been specified). The probability of a given dataset is then:
P ( x
1
,  , x n
)
 p k
R
RR
( 1
 p
RR
) n
R
 k
R p k
N
NR
( 1
 p
NR
) n
N

1-p
RR k
N where k
R and n
R is the number of rainy days and total days when it rained the previous day and k
N and n
N are similarly defined for cases where it didn’t rain the previous day. The ML estimate for the two parameters then become: p
RR
 k
R
/ n
R
,
NR
 k
N
/ n
N
R
N p
NR
1-p
NR

Exercise 4: Independence, Markov chains and precipitation at Blindern
(contd).
Check if the rain status at Blindern is time dependent or not. Code: http://folk.uio.no/trondr/nve_course_3/blindern.R
. Data: http://folk.uio.no/trondr/nve_course_3/blindern_terkslet.txt
.
a) For the Blindern data, what is the estimate for p (rain probability in the zero hypothesis) and p
NR and p
RR
(alternative hypothesis). From the last, what seems to be the nature of the rain status one day and the next?
p=16.3% (zero hypothesis) , p
NR
=12.9% og p
RR
=33.9% (alternative hypothesis).
Rain one seems to increase the probability of rain the next day quite a bit, but not so much as to make rain more likely than no rain, it seems.
b) p’=16.3% (a little difference in the further digits).
c) 2*(l a
-l
0
)=138.

2 (0.95)=3.84 => rejecting the zero hypothesis with 95% confidence. P value=7.6*10 -32 . In other words, no, the rains tate is not independent from day to day! d) 95% confidence interval for p
RR
95% confidence interval for p
RR
: (30.1%,37.7%)
: (11.7%,14.1%) e) For p, theoretic: (15.1%, 17.5%)
For p, bootstrapped: (15.1%, 17.5%)
1-p
RR
For p’, bootstrapped: (14.8%, 17.8%) . It makes sense that p’ is more uncertain, since it comes from a model with more parameters and since this model suggests dependent data (thus less independent information). p
RR
R
N p
NR
1-p
NR

Exercise 5: Medical example translated to the language of Bayesian statistics. (PS: probabilities instead of probability densities and sums instead of integrals). Reminder: 0.1% of the population has a certain sickness. There is a test that always tests positive when the subject has the sickness and only tests positive 1% of the time when the subject is healthy.
a) Identify what is data outcomes and what is the thing we want to do inference on (the parameter, if you like).
Data = test outcome (positive or negative test)
Parameter = Sickness status (sick or healthy) b) What is the prior distribution?
The prior is pre-knowledge concerning the parameter, i.e. the sickness status. Our prior knowledge is that P(sickness)=0.1%. Pr(healthy)=99.9%. c) What is the likelihood?
Pr(test outcome|sickness status), giving a total of 4 combinations:
Pr(positive test | sick)=100% Pr(positive test | healthy)=1%
Pr(negative test | sick)=0% Pr(negative test | healthy)=99% d) Describe and calculate (using the law of total probability) the prior prediction distribution
(marginal data distribution):
The prior predictive distribution will here be Pr(test outcome) unconditioned on the sickness status.
Pr(positive test)=Pr(positive test | sick)Pr(sick)+Pr(positive test | healthy)Pr(healthy)=
100%*0.1%+1%*99.9%=1.099%
P(neg)=P(neg|sick)P(sick)+P(neg|healthy)P(healthy)=0%*0.1%+99%*99.9%=98.901%.
Alt: P(neg)=100%-P(pos))=98.901%.

Exercise 5: Medical example translated to the language of Bayesian statistics. (PS: probabilities instead of probability densities and sums instead of integrals). Reminder: 0.1% of the population has a certain sickness. There is a test that always tests positive when the subject has the sickness and only tests positive 1% of the time when the subject is healthy.
d) What is the posterior distribution, given a positive or a negative test?
Positive test: Pr(sick|pos)=Pr(pos|sick)Pr(sick)/Pr(pos)=100%*0.1%/1.099%=9.099%
Pr(healthy|pos)=100%-Pr(sick|pos)=90.9%
Negative test: Pr(sick|neg)=Pr(neg|sick)Pr(sick)/P(neg)=0%
P(healthy|neg)=100%-P(sick|neg)=100%

