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Problem 1 Computing joint probabilities and correlations of a time series
Suppose that the total streamflows (in m3
/month) Qt and Qt+1 over two successive
months t and t+1 are related by the following equation:
Qt+1 = 0.6Qt + et t = 1....10
In deterministic applications such time series expressions are solved one step at a
time, starting with a specified value of the initial value of the dependent (in this case,
the initial streamflow Q1) and using a specified series of et values (for t = 10). The
resultis a series of Qt values for t > 1.
When Q1 and the et values are uncertain, we can use the time series equation to derive
the mean, variance, and covariance of the streamflow for t > 1. These give a
probabilistic description of the temporally variable streamflow.
Suppose that the random excitation values e1 … e10 are independent normally
distributed random variables, each with mean 4.0 and variance 0.64 (standard
deviation 0,8). In addition, suppose that the initial streamflow Q1 is normally
distributed with mean 10.0 and variance 1.0 (standard deviation 1.0). Assume that Q1
and the et values are independent.
Use the definitions of variance and correlation to evaluate the correlation between the
two successive streamflows Q2 and Q3. First use the defining time series difference
equation and the properties of the mean and variance of a linear function to find the
means and variances of Q2 and Q3. Then use the same equation to derive an
expression for the covariance and correlation between Q2 and Q3. Are Q2 and Q3
independent? Why? What do you think you would get for the means and variances of
Q9 and Q10. Why?
Use MATLAB to determine the joint probability that Q2 and Q3 are both greater than
12.0. Also, plot a typical replicate of streamflow vs. time.
Some relevant MATLAB functions: normrnd, sum
Computing and Data Analysis for Environmental Applications

Solution:
To find the mean and variance of Q2 and Q3: Q2 =
0.6Q1 + e1
E[Q2] = 0.6(10) + 4 =10
Q3 = .6(.6Q1 + e1)+e2
E[Q3] = .36(10)+.6(4)+4 =10
If the Qi’s are independent:
Var[a1Q1+a2Q2+…+anQn] = a1
2
Q1
2
+…+an
2
2
Qn
Apply to the difference equations for Q2 and Q3 :
Var[Q2] = (.6)2
(1)2
+(.8)2
= 1
Var[Q3] = (.36)2
(1)2
+(.6)2
(.8)2
+(.8)2
= 1
Find the correlation between Q2 andQ3:
Correl(Q2,Q3) = Cov(Q2,Q3)/(Var[Q2]*Var[Q3])
e2 )(Q2-Q2 )]
Cov(Q2,Q3) = E[(Q2-Q2 )(Q3-Q3 )]
Q3- Q3= 0.6(Q2- Q2 )+(e2 e2 )
E[(Q2-Q2 )( Q3-Q3 )] = 0.6*E[(Q2-Q2 )2
]+E[(e2
= 0.6*Var[Q2] + Cov(Q2,e2)
What is Cov(Q2,e2)?
e2 )]+
Note: Q2- Q2 = 0.6(Q1- Q1)+(e1 e1) Then:
Cov(Q2,e2) = E[ (Q2- Q2 )( e2
E[ (e1 e1 )(e2
e2 )] = 0.6* E[(Q1-Q1)(e2
e2 )]
From the problem statement, Q1 is independent of the ei’s, and the ei’s are independent of each
other, Therefore:
Cov(Q2,e2) = E[ (Q2-Q2 )(e2 e2 )] = 0
And:
Cov(Q2,Q3) = 0.6*Var[Q2]+Cov(Q2,e2) = 0.6*Var[Q2] = 0.6
Correl(Q2,Q3) = 0.6/(1*1) = 0.6

MATLAB solution to second part of problem.
Problem 2 The Central Limit Theorem(solution)
The Central Limit Theorem states that the distribution of a linear function of independent
random variables approaches the normal distribution as the number of variables increases. Use
MATLAB to demonstrate this by carrying out the following virtual experiment:
Generate a random array of 1000 replicates (rows) each consisting of 32 independent
identically distributed sample values (columns), using one of the non-normal random number
generators provided in the MATLAB statistics toolbox (e.g. exponential, uniform, lognormal,
etc.). Select your own values for all required distributional parameters. Plot histograms and
CDFs (over all replicates) for the sample mean:
N
i 1
1
N
xi
mx
For N= 1, 4, 16, and 64. Plot all 4 CDFs on the same axis using the MATLAB normplot
function, which displays normally distributed data as a straight line.Your CDFs should more
closely approach a straight line as Nincreases, indicating convergence to a normal probability
distribution. The approach to normality should also be apparent in your 4 histograms (try
plotting all four on one page, using the MATLAB subplot function)
Some relevant MATLAB functions: exprnd, chi2rnd, lognrnd, hist,
normplot, mean, subplot.

% Problem Set 5
% Problem 1
clear all
close all
nrep=1000000;
emu=4;
esigma=.8;
qmu=10;
qsigma=1;
% create a matrix of random e's
emat=normrnd(emu,esigma,nrep,9);
% create the Q matrix
Qmat = zeros(nrep,10);
Qmat(:,1) = normrnd(qmu,qsigma,nrep,1);
for i=2:10
Qmat(:,i) = 0.6*Qmat(:,i-1)+emat(:,i-1);
End
% to get the correlation, need mean of (Q2-Q2bar)*(Q3-Q3bar) divided by
% sqrt [var(Q2)*var(Q3)]
Q2bar=mean(Qmat(:,2));
Q3bar=mean(Qmat(:,3));
numerator = mean((Qmat(:,2)-Q2bar).*(Qmat(:,3)-Q3bar));
denominator = sqrt(var(Qmat(:,2))*var(Qmat(:,3)));
correl = numerator/denominator
covariance = numerator

vect = Qmat(:,2)>12 & Qmat(:,3)>12;
jointprob = sum(vect)/nrep
plot(1:10,Qmat(5,:),'-*')
xlabel('Time')
ylabel('Streamflow')
title('Sample Streamflow vs. Time')
% One set of results:
%correl =
% 0.6007
% covariance =
% 0.6012 %
jointprob =
% 0.0057

% Problem Set 5
% Problem 2
clear all
close all
nrep=1000;
mat=exprnd(5,nrep,64);
m1=mat(:,1);
m4=sum(mat(:,1:4),2)/4;
m16=sum(mat(:,1:16),2)/16;
m64=sum(mat(:,1:64),2)/64;
figure
subplot(2,2,1), hist(m1)
figure
normplot(m1)
hold normplot(m4)
normplot(m16)
normplot(m64)
legend('N=1','N=4','N=16','N=32')

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