Advanced High-School Mathematics

SECTION 6.1 Discrete Random Variables 335

E(X) =

max∑{n,k}

m=0

m

Än

m

äÄN−n

k−m

ä

ÄN

k

ä.

We can calculate the above using simple differential calculus. We
note first that

[d

dx

(x+ 1)n

]

(x+ 1)N−n=n(x+ 1)N−^1 =

n

N

d

dx

(x+ 1)N.

Now watch this:

∑N

k=0

(∑n

m=0

m

(

k

m

)(

N−n

k−m

))

xk =

∑n

m=0

m

(

n

m

)

xm·

N∑−n

p=0

(

N−n

p

)

xp

Ç

this takes

some thought!

å

=

ñ

xdxd(x+ 1)n

ô

(x+ 1)N−n

= xn

N

d

dx

(x+ 1)N

= Nn

∑N

k=0

k

(

N

k

)

xk;

equating the coefficients ofxkyields

∑n

m=0

m

Ñ

n

m

éÑ

N−n

k−m

é

=

nk

N

Ñ

N

k

é

.

This immediately implies that the mean of the hypergeometric distri-
bution is given by

E(X) =

nk

N

.

Turning to the variance, we have

E(X^2 ) =

∑n

m=0

m^2

Än

m

äÄN−n

k−m

ä

ÄN

k

ä.

Next, we observe that