Exercise 5: Medical example translated to the language of Bayesian statistics. (PS: probabilities instead of probability densities and sums instead of integrals). Reminder: 0.1% of the population has a certain sickness. There is a test that always tests positive when the subject has the sickness and only tests positive 1% of the time when the subject is healthy.
f) A little risk analysis: The cost (C) for society (patient time as well as medical staff and equipement) of a medical operation is 10000kr. The benefit (B) is 100000kr if the subject has the sickness and 0kr if not. Assume you have tested positive. Find the expected benefit (=-risk): E(B-C|positive test). If you test positive, will the oepration pay off in total?
C=10.000kr, B=100.000kr if sick, B=0kr if healthy.
E(B-C|pos)=(B-C|syk)Pr(sick|pos)+(B-C|healthy)Pr(healthy|pos)= (B|sick)*Pr(sick|pos)-C=
100.000kr*Pr(sick|pos)-C=100.000kr*9.099%-10.000kr= -901kr.
Thus an operation now does not pay off (in expectancy). g) Assume the first test to be positive. What is the posterior prediction distribution for a new test, now?
Pr(new pos|pos)=Pr(new pos|sick,pos)*Pr(sick|pos)+Pr(new pos|healthy,pos)*
Pr(healthy|pos)=100%*9.099%+1%*90.9%=10.001%
Pr(new neg|pos)=1-Pr(new pos|pos)=90% g) A new test is performed and again gives a positive outcome. What is now the posterior probability that you have the sickness? Will an operation now pay off in expectancy?
Pr(sick|new pos, pos)=Pr(new pos | sick,pos)P(sick|pos)/P(new pos|pos)=100%*9.099%/10.001%=90.9%
E(B-C|ny pos,pos)=(B|sick)*Pr(sick|new pos,pos)-K=100.000kr*90.9%-10.000kr=80.917kr
Now an operation will pay off!

Exercise 6: Expectation value for mean discharge at Hølen – Bayesian analysis http://folk.uio.no/trondr/nve_course_3/hoelen3.R
Assume that the data is normally distributed, f ( x i
|

,

)
Then the overall mean will also be normally distributed f
~
( x |
N

(
,


,

2
)
)
~ N (

,

2
/ n )
(this will approximately be the case even if the data itself is not normally distributed).
We have a vague normal prior for the data expectancy, where f (

| x )

~
0
N x


2
2




0
2

/
2 n
/ n
,


2
2

 
2
2
/
/ n n

N (

( x ),

( x ))

N

(
:
 f
(
(
D ),


)
(
~
D ))
N (

0
,

2
)
=

=10. Assume we know that the standard deviation of the individual values is

=2.83.
a) What are the expectancy and standard deviation of the posterior distribution of
 in this case? Can you from this give an estimate of the data expectancy,

? If so, is that estimate far from the estimate you got from exercise 2? mu.D
11.669 now (11.670 in exercise 2). Not a big difference.
[1] 11.66884
tau.D
[1] 0.3068121
b) Make a 95% credibility band for the discharge expectancy. (Tip: 95% of the probability of a normal distribution is less than 1.96 standard deviations from the expectancy). Was this much different from the 95% confidence interval in exercise 2? Can you from this conclude concerning the assumption

=10m 3 /s? c(mu.D-1.96*tau.D,mu.D+1.96*tau.D)
[1] 11.06749 12.27019
Now, 11.07-12.27, before 11.06-12.28. Not a big difference (That the interval was a little wider for the t test is reasonable, since we did not assume we knew

=2.83 there).
Strictly speaking you can’t conclude anything concerning the assumption

=10m 3 /s. It’s not the most probable value, but there is no one-to-one relationship between model testing and uncertainty intervals, as there is in frequentist methodology.

