Environmental Engineering Reference
In-Depth Information
In cases where
x
i
are independent (uncorrelated)
random variables, the mean and variance of
y
simplify
to
and in accordance with Equation (10.147) the variance
of
c
, denoted by
σ
2
, is given by
2
2
2
2
2
2
2
2
2
σ
=
( . )
0 4
σ
+
( . )
0 6
σ
=
( . ) ( .
0 4
1 94
)
+
( . ) ( .
0 6
2 31
)
c
n
∑
(mg/L)
2
=
2 52
.
µ
=
a
+
a
µ
(10.146)
y
0
i
i
i
=
1
where
σ
c
=
. . mg/L
. Hence the weighted-
average concentration has a mean and standard devia-
tion of 6.19 and 1.59 mg/l, respectively. It is noteworthy
that the standard deviation of the weighted average is
less than both of the standard deviations of the concen-
trations being averaged.
2 52
=
1 59
n
∑
2
2
2
σ
=
a
σ
(10.147)
y
i
i
i
=
1
The relationships given in Equations (10.144-10.147)
are valid for both addition and subtraction, since sub-
traction simply corresponds to a negative coefficient.
For example, the variance of the sum of two indepen-
dent random variables is equal to the variance of the
difference of the random variables, and in both cases,
the resulting variance is the sum of the variances of the
independent random variables.
10.11.2 Multiplication
Consider the case of multiplication of two random vari-
ables,
x
1
and
x
2
, such that
y
=
x x
(10.148)
EXAMPLE 10.27
1
2
It can be shown that the mean and variance of
y
,
denoted by
μ
y
and
σ
2
, respectively, are given by (Kot-
tegoda and rosso, 1997)
Two sets of concentration measurements are collected
in a lake. The first set consists of 20 measurements, and
statistical analyses indicate that these samples were
drawn from a population distribution having a mean
and standard deviation of 5.83 and 1.94 mg/l, respec-
tively. The second set of data consists of 30 measure-
ments drawn from a population distribution having a
mean and standard deviation of 6.43 and 2.31 mg/l,
respectively. Assuming that both sets of measurements
are independent of each other, determine the mean
and standard deviation of a weighted average of these
data.
µ
y
=
µµ ρ σσ
+
2
(10.149)
1
2
12
1
σ
2
=
σ µ σ µ
2
2
+
2
2
+
2 12
ρ σσ µµ ρ σ σ
−
1
2
2
2
y
1
2
1
2
+
E x
[(
−
µ
) (
2
x
−
µ
) ] 2
2
+
µ
E x
[(
−
µ
)(
x
−
µ
) ]
2
i
1
2
2
1
1
1
2
2
+
2
µ
E x
[(
−
µ
)(
y
−
µ
)]
1
1
2
(10.150)
where
μ
i
and
σ
i
2
are the mean and variance of
x
i
, respec-
tively,
ρ
ij
is the correlation coefficient between
x
1
and
x
2
,
and
E
[·] denotes the expected value.
In cases where
x
i
are independent (uncorrelated)
random variables, the mean and variance relationships
are considerably more simple. In the case of multiplica-
tion of
n
independent random variables,
Solution
From the given data:
N
1
= 20,
μ
1
= 5.83 mg/l,
σ
1
= 1.94 mg/l,
N
2
= 30,
μ
2
= 6.43 mg/l, and
σ
2
= 2.31 mg/l. If
c
is the weighted average of the con-
centration measurements, then
n
∏
y
=
x x x
⋅
⋅
x
=
x
(10.151)
N
N N
N
N N
1
2
3
n
i
1
2
c
=
c
+
c
1
2
i
=
1
+
+
1
2
1
2
20
20 30
30
20 30
=
c
+
c
=
0 4
.
c
+
0 6
.
c
where
x
i
are independent random variables, the mean
and variance of
y
are given by
1
2
1
2
+
+
where
c
1
and
c
2
are concentrations drawn from the
first and second data sets, respectively. In accordance
with Equation (10.146), the mean of
c
, denoted by
μ
c
, is
given by
n
∏
1
(10.152)
µ
=
µ
y
i
i
=
n
n
∏
∏
(
)
−
σ
2
=
µ σ
2
+
2
µ
2
(10.153)
y
i
i
i
µ
c
=
0 4
.
µ
+
0 6
.
µ
=
0 4 5 83
. ( .
)
+
0 6 6 43
. ( .
)
=
6 19
. mg/L
i
=
1
i
=
1
1
2
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