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The main points i the stochastic hydrology are listed below:Method of Maximum Likelihood, Likelihood Function, Estimates of Parameter, Chebyshev Inequality, Irrespective of Probability Distribution, Moments and Expectation, Covariance, Degree of Association
Typology: Study notes
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L = f(x 1
;θ 1
;θ 2
…θ m
) x f(x 2
;θ 1
;θ 2
…θ m
) x f(x n
;θ 1
;θ 2
… θ m
estimated
3
1 1
n
i m i
f x θ θ
=
i
i
θ
Obtain the maximum likelihood estimates of the parameter
β in the pdf
4
2 2
x
β
β
β
π
−
( )
2 2 2
1 2
2
1
2
1
2 2 2
1 2
/
2
1
/2 (^) /2 2
1
n n i i n i i x x^ x
n
n n x
n n
i i
n^ x
n n^ n n
i i
L x e x e x e
x e
x e
β β^ β
β
β
β β β
β β β β
π π π
β
β
π
β π
=
=
− −^ −
−
=
−
=
selected at random from any probability distribution will
deviate more than kσ from mean μ with a probability
less than or equal to 1/k
2 .
(places an upper bound on the probability for deviation
from the mean.)
6
2
⎡ (^) − μ ≥ σ⎤ ≤
The mean annual stream flow of a river is 135 Mm
3 and
standard deviation is 23.8 Mm
3 .What is the maximum
probability that the flow in a year will deviate more than
45 Mm
3 from the mean.
Applying Chebyshev inequality,
kσ = 45
k x 23.8 = 45
k = 1.
2
7
2
k
2
⎡ − μ ≥ σ⎤ ≤
or Cov(X, Y)
= Cov(X, Y) = 0, if X and Y are independent
9
( )( )
( )( )
1,
( , ) x y
x y
x y f x y dx dy
E x y
μ μ μ
μ μ
∞ ∞
−∞ −∞
= − −
⎡ ⎤ = − −
⎣ ⎦
∫ ∫
( )( )
1
,
1
n
i i
i
X Y
x x y y
s
n
=
− −
=
−
∑
between two rvs X and Y
= 0, if X and Y are independent
10
,
,
X Y
X Y
X Y
σ
ρ
σ σ
=
,
,
X Y
X Y
X Y
s
r
s s
=
-1 < ρ X,Y
12
stochastic dependence
x
y
r =0.
x
y
r =0.
linear dependence
13
dependence
dependence is non linear,
a high correlation
coefficient can result
x
y
r =0.
functionally related
x
y
r =
Consider Y = aX+b ; perfect linear relation
Substitute Y = aX+b
15
,
,
X Y
X Y
X Y
σ
ρ
σ σ
=
( )
( [^ ]^ [^ ]^ [^ ])
[ ] [ ] ( )
2
, 2
, (^2 )
2
2 2
2
2
2 2
X Y
X Y
X Y
X Y
X Y
E XY E X E Y
E aX bX E X E aX b
σ
ρ
σ σ
σ σ
σ σ
=
−
=
⎡ (^) + ⎤− +
⎣ ⎦
=
ρ = +1 if there is a perfect relationship in between X and Y
Correlation coefficient is a measure of linear dependence
16
2 2 2
2 2
2 2 2 2
2 2
2 2 2
2 2 2
1
X Y
X Y
X
X X
aE X bE X a E X bE X
a E X E X
a
a
σ σ
σ σ
σ
σ σ
⎡ ⎤ (^) + − −
⎣ ⎦
=
⎡ ⎤ (^) −
⎣ ⎦
=
= =
2 2 2
Y X
Q σ = a σ
Example-3 (contd.)
18
Mean,
Therefore mean, = 1627/
= 108.5 cm
Variance,
Standard deviation, s x
= 15.811 cm
1
n
i
i
x
x
n
=
=
1
n
i
i
x
=
x
( )
2
(^2 )
250
1 15 1
n
i
i
x
x x
s
n
=
−
= = =
− −
∑
Example-3 (contd.)
19
Mean,
Therefore mean, = 575/
= 38.33 cm
Variance,
Standard deviation, s y
= 10.841 cm
1
n
i
i
y
y
n
=
=
1
n
i
i
y
=
y
( )
2
(^2 )
1 15 1
n
i
i
y
y y
s
n
=
−
= = =
− −
∑
Example-3 (contd.)
21
Correlation coefficient,
( )( )
1
,
1
15 1
n
i i
i
X Y
x x y y
s
n
=
− −
=
−
=
−
=
∑
,
,
15.811 10.
X Y
X Y
X Y
s
r
s s
=
=
×
=
(x i
, y i
) are observed values
is predicted value of x i
Error,
Sum of square errors
22
ˆ i
y x
y
Best fit line
ˆ i i
y = a + bx
ˆ
i i i
e = y − y
( )
{ (^ )}
2 2
1 1
2
1
ˆ
n n
i i i
i i
n
i i
i
e y y
M y a bx
= =
=
= −
= − +
∑ ∑
∑
Estimate the parameters a, b such that
the square error is minimum
i
y
i
y