The prior predictive (marginal) distribution is: f ( x ) ~ N (

0
,

2  
2
/ c) We will now test the assumption

=10. Compare the prior predictive distribution under our model (free-ranging

), with the prior predictive n ) distribution when

=10 (see the expression for the likelihood). What does this suggest?
> d1=dnorm(mean(Q),mu.0,sqrt(tau^2+sigma^2/n))
> d1
[1] 0.03932351
>
> # Probability density when mu=10
> d0=dnorm(mean(Q),10,sigma/sqrt(n))
> d0
[1] 4.822864e-07 d1 = prior predictive distribution density under free-ranging

.
d0 = prior predictive distribution density when

=10.
d1>>d0 suggest that we have evidence for our free-ranging model rather than for the model

=10.

d) Calculate the posterior model probability for model 0 (

=10) and model 1
(free-ranging

).
Pr( M | D )

 f f
( D
( D
|
|
M ) Pr( M )
M ' ) Pr( M ' )
Assume the prior model probability for each model to be 50%. What is the conclusion?
> p0.D=d0*p0/(d0*p0+d1*p1)
> p1.D=d1*p1/(d0*p0+d1*p1)
> c(p0.D,p1.D)
[1] 1.226443e-05 9.999877e-01
99.9988% probability for model 1 (free-ranging

) vs modell 0 (

=10). Very probable that
 
10, in other words.

e) Make a plot of the prior predictive distribution given different data outcomes and compare that to model 0. What data outcomes (if any) would be evidence for model 0, i.e. for

=10?
> c(min(x[d0>d1]),max(x[d0>d1]))
[1] 9.19 10.81
If we got a data mean between 9.19 and 10.81, the prior predictive probability density is larger for model 0 than for model 1. Thus such an outcome would be evidence for model 0 (

=10) rather than for model 1 (free-ranging

).

Will look at the danger for the discharge to go above a fixed threshold. The station
Gryta has discharge>1.5m
3 /s y=27 times in the course of t=44 years. Assume such events to be independent in time, i.e. a Poisson process. Instead of rate, we will use the expected repeat period, T=1/

(where
 is the rate). When one has data over a time period, t, the probability for y events will then be (Poisson y distribution):
P ( y event in time t | T )

( t / T y !
) e
 t / T the parameter, T: f ( T )


(

)
T
  
1 e
 
/ T
The prior predictive (marginal) data distribution is then:
P ( y events in t )



  y y

1

 p y
( 1
 p )

where p
 t
 t

(this is the so-called negative binomial distribution).

f ( T )

 

(

)
T
  
1 e
 
/ T
P ( y event in time
P ( y events in time t | T ) t )




/ T ) y

( t e
 t / T y !
 y y

1

 p y
( 1
 p )

where p
 t
 t

The station Gryta has had discharge>1.5m
3 /s y=27 times in the course of t=44 years.
a) Plot the prior distribution for T and the prior predictive data distribution while using

=

=1 in the prior.
The prior has a peak around T=1, but is fairly wise. So is the prior predictive distribution. y=27 is not a particularly expectational outcome according to this model.

P ( y event in time
P ( y events in time t t )
|

T )





( t y y
/ T ) y e
 t / T y !
1

 p y ( 1
 f ( T )

 

(

)
T
  
1 e
 
/ T p )

where p
 t
 t

The station Gryta has had discharge>1.5m
3 /s y=27 times in the course of t=44 years.
b) Use Bayes formula to express the posterior distribution for T. f ( T | y )
 f ( y | T ) f ( T )
 f ( y )
( t / T )

 y !
 y
 e
 t y y
/ T
 

(


1

 p y
)
T
( 1

  
1 e
 
/ T p )

(
  t )
  y

(
  y )
T
   y

1 e

(
  t ) / T
~

(

*
   y ,

*
   t )


P ( y event in time
P ( y events in time t | t )
T

)




( t

) y
/ y
T y y !

1

 e
 t p y
/ T
( 1
 f ( T )

 

(

)
T
  
1 e
 
/ T p )

where p
 t
 t

The station Gryta has had discharge>1.5m
3 /s y=27 times in the course of t=44 years.
b) Use Bayes formula to express the posterior distribution for T. Plot this for Gryta when

=

=1. Compare to the prior distribution. Try also

=

=0.5 and even

=

=0 (noninformative) . Were there large differences in the posterior probability distribution?
Compare these results with the frequentist result: T
ML
T | y ~

(

*
   y ,

*
   t )
=t/y=1.63 years.
Posterior with prior
Posterior using different priors
Zoomet version
A peak (modus) is found around 1,55 years, relatively near the ML estimate. Expectation

/(

-1)=1.67years, median=1.63years, which is even closer.

