PROBABILITY
A biased die has probabilitiesp/ 2 ,p,p,p,p, 2 pof showing1, 2, 3, 4, 5, 6respectively. Find
(i)the mean,(ii)the second moment and(iii)the variance of this probability distribution.By demanding that the sum of the probabilities equals unity we requirep=2/13. Now,
using the definition of the mean (30.46) for a discrete distribution,
E[X]=∑
jxjf(xj)=1×^12 p+2×p+3×p+4×p+5×p+6× 2 p=
53
2
p=53
2
×
2
13
=
53
13
.
Similarly, using the definition of the second moment (30.50),
E[X^2 ]=∑
jx^2 jf(xj)=1^2 ×^12 p+2^2 p+3^2 p+4^2 p+5^2 p+6^2 × 2 p=
253
2
p=253
13
.
Finally, using the definition of the variance (30.48), withμ=53/13, we obtain
V[X]=∑
j(xj−μ)^2 f(xj)=(1−μ)^212 p+(2−μ)^2 p+(3−μ)^2 p+(4−μ)^2 p+(5−μ)^2 p+(6−μ)^22 p=(
3120
169
)
p=480
169
.
It is easy to verify thatV[X]=E
[
X^2
]
−(E[X])^2 .
In practice, to calculate the moments of a distribution it is often simpler to usethe moment generating function discussed in subsection 30.7.2. This is particularly
true for higher-order moments, where direct evaluation of the sum or integral in
(30.50) can be somewhat laborious.
30.5.5 Central momentsThe varianceV[X] is sometimes called thesecond central momentof the distribu-
tion, since it is defined as the sum or integral of the probability density function
multiplied by thesecondpower ofx−μ. The origin of the term ‘central’ is that by
subtractingμfromxbefore squaring we are considering the moment about the
mean of the distribution, rather than aboutx= 0. Thus thekthcentralmoment
of a distribution is defined as
νk≡E[
(X−μ)k]
={∑
j(xj−μ)kf(x
j) for a discrete distribution,
∫
(x−μ)kf(x)dx for a continuous distribution.(30.52)It is convenient to introduce the notationνkfor thekth central moment. Thus
V[X]≡ν 2 and we may write (30.51) asν 2 =μ 2 −μ^21. Clearly, the first central
moment of a distribution is always zero since, for example in the continuous case,
ν 1 =∫
(x−μ)f(x)dx=∫
xf(x)dx−μ∫
f(x)dx=μ−(μ×1) = 0.