Biomedical Engineering Reference
In-Depth Information
Standard Deviation
The variance of the Poisson distribution is given by Eq. (E.8), with the
P
n
defined
by Eq. (E.23). As before, it remains to find the expected value of
n
2
. Again, manip-
ulating the index of summation
n
, we write, using Eq. (E.23),
∞
∞
∞
n
2
n
n
2
n
µ
µ
n
2
P
n
≡
e
-
µ
e
-
µ
=
(E.24)
n
!
n
!
n
=
0
n
=
0
n
=
1
∞
∞
n
-1
(
n
- 1)!
=
n
n
µ
(
n
+1)
µ
e
-
µ
µ
e
-
µ
µ
=
(E.25)
n
!
n
=
1
n
=
0
n
µ
∞
n
n
n
!
+
µ
= e
-
µ
µ
2
+
µ
.
=
µ
(
µ
+1)=
µ
(E.26)
n
!
n
=
0
Substitution of this result into Eq. (E.8) gives for the variance
2
2
+
2
σ
=
µ
µ
-
µ
=
µ
.
(E.27)
We obtain the important result that the standard deviation of the Poisson distribu-
tion is equal to the square root of the mean:
σ
=
√
µ
(E.28)
.
Normal Distribution
We begin with Eq. (E.23) for the Poisson
P
n
and assume that
µ
is large. We also
assume that the
P
n
are appreciably different from zero only over a range of values
of
n
about the mean such that
|
n
-
µ
|
µ
. That is, the distribution of the
P
n
is
relatively narrow about
µ
;andboth
µ
and
n
are large. We change variables by
writing
x
=
n
-
µ
. Equation (E.23) can then be written
P
x
=
µ
µ
+
x
e
-
µ
x
e
-
µ
µ
!(
µ
+1)(
µ
+2)···(
µ
+
x
)
,
µ
µ
µ
(
µ
+
x
)!
=
(E.29)
with
|
|
µ
. We can approximate the factorial term for large
µ
by means of the
Stirling formula,
x
=
2
πµµ
µ
e
-
µ
,
µ
!
(E.30)
giving
x
µ
P
x
=
√
2
(E.31)
πµ
(
µ
+ 1)(
µ
+2)
···
(
µ
+
x
)
1
=
.
(E.32)
1+
1
µ
1+
2
µ
1+
x
µ
√
2
πµ
···