P ( y event in time
P ( y events in time t t )
|

T )





( t y y
/ T ) y !
1

 y e
 t / T p y ( 1
 f ( T ) p )

where p

 t
 

(
 t
)
T
  
1 e
 
/ T
 
The station Gryta has had discharge>1.5m
3 /s y=27 times in the course of t=44 years. c) Find the 95% credibility interval by calculating the 2.5% and 97.5% quantile of the posterior distribution, when the prior is specified by

=

=1. (R-tip: The 2.5% quantilen of the inverse gamma distribution is one over the 97.5% quantile of the gamma distribution and vice versa.) Compare with the 95% confidence interval in the frequentist analysis.
     
=

=1 is used.
Result: (1.15, 2.42)
Asymptotic classic 95% confidence: (1.01,2.24). Not very different, though the Bayesian analysis suggest that T has a skewed distribution, so higher values are more reasonable.

P ( y event in time t | T )

( t /
 y
T
!
1
)

 y e
 t / T ( T

 
  
1 e
 
/ T
P ( y events in time t )



 y y p y ( 1
 p )

where f p
) t

(
 t
)
T


  d) Can you find the posterior prediction distribution for new floods at Gryta the next
100 years? If so, plot that. Compare to the Poisson distribution with the ML estimate inserted. Why is that distribution sharper than the Bayesian posterior prediction distribution?
Since the posterior distribution for T comes form the same family of distributions as the prior distribution, we can just replace
 with

*=

+y and
 with

*=

+t in the predictive distribution.
P ( y events in time t )




*
 y ny y ny

1

 p y
( 1
 p )

where p
 t ny t ny
 
*
The Poisson distribution with a fixed parameter T is sharper than the posterior prediction distribution
Simply because we do not know the exact value for T.
Thus T has a distribution rather than a fixed value. Using the ML estimate doesn’t take the parameter uncertainty into account.

P ( y event in time
P ( y events in time t | T ) t )





 y
( t y
/ T ) y e
 t / T
 y !
1

 p y
( 1
 f p )

where p
( T )
 t

 t
 

(

)
T
  
1 e
 
/ T
 e) Do a simple MCMC run to fetch samples from the posterior distribution for T.
Find how many samples are need before the process stabilizes (burn-in) and approximately how many samples are need before you get independent sample
(spacing).
Burn-in=20-40, spacing=10

P ( y event in time
P ( y events in time t t )
|

T )





( t y y
/ T ) y e
 t / T y !
1

 p y ( 1
 f ( T )

 

(

)
T
  
1 e
 
/ T p )

where p
 t
 t
 f) Fetch 1000 approximately independent samples after burn-in. Compare this result with the theoretical posterior distribution (histogram and QQ plot).
Looks good in my run.
g) Perform an new MCMC sampling, this time with the prior distribution
f(T)=lognormal(

=0,

=2). (This cannot be solved analytically). Compare with the samples you got in f.
Relatively independent
Pretty much the same, but maybe a few deviations in the upper tail?

Extra exercise 5: Extreme value analysis at Bulken (about 120 years of data).
Code: http://folk.uio.no/trondr/nve_course_3/bulken_extreme.R
Data: : http://folk.uio.no/trondr/nve_course_3/bulken_max.txt
Will use the Gumbel distribution as model for the yearly extreme values (floods): f ( x |

,

)

1 e

( x
 
) /
  e

( x
 
) /

 a) Make an extreme value plot, by ordering the yearly maximal discharges and plot them against expected repeat time interval where n=number of years and i is an index running from 1 to n t i
 i

0 .
44 b) Fit the Gumbel distribution to the data using the two first l-moments (see the
 ˆ
1 code), parameters relate to the l-moments as

=

Calculate

1 n j n 

1 x j

1
 and
 x

1 and
,
 ˆ
2

2
. (Extra: Compare with what you get from DAGUT). The

2 and as n ( n
1

1 ) j n 

1
(( j

1 )

