Advanced High-School Mathematics

336 CHAPTER 6 Inferential Statistics

x^2



d

2

dx^2

(x+ 1)n



(x+ 1)N−n = n(n−1)x^2 (x+ 1)N−^2

= x^2

n(n−1)

N(N−1)

d^2

dx^2

(x+ 1)N.

Next comes the hard part (especially the first equality):

∑N

k=0

(∑n

m=0

m(m−1)

(

n

m

)(

N−n

k−m

))

=

∑n

m=0

(m−1)

(

n

m

)

xm·

N∑−n

p=0

(

N−n

p

)

xp

=

ñ

x^2 d

2

dx^2 (x+ 1)

n

ô

(x+ 1)N−n

= x^2 Nn((nN−−1)1) d

2

dx^2 (x+ 1)

N

= Nn((nN−−1)1)

∑N

k=0

k(k−1)

(

N

k

)

xk.

Just as we did at a similar juncture when computingE(X), we equate
the coefficients ofxk, which yields the equality

∑n

m=0

m(m−1)

Ñ

n

m

éÑ

N−n

k−m

é

=

n(n−1)k(k−1)

N(N−1)

Ñ

N

k

é

.

The left-hand sum separates into two sums; solving for the first sum
gives

∑n

m=0

m^2

Ñ

n

m

éÑ

N−n

k−m

é

=

n(n−1)k(k−1)

N(N−1)

Ñ

N

k

é

+

∑n

m=0

m

Ñ

n

m

éÑ

N−n

k−m

é

=

n(n−1)k(k−1)

N(N−1)

Ñ

N

k

é

+

nk

N

Ñ

N

k

é

,

which implies that

E(X^2 ) =

n(n−1)k(k−1)

N(N−1)

+

nk

N

.

Finally, from this we obtain the variance of the hypergeometric distri-
bution: