PENELOPE 2003 - OECD Nuclear Energy Agency

1.3. Monte Carlo integration 19
Consider the integral
I =
∫ b
which we recast in the form of an expectation value,
a
F (x) dx, (1.63)
∫
I = f(x) p(x) dx ≡ 〈f〉, (1.64)
by introducing an arbitrary PDF p(x) and setting f(x) = F (x)/p(x) [it is assumed that
p(x) > 0 in (a, b) and p(x) = 0 outside this interval]. The Monte Carlo evaluation of the
integral I is very simple: generate a large number N of random points x i from the PDF
p(x) and accumulate the sum of values f(x i ) in a counter. At the end of the calculation
the expected value of f is estimated as
f ≡ 1 N
N∑
f(x i ). (1.65)
i=1
The law of large numbers says that, as N becomes very large,
f → I (in probability). (1.66)
In statistical terminology, this means that f, the Monte Carlo result, is a consistent
estimator of the integral (1.63). This is valid for any function f(x) that is finite and
piecewise continuous, i.e. with a finite number of discontinuities.
The law of large numbers (1.66) can be restated as
1
〈f〉 = lim
N→∞ N
N∑
f(x i ). (1.67)
i=1
By applying this law to the integral that defines the variance of f(x) [cf. eq. (1.16)]
we obtain
∫
var{f(x)} = f 2 (x) p(x) dx − 〈f〉 2 , (1.68)
var{f(x)} = lim
⎧
⎨
N→∞ ⎩
1
N
[
N∑
1
[f(x i )] 2 −
i=1
N
2
⎫
N∑ ⎬
f(x i )]
⎭ . (1.69)
i=1
The expression in curly brackets is a consistent estimator of the variance of f(x). It is
advisable (see below) to accumulate the squared function values [f(x i )] 2 in a counter
and, at the end of the simulation, estimate var{f(x)} according to eq. (1.69).
It is clear that different Monte Carlo runs [with different, independent sequences of
N random numbers x i from p(x)] will yield different estimates f. This implies that the
outcome of our Monte Carlo code is affected by statistical uncertainties, similar to those
found in laboratory experiments, which need to be properly evaluated to determine the
“accuracy” of the Monte Carlo result. For this purpose, we may consider f as a random