( n
 j ) ) x
( j )
2
/log(2),

=

Sorted data
1
-0.57721

.
> c(mu.lmom, beta.lmom)
DAGUT: Same result, essentially… [1] 305.22055 67.96247

Extra exercise 5 – contd.
f ( x |

,

)

1
 e

( x
 
) /
  e

( x
 
) /
 c) Make an extreme value plot (extreme value against repetition time) given the l-moment estimates together with the data.
Looks ok.
d) Do numerical ML estimation of the parameters.
> c(mu.ml, beta.ml)
[1] 304.49483 74.13713
e) Make an extreme value plot given the ML estimates.
Fairly much the same as the l-moment estimate for small values. Slight
Disagreement for larger values.

Extra extercise 5 – contd.
f ( x |

,

)

1
 e

( x
 
) /
  e

( x
 
) /
 f) (PS: If this seems mystic, stop here.) Perform a Bayesian analysis with flat prior. Sample 1000 MCMC samples (burn-in=1000, spacing=10). Compare estimated parameters with previous results. Look at the uncertainty of the parameters .Also look at the MCMC samples themselves. Has the chain of samples stabilized (is the burn-in long enough) and are the dependencies in the samples (spacing).
The histogram suggest that mu is somewhere between 290 and 320 and that the posterior distribution is approximately normal.
Peaks found around mu=305, beta=75 (modus estimate).
Not to different from other estimates.
# Expectation estimate::
> c(mean(mu.mcmc),mean(beta.mcmc))
[1] 304.81509 74.86763
# Median estimate:
> c(median(mu.mcmc),median(beta.mcmc))
[1] 305.16144 74.93275

Extra exercise 5 – contd.
f ( x |

,

)

1
 e

( x
 
) /
  e

( x
 
) /
 g) Use also the posterior prediction distribution to make an extreme value plot
(with the parameter uncertainty taken into account). Compare with previous results.
Fairly much the same as the
ML estimate.

http://folk.uio.no/trondr/nve_course_3/gryta.R
a) Plot the data, both on the original scale and on log-scale.
Seems like the points are curving upwards.
Looks fairly linear

Exercise 8 - contd b) Run a linear regression on log(discharge) against log(stage). Interpret the result. Is there a significant relationship between stage and discharge?
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) 0.93691 0.03870 24.21 2.74e-16 *** lh 2.53572 0.03841 66.03 < 2e-16 ***
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
Residual standard error: 0.0492 on 20 degrees of freedom
Multiple R-squared: 0.9954, Adjusted R-squared: 0.9952
F-statistic: 4359 on 1 and 20 DF, p-value: < 2.2e-16
Stage is significant, p-value<2*10^-16.
R 2 =99.54%. Good fit. c) What is the formula for discharge vs stage on the original scale? Plot this.
Q(h)=exp(0.937)*h 2.536
Plot this:
Looks good

Exercise 8 - contd.
d) Check if anything is wrong with the residuals (trend or non-normality).
No visible trend. Suggestions of a heavy upper tail, but that’s it. e) Extra: Perform a linear regression on the original scale and make a plot of this also. (PS: R code not included).
Not a big success… 

Exercise 9:
The dataset http://folk.uio.no/trondr/nve_course_3/Env-Eco.csv
contains species counts in various lakes together with some environmental variables, namely height, substrate diameter, water velocity, dissolved oxygen and water temperature. We want to see if the species richness is connected to any of the environmental variables according to this dataset. A species count can range from 0 but does not have an upper limit. Thus we will use Poisson regression
(glm(family=poisson()) in R). The model looks like this: y i
~ poisson (

( x i
)) where log(

( x i
))
 
