Constructing multivariate distributions for soil parameters - Risk and Reliability in Geotechnical Engineering - page 12

1.2.3.1 Method of moments
The model parameters μ and σ can be estimated by the sample mean ( m ) and sample stan-
dard deviation ( s ):

n n

1 1

∑ ∑

( ) ≡

2

µ ≈ Y

() k

≡ m σ ≈ Y

() k

− ms (1.13)

n n − 1

k = 1 k = 1

The sample mean and sample standard deviation obtained from simulated data are 101.84
and 23.82, respectively. They are different from the theoretical values of μ = 100 kPa and
σ = 20 kPa, because of the statistical uncertainty. We can obtain a comprehensive view
of statistical uncertainty by simulating 1000 different sets of values. Each set consists of
10 numbers. Equation 1.13 is applied to each set, thus producing 1000 sample means and
1000 sample standard deviations. Figure 1.8 s hows the empirical histograms for ( m − μ)/(s/
n 0.5 ) and ( n − 1) s 2 /σ 2 . The sample mean m can be as small as 81.43 and as large as 120.36,
although μ = 100 kPa. Note that s 2 is not normally distributed, although the underlying data
are normally distributed. It can be proved theoretically that ( n − 1) s 2 /σ 2 is χ-squared distrib-
uted. These “sampling distributions” are discussed in more detail below.

1.2.3.2 Percentile method
The percentile has been defined in Equation 1.9 . The median of Y is the 0.5-percentile of
Y. For the normal distribution, the mean value μ is identical to the median (0.5-percentile)
by substituting Φ −1 (0.5) = 0 in Equation 1.9 . As a result, one can estimate μ by estimating
the 0.5-percentile. A simple illustration of the sample value of 0.5-percentile is given. Table
1.2 shows the sorted Y data, from the smallest to the largest. The third column contains
the simulated data sorted in an ascending order. The fourth column contains the rank. The
smallest number is rank 1. The second smallest number is rank 2 and so forth. The sample
size n = 10; hence, the sample median can be taken as the average of the Y values with ranks
5 and 6, namely sample median = (90.52 + 105.41)/2 = 97.97. It is not the same as the actual
median (=100) because of the statistical uncertainty.

(a) 400 (b) 300

350
300

250

200

250

200 150

150

100

100

50

50
0 0 0

-4 -2 0

( m -µ)/( s /n 0.5 )

2 4 10 20 30

( n -1)/( s 2 / s 2 )

Figure 1.8 Histograms of ( m − μ )/( s / n 0.5 ) and ( n − 1) s 2 / σ 2 and the sampling distributions.