A Gaussian pdf with unit variance is given by:
fx=12πⅇ-x22
f
x
1
2
x
2
2
(1)
The probability that a signal with a pdf given by
fx
f
x
lies above a given threshold
xx is given by the Gaussian Error
Integral or
QQ function:
Qx=∫x∞fudu
Q
x
u
x
f
u
(2)
There is no analytical solution to this integral, but it has a
simple relationship to the error function,
erfx
erf
x
, or its complement,
erfcx
erfc
x
, which are tabulated in many books of mathematical
tables.
erfx=2
π
∫0xⅇ-u2du
erf
x
2
u
0
x
u
2
(3)
and
erfcx=1−erfx=2π∫x∞ⅇ-u2du
erfc
x
1
erf
x
2
u
x
u
2
(4)
Therefore,
Qx=12erfcx2=121−erfx2
Q
x
1
2
erfc
x
2
1
2
1
erf
x
2
(5)
Note that
erf0=0
erf
0
0
and
erf∞=1
erf
1
, and therefore
Q0=0.5
Q
0
0.5
and
Qx→0
Q
x
0
very rapidly as
xx becomes large.
It is useful to derive simple approximations to
Qx
Q
x
which can be used on a calculator and avoid the need
for tables.
Let
v=u−x
v
u
x
, then:
Qx=∫0∞fv+xdv=12π∫0∞ⅇ-v2+2vx+x22dv=ⅇ-x222π∫0∞ⅇ-vxⅇ-v22dv
Q
x
v
0
f
v
x
1
2
v
0
v
2
2
v
x
x
2
2
x
2
2
2
v
0
v
x
v
2
2
(6)
Now if
x≫1
≫
x
1
, we may obtain an approximate solution by replacing
the
ⅇ-v22
v
2
2
term in the integral by unity, since it will initially
decay much slower than the
ⅇ-vx
v
x
term. Therefore
Qx<ⅇ-x222π∫0∞ⅇ-vxdv=ⅇ-x222πx
Q
x
x
2
2
2
v
0
v
x
x
2
2
2
x
(7)
This approximation is an upper bound, and its ratio to the true
value of
Qx
Q
x
becomes less than
1.11.1 only when
x>3
x
3
, as shown in
Figure 1. We
may obtain a much better approximation to
Qx
Q
x
by altering the denominator above from (
2πx
2
x
) to (
1.64x+0.76x2+4
1.64
x
0.76
x
2
4
) to give:
Qx≈ⅇ-x221.64x+0.76x2+4
Q
x
x
2
2
1.64
x
0.76
x
2
4
(8)
This improved approximation gives a curve indistinguishable from
Qx
Q
x
in
Figure 1 and its ratio
to the true
Qx
Q
x
is now within
±0.3%
±
0.3
%
of unity for all
x≥0
x
0
as shown in
Figure 2. This
accuracy is sufficient for nearly all practical problems.