0
 
1 x
1 , i
    k x k , i where y i is the response (number of species) for site number i and x j,i is covariate number j, site number i. Will log-transform the covariates diameter, velocity and dissolved oxygen, since these are strictly positive of nature.
Code: http://folk.uio.no/trondr/nve_course_3/env_eco.R
a) Fetch the data and take a glance at it. .... ok b) Run a Poisson fit with only a constant term (no covariates). Compare the result with the mean species number.
Estimate glm(formula = species.number ~ 1, family = poisson(), data = env.eco)
Deviance Residuals:
Min 1Q Median 3Q Max
-3.2221 -0.8728 -0.4149 0.6673 3.0022
Coefficients:
Estimate Std. Error z value Pr(>|z|)
(Intercept) 3.03770 0.05852 51.91 <2e-16 ***
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
(Dispersion parameter for poisson family taken to be 1)
Null deviance: 35.513 on 13 degrees of freedom
Residual deviance: 35.513 on 13 degrees of freedom
AIC: 105
Number of Fisher Scoring iterations: 4
# Compare species mean with result
# from regression:
> exp(3.0377)
[1] 20.85722
> mean(env.eco$species.number)
[1] 20.85714
Pretty much the same…

Exercise 9 – contd: c) Try each covariate in term. Are any statistically significant (according to the score test reported for each individual covariate)? If so, which is most significant (has the lowest p-value)? We have now started a stepwise-up approach to model building.
Estimate Std. Error z value Pr(>|z|) height 0.0000847 0.0001283 0.66 0.509 log(diameter) 0.05215 0.03479 1.499 0.134 log(depth) -0.09970 0.09059 -1.101 0.271 log(velocity) 0.07223 0.05012 1.441 0.15 log(dissolved.oxygen) 1.450256 0.476236 3.045 0.00232 ** water.temperature
-0.000748 0.0120500 -0.062 0.95
Only dissolved oxygen is statistically significant (p value < 5%) all on it’s own.
d) Use the best covariate and then use that as a basis to see if there are any more significant variables (significance level 5%). If so add the most significant and see if any more can be added.
If not, stop there. What does the resulting model say?
Estimate Std. Error z value Pr(>|z|) height 3.852e-06 1.347e-04 0.029 0.97718 log(diameter) 0.05213 0.03604 1.446 0.1481
No more statistically significant covariates. Stop here.
log(depth) -0.11330 0.09501 -1.192 0.23309 log(velocity) 0.03894 0.05306 0.734 0.46297 water.temperature
-0.001036 0.011768 -0.088 0.92985
Only dissolved oxygen seems to matter. The more the merrier
(species-wise).

Exercise 9 - contd: e) Now start with a large model containing all possible covariates. If any covariates are not significant, remove the most insignificant and start over again. What model do you end up with now? If there are any differences between the results now and for the stepwise-up approach, why do you think that happened?
Full model: height most insignificant (highest p-value)
Diameter, depth, velocity, oxygen, temperature: temperature most insignificant
Diameter, depth, velocity, oxygen: velocity most insignificant
Diameter, depth, oxygen: No more insignificant covariates!
Result:
Estimate Std. Error z value Pr(>|z|)
(Intercept) 0.75258 1.13125 0.665 0.5059 log(diameter) 0.09564 0.04306 2.221 0.0263 * log(depth) -0.25871 0.11335 -2.283 0.0225 * log(velocity) 0.05423 0.05536 0.980 0.3273 log(dissolved.oxygen) 1.30535 0.49974 2.612 0.0090 **
Depth has negative impact on species richness, while substrate diameter has a positive influence, it seems. The reason why we didn’t arrive at this model when going stepwise-up might be because (log) diameter and depth are also positively correlated, correlation=47%. Thus if only one covariate , say diameter, is added then the effect will be that it gets a regression f actor
0.096 directly from the change in diameter but also a factor 0.47*(-0.26)) from the depth, since increasing diameter means increasing depth on average. The result in an effect too small to be detectable.

Perform a power-law regression on discharge as a function of stage at station Gryta, that is do a linear regression of log(discharge) against log(stage-zero plane), now
 with unknown zero-plane. I.e. do regression on the model log( Q i
)
 a
 b log( h i
 h
0
)
  i
 i
~ N(0,

) q i or if you will Q i

C ( h i
 h
0
) * E i
E i
~ log N ( 0 ,

), C
Code: http://folk.uio.no/trondr/nve_course_3/gryta2.R
 e a a) Perform a linear regression of log-discharge on log(stage-h
0
) for a set of candidate values for h
0
. Look at the likelihood as a function of these candidate values. What is the best estimate for h
0 is the estimate for a, b, C and

?
? What h0[l==max(l)]
[1] 0.08
ML estimate: h
0
=0.08
c(a,b,exp(a),sigma)
[1] 0.87241912 1.96226749 2.39269207
0.04131054

b) A test called the likelihood-ratio test says that the zero-hypothesis is rejected with 95% confidence when 2(l full
-l
0
)>3.84 (PS: for one parameter). Test if h
0
=0.
2*(l.a-l.0)
[1] 7.690884
2(l a
-l
0
)>3.84
Reject h
0
=0 with 95% confidence.

