Exercises 1: Execute step-by-step the R code at
http://folk.uio.no/trondr/nve_course_3/hoelen1.R
Take a look at the yearly discharge means at Hølen.
This code should give answers to the following:
a) Take a look at the histogram together with the normal
distribution having the same mean(expectancy) and
standard deviation. Do the data look approximately
normally distributed?
b) Make a QQ plot to further investigate whether the data is
approximately normally distributed.
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).
d) Repeat a-c for daily discharges also (found at
http://folk.uio.no/trondr/nve_course_3/TrendDognHoele
n.txt). What conclusions change and why?
Exercise 2: Is the expectancy of the yearly mean discharges at
Hølen equal to 10m3/s?
http://folk.uio.no/trondr/nve_course_3/hoelen2.R
a)
b)
c)
d)
e)
Estimate the expectancy.
Check the assumption expectancy=10m3/s using a t-test (takes
the variance uncertainty into account). Use preferably a
significance level of 5% (confidence level of 95%).
Show the data together with the confidence interval. Should
we worry that most of the data is outside this interval? Is it
95% probability that the real expectancy lies within this
specific interval?
Would we be justified in using the techniques in a-c for daily
data also?
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=10m3/s?
Extra exercise 1
Will illustrate the law of large numbers and the central limit theorem
Code: http://folk.uio.no/trondr/nve_course_3/lawoflargenumbers.R
a) Draw n=10000 samples from the Poisson distribution with rate (and
thus expectancy) =0.3. Calculate the mean and see how much or
little this differs from the expectancy. Repeat a couple of times. Why
are the results as they are?
b) Look at the histogram and compare with the probability distribution.
c) Look at the distribution of means of n=10000 samples from the
Poisson distribution. Sample N=1000 such means. Look at the
histogram and compare with the normal distribution N (  ,  / n )
which the central limit theorem says the distribution of the mean
should start to resample. Check with a QQ plot.
d) What happens with the mean when only a few samples, n=10, are
used. Increase N to 100000. Is there a noticeable difference
between the distribution of the means and the normal distribution?
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)
b)
c)
d)
e)
f)
g)
h)
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?
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.
What is the probability for sum<=4?
What is the probability for equal outcome on the two dice?
What is the probability of getting equal outcomes and sum<=4?
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.
Both from the rule for conditional probability and from the list of
outcomes where sum<=4, what is the probability of equal dice
given that sum<=4
Calculate the probability from sum<=4 given two equal dice, both
from a list of possible outcomes and from Bayes formula.
Extra exercise 3:
At Blindern, there is a 33.9% chance of rain any given
day, given that it rained yesterday. There is a 12.9%
chance of rain given that it didn’t rain the previous day.
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: The law of
total probability gives Pr(rain today)=
Pr(rain today|rain yesterday)Pr(rain yesterday)+
P(rain today| no rain yesterday)Pr(no rain yesterday).
b) Why is the chance of rain yesterday given that it
rained today also 33.9% (Tip: Bayes formula)
Extra exercise 4 – conditional
probabilities
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%
were dragon-infested.
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
Here be 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?
or
(Hint: The law of total probability)
Show using Bayes formula that the
probability of there being a dragon in
the area if you actually detected one,
is 100%.
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
Dragon
detected
Dragon
detected
and in the
area
Dragon in
the area
No
dragon
No
dragon
Exercise 3: Analysis of repeat period for a given threshold
value in continuous time.
Will look at the danger for the discharge to go above a fixed threshold. The station
Gryta has discharge>1.5m3/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 ) e
y
t /T
y!
The likelihood is maximal at Tˆ  t . (For those with analytical optimization
y
experience, show this).
a) What is the ML-estimate for T in this particular case?
b) From asymptotic theory, one has that:
-1
ˆ ~ N(  , I( ) ) where
 l ( )
2
I ( )   E

