216
DOC.
7
PROBABILITY CALCULUS
~(f)df
I.
=
fdA~
LF(so)(dF\
i-Ai
IiI
(dAJ5J
I
-8v7
,®df
-fda
+
_
L
L
Integrated
by
parts,
this
becomes
o
%
=
"
\dA-
*-F(S0)
dF
dA
p(-A\fz).
\[Z
'S
-
JdA"
0
ms0)
+
(dF_
KdASS.
p(-Ayfz). v/Z.
Now,
since
by
the
assumption,
J/p(f)df=0,
we
get,
if
we
introduce
A
/z
=
f
as a
variable,
(9)
s.
1
2Z
(dF_
v^A/S
•f-
(8)
and
(9)
inserted
in
(6) yield
the differential
equation
SF
+
f2-
=
0,
J
ds
the
solution
of
which,
(10)
F
=
const.e
expresses
the
Gaussian law
of
errors.