Extra exercise 6:
Will de-trend daily data from Hølen using linear regression. Then a relatively naive
ARMA time series analysis will be performed.
Code: http://folk.uio.no/trondr/nve_course_3/hoelen_arima.R
a) Plot the data b) De-trend the data using linear regression on time+trigonometric functions over time (to catch the seasonality).
Doesn’t look like we have removed time-dependency…

Extra exercise 6 - contd.
c) Take a look at the residuals (i.e. the de-trended data). Is there auto-correlations there? Can we trust model selection using linear regression here?
Auto-correlation: Yes!
Can we trust the model selection? No! Severely autcorrelated residuals. d) Take a look at the partial auto-correlation also.
Would an AR- or MA-model be expected to perform best?
PACF terms die out rapidely. Seems like AR(2) might perform well, but will do things step-wise.
ACF
PACF e) Fit an AR(1) model (PS: pacf suggests AR(2) is better). Check whether the estimated parameter is approximately the same as the first term in the autocorrelation?
AR(1) c oefficient=0.9472, pretty much equal to the first value in the pacf and acf plots.
Coefficients: ar1 intercept
0.9472 0.0000
s.e. 0.0018 0.0252

Extra exercise 6 - contd.
e) Make analytic plots of the AR(1) residuals. What does these suggest?
Suggests that a lot of the time dependencies have been explained, but that there is still some residual time dependency in the data. f) Also use an ARMA(1,1) model. Take another look at the residuals. What do they seem to say now?
That this looks good?

Extra exercise 7:
BUse the Kalman filter to interpolate over holes at the station Farstad, year 1993. Will use the simplest type of continuous time Markov chain, namely the Wiener process (random walk). The data has varying time resolution and will be log transformed before use.
The system (updating) equation is: x i
| x i

1
~ N ( x i

1
,
 t i
 t i

1
)
The observation equation is: y i
| x i
~ N ( x i
,

)
In total, only two parameters,
 and

.
a) Plot the data.
Make artifical holes: 1993-02-01 00:00:00 - 1993-02-03 00:00:00
1993-03-01 12:00:00 - 1993-03-15 12:00:00
1993-08-01 00:00:00 - 1993-08-01 18:00:00
1993-09-01 00:00:00 - 1993-09-02 00:00:00
1993-10-01 00:00:00 - 1993-11-01 00:00:00
Plot the data in the holes together with the rest of the data.

Extra exercise 7- contd:
Use the Kalman filter to interpolate over holes at the station Farstad, year 1993. Will use the simplest type of continuous time Markov chain, namely the Wiener process (random walk). The data has varying time resolution and will be log transformed before use.
The system (updating) equation is: x i
| x i

1
~ N ( x i

1
,
 t i
 t i

1
)
The observation equation is: y i
| x i
~
In total, only two parameters,
 and

.
N ( x i
,

) b) Take a glance at the Kalman filter to see if you recognize what the various pieces are doing.
The method is follows slavishly, but the first instance needs to be handled some way. I use a prior x
1
~N(0,10). We have a one-dimensional state space, so all vectors and matrices become scalars. Notice also what happens when there is missing data: posterior= prior c) Run an ML optimization and check the results. What does this say about the observational noise parameter,

? sigma.ml=0.0330157805 tau.ml=0.0000153492
Tau is estimated as vanishingly low. Could thus use the observation equation y i
=x i observations themselves as a random walk.
and handled the d) Perform a Kalman smoothing Kalman using the ML estimated parameters. Look at the end result.

Extra exercise 7: e). Check how the interpolation has functioned in the holes. What kind of interpolation is this (on the log-scale)? What do you in addition get?
Hva slags interpolasjon er dette (på log-skala). Lineær-interpolasjon.
What do you get in addition? Uncertainty!