2
is Fisher' s informatio
n matrix.
Construct a 95% confidence interval using I (T )  t / T and that the normal
distribution has 95% of it’s probability mass within +/-1.96 standard deviations
from the expectancy (Insert the ML estimate for T).
3
Exercise 4: Independence, Markov chains and
precipitation at Blindern
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 xi, where i denotes the day index. Since we need to condition
on the previous state, we set aside the first day, x0. The number of days
except this, is denoted n.
Zero hypothesis: The rain status (0 or 1) is independent from day to day
(Bernoulli 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:
k
nk
where k=number of days with rain. This
P ( x1 ,  , x n )  p (1  p )
will thus be the likelihood-function of this model. The ML estimate of p is
then pˆ  k / n
Alternative hypothesis: The rain status (0 or 1) is a Markov chain. There is a
transition probability from no rain to rain pNR and a transition probability for
rain given rain the previous day also, pRR. (The two other probabilities are
given as the negation of the two that have been specified). The probability of 1-pRR
a given dataset is then: P ( x ,  , x )  p k R (1  p ) n R  k R p k N (1  p ) n N  k N
1
n
RR
RR
NR
NR
where kR and nR is the number of rainy days and total days when it rained the
previous day and kN and nN are similarly defined for cases where it didn’t rain
the previous day. The ML estimate for the two parameters then become:
pˆ RR  k R / n R , pˆ NR  k N / n N
pRR
R
pNR
N
1-pNR
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 pNR and pRR (alternative hypothesis). From the last, what seems
to be the nature of the rain status one day and the next?
b) From extra exercise 2a, one has that the marginal probability of rain under
the alternative hypothesis p’=pNR/(1-pRR+pNR). Compare the estimate of this with
the estimate of p from the zero hypothesis.
c) The likelihood ratio test check a zero hypothesis
against an alternative (more
2
advanced) hypothesis by using that 2 ( l a  l 0 ) ~   k under the zero hypothesis,
where la and l0 is the logarithm of the maximal likelihood for alternative and zero
hypothesis respectively and  2 is the so-called chi-square distribution (the
distribution of the square of something that is standard normal) and k is the
difference in the number of parameters. Calculate the p value and check if one
can reject time independence with 95% confidence. i.e. check whether the rain
status can be independent from day to day.
d) Assume that the ML estimators for pNR and pRR are approximately normally
1-pRR
distributed (asymptotic theory) and var( pˆ RR )  p RR (1  p RR ) / n R and similarly
for pNR (and p). What is the 95% confidence interval for these two parameters?
e) Run a parametric bootstrap to look at the uncertainty (95% confidence
interval) in p (zero hypothesis) and p’ (Markov chain). Compare the 95%
confidence intervals for p and p’ from bootstrap and also the methodology in d as
applied to p.
pRR
R
pNR
N
1-pNR
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)
b)
c)
d)
e)
f)
g)
h)
Identify what is data outcomes and what is the thing we want to do inference
on (the parameter, if you like).
What is the prior distribution? (Tip: What is it we want to do inference on?)
What is the likelihood? (Tip: How is it the data is produced conditioned on what
we want to do inference on)
Describe and calculate (using the law of total probability) the prior prediction
distribution (marginal data distribution).
What is the posterior distribution, given a positive or a negative test?
A little risk analysis: The cost (C) for society (patient time as well as medical staff
and equipment) 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 operation pay off in total?
Assume the first test to be positive. What is the posterior prediction distribution
for a new test, now?
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?
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 |  ,  ) ~ N (  ,  )
Then the overall mean will also be normally distributed f ( x |  ,  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, : f (  ) ~ N (  0 ,  2 )
where 0==10. Assume we know that the standard deviation of the individual
values is =2.83.
It can be shown that the posterior distribution is:
2
2
 x 2    2 / n


/n
0
f ( | x ) ~ N 
,
2
2
  2  2 /n
  /n

a)
b)

  N (  ( x ),  ( x ))  N (  ( D ),  ( D ))


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?
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 =10m3/s?
Exercise 6 – contd: Expectation value for mean discharge at Hølen – Bayesian
analysis
2
2
The prior predictive (marginal) distribution is: f ( x ) ~ N (  0 ,    / n )
c) We will now test the assumption =10. Compare the prior predictive
distribution under our model (free-ranging ), with the prior predictive
distribution when =10 (see the expression for the likelihood). What does
this suggest?
d) Calculate the posterior model probability for model 0 (=10) and model 1
(free-ranging ).
Use Pr( M | D ) 
f ( D | M ) Pr( M )

f ( D | M ' ) Pr( M ' )
and assume the prior model probability for each model to be 50%. What is
the conclusion?
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?
Exercise 7: Bayesian analysis of repeat period for a given
threshold value in continuous time.
Will look at the danger for the discharge to go above a fixed threshold. The station
Gryta has discharge>1.5m3/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 yy events will then be (Poisson
distribution): P ( y event in time t | T )  ( t / T ) e  t / T
y!
Assume an inverse gamma distribution (which is mathematically convenient) for
the parameter, T: f (T )   T   1e   / T
 ( )
The prior predictive (marginal) data distribution is then:
   y  1 y

 p (1  p ) where
P ( y events in t )  
y


p
(this is the so-called negative binomial distribution).
t
t
Exercise 7 (contd.): R code is found at
http://folk.uio.no/trondr/nve_course_3/gryta_extreme.R
P ( y event in time
P ( y events in time
t |T ) 
(t / T )
y
e
t /T
f (T ) 


 ( )
T
  1   / T
e
y!
   y  1 y
t

 p (1  p ) where p 
t )  
y
t

3 
The station Gryta has had discharge>1.5m /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.
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 (non-informative) . Were there
large differences in the posterior probability distribution? Compare these results with the frequentist
result: TML=t/y=1.63 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% quantile 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.
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?
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).
f)
Fetch 1000 approximately independent samples after burn-in. Compare this result with the
theoretical posterior distribution (histogram and QQ plot).
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.
Exercise 8:
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, h0=0)
Code: http://folk.uio.no/trondr/nve_course_3/gryta.R
Data: http://folk.uio.no/trondr/nve_course_3/gryta.txt
a) Plot the data, both on the original scale and on log-scale.
b) Run a linear regression on log(discharge) against
log(stage). Interpret the result. Is there a significant
relationship between stage and discharge?
c) What is the formula for discharge vs stage on the original
scale? Plot this.
d) Check if anything is wrong with the residuals (trend or
non-normality).
e) Extra: Perform a linear regression on the original scale and
make a plot of this also. (PS: R code not included).
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 x1, i     k x k , i
where yi is the response (number of species) for site number i and xj,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.
b)
Run a Poisson fit with only a constant term (no covariates). Compare the result with the
mean species number.
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 pvalue)? We have now started a stepwise-up approach to model building.
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?
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 stepwiseup approach, why do you think that happened?
Exercise 10:
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
q i  log( Q i )  a  b log( h i  h 0 )   i
or if you will
Q i  C ( hi  h0 ) * E i
 i ~ N(0,  )
E i ~ log N ( 0 ,  ), C  e
a
Code: http://folk.uio.no/trondr/nve_course_3/gryta2.R
Data: http://folk.uio.no/trondr/nve_course_3/gryta.txt
a) Perform a linear regression of log-discharge on log(stage-h0)
for a set of candidate values for h0. Look at the likelihood as
a function of these candidate values. What is the best
estimate for h0? What is the estimate for a, b, C and ?
b) A test called the likelihood-ratio test says that the zerohypothesis is rejected with 95% confidence when
2(lfull-l0)>3.84 (PS: for one parameter). Test if h0=0.
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):
1  ( x   ) /   e ( x  ) / 
f (x | ,  ) 
e
a)

Make an extreme value plot, by ordering the yearly maximal discharges and plot
them against expected repeat time interval t  n  0 .12 where n=number of years
i  0 . 44
and i is an index running from 1 to n
Fit the Gumbel distribution to the data using the two first l-moments (see the
code), 1 and 2. (Extra: Compare with what you get from DAGUT). The
parameters relate to the l-moments as = 2/log(2), = 1-0.57721. Calculate 1
and 2 and as
i
b)
1
ˆ1 
n
c)
d)
e)
f)
g)
n

j 1
x j  x , ˆ2 
1
n ( n  1)
n
 (( j  1)  ( n 
j )) x( j )
Sorted data
j 1
Make an extreme value plot (extreme value against repetition time) given the lmoment estimates together with the data.
Do numerical ML estimation of the parameters.
Make an extreme value plot given the ML estimates.
Perform a Bayesian analysis with flat prior. Sample 1000 MCMC samples (burnin=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).
Use also the posterior prediction distribution to make an extreme value plot (with
the parameter uncertainty taken into account). Compare with previous results.
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
Data: http://folk.uio.no/trondr/nve_course_3/TrendDognHoelen.txt
a) Plot the data.
b) De-trend the data using linear regression on time+trigonometric
functions over time (to catch the seasonality).
c) Take a look at the residuals (i.e. the de-trended data). Is there autocorrelations there? Can we trust model selection using linear
regression here?
d) Take a look at the partial auto-correlation also. Would an AR- or MAmodel be expected to perform best?
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?
f) Make analytic plots of the AR(1) residuals. What does these suggest?
g) Also use an ARMA(1,1) model. Take another look at the residuals.
What do they seem to say now?
Extra exercise 7:
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 ~ N ( x i ,  )
In total, only two parameters,  and .
Code: http://folk.uio.no/trondr/nve_course_3/farstad.R
Data: http://folk.uio.no/trondr/nve_course_3/farstad_1993.txt
a)
Plot the data. The time series has an unmarked hole between May and June which needs to be marked.
In addition, we will make artificial 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.
b)
Take a glance at the Kalman filter to see if you recognize what the various pieces are doing.
c)
Run an ML optimization and check the results. What does this say about the observational noise
parameter, ?
d)
Perform a Kalman smoothing Kalman using the ML estimated parameters. Look at the end result.
e)
Check how the interpolation has functioned in the holes. What kind of interpolation is this (on the logscale)? What do you in addition get?
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.
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 2dimensional AR(1) process (thus stationary), correlated noise (so that completition is possible),
equal auto-correlation, variance and seasonal trends but individual expectancy.
The system equation then becomes:
x i | x i 1 ~ N ( a x i 1  (1  a )  ( t ),  )
where
 2
  e  S 1 sin( 2  t / 365 )  C 1 cos( 2  t / 365 ) 
 and   
 ( t )  
  2


S
sin(
2

t
/
365
)

C
cos(
2

t
/
365
)
1
1
 h


2
 

2



The observational equation is y i , e / h | x i , e / h ~ N ( x ,  )
In total eight parameters: a,e,h,,,S1,C1 and .
Code: http://folk.uio.no/trondr/nve_course_3/kompletering.R
Data: http://folk.uio.no/trondr/nve_course_3/etna_hoelervatn.csv
a)
Plot the data. Make artificial holes (specified in the R code). 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)
Perform a Kalman smoothing with the ML estimated parameters. Take a look at the full dataset.
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?
e)
Test if the cross-correlation is zero (thus no reason to use one station to complete the other),
using the likelihood ratio test.
i ,e / h