Extra exercise 7: f). Criticize the model and see if you understand how the fit and uncertainty is as it is. If you feel creative, try to come up with a better model.
With an expectation equal to the previous value, it may not come as a surprise that when you interpolate you get a compromise between the values around the hole. The model relates only to actual discharges, not to the time derivates of that (the Wiener process doesn’t even have time derivatives!). That the uncertainty bubbles out more the larger the holes, is not very unreasonable.
That they can grow beyond limits for larger and larger holes may however be not so reasonable.
(Would need a stationary process for that).
Thus the change (derivative) does not depend on the direction the process was taking before the hole.
A model with a continuous derivative might do better if such a behaviour is wanted. See for instance the hole in February.
Seasonailty is almost certain to be present and should form part of the description.
A model that incorporates a simple hydrological model can be expected to perform much much better.
However for such a model, a more sophisticated method must be used (extended Kalman filter or particle filter).

Extra exercise 8:
Use the Kalman filter to interpolate over holes at the stations Etna and Hølervatn, 2010-2011. The data has equi-distant time resolution
(days) and will be log-transformed before the analysis. Use a 2-dimensional AR(1) process (thus stationary), corr-correlated noise (so that completition is possible), equal auto-correlation, variance and seasonal trends but individual expectancy.
The sytem equatin then becomes: x i
| x i

1
~ N ( a x i

1

( 1
 a )

( t ),

) der

( t )




 e h


S
1
S
1 sin( sin(
2
2

 t t
/
/
365 )
365 )
The observational equation is y i


, e /
C 1 cos( 2
 t / 365 ) h
C 1
| x i , cos( 2
 e / h
~ N ( t x i
/
, e /
365 )
,

) h

 og

In total eight parameters: a,  e
,  h
,  ,  ,S
1
,C
1 and  .




2
 2

 2
2

 a) Plot the data. Make artificial holes (specified in the R coden). Plot the data in the holes together with the remaining data.
b) Perform an ML optimization (using the included Kalman filter to calculate the likelihood) and check the resulting parameter estimates. What does the auto-correlation means here? What does the cross-correlation imply? c(a.ml,mu.e.ml,mu.h.ml,sigma.ml,rho.ml,tau.ml,S1.ml,C1.ml)
[1] 0.94775606 0.16454854 -0.03336243 0.26462117 0.96971749 0.03952556 -0.07919922 -0.15421623
Tau is not at all zero here. That means there is some uncertainty associated with the measurements. This can be seen as a sign that the model can be improved. Strong auto- and cross-correlation.. a^30=0.5 means a characteristic time of about a month. Cross korrelasjonen of 97% means that completion is possible and can be expected to work fairly well.

Extra exercise 8 – contd: c) Perform a Kalman smoothing with the ML estimated parameters. Take a look at the full dataset.
Most of the holes are small compared to the whole series, Can however see some uncertainty bubling up in the larger holes.
Also some uncertainty concerning the alrger measured values, since there is a bit of observational noise.

Extra exercise 8 – contd: d) Check how the completion worked in and around the holes. What happens when both stations are missing data at the same period? A couple of completions worked less well than the others. Why?
Etna:
Here there are holes at Hølenvatn also.
Hølenvatn:
The completion worked less well because the completion series (Etna) does not contain this extra peak. The uncertainty of the model seems to small and may suggest that better models might be needed.
Here there are holes at Etna also. Thus the uncertainty bubbles out.

Extra exercise 8 – contd: e) Test if the cross-correlation is zero (thus no reason to use one station to complete the other), using the likelihood ratio test.
# test with 5% significance level (95% confidence): c(2*(l.a-l.0), qchisq(0.95,1))
[1] 521.444714 3.841459
The likelihood difference is larger than what we can expect from data produced by the zero-hypothesis for 95% of the cases. Thus the zero-hypothesis is rejected with 95% confidence (and we could easily og higher).
pchisq(2*(l.a-l.0),1,lower=F)
[,1]
[1,] 2.05223e-115
The p value is extremely low! Thus it’s very very plausible that there is cross correlation between the two series. Since the model takes time correlation into acount, this result is *not* and artefact of the time series aspect of the data.