derivation of the mean of geometric distributon
Answers
Answer:
Derivation of the Mean and Variance of a Geometric Random Variable
Brett Presnell
Suppose that Y ∼ Geom(p), i.e., Y has probability function
p(y) = q
y−1
p, y = 1, 2, . . . .
Here is an alternative to the derivation of the mean of Y given in the course textbook. It basically
depends on the simple trick of writing y =
Py
k=1 1 and exchanging the order of summation.
E(Y ) = X
y
yp(y) = X∞
y=1
yqy−1
p = p
X∞
y=1
yqy−1
= p
X∞
y=1 " X
y
k=1
1
!
q
y−1
#
= p
X∞
y=1 X
y
k=1
q
y−1
!
(the trick)
= p
XX
1≤k≤y<∞
q
y−1 = p
X∞
k=1
X∞
y=k
q
y−1
(exchange order of summation)
= p
X∞
k=1
X∞
y=k
q
k−1
q
y−k
= p
X∞
k=1
q
k−1
X∞
y=k
q
y−k
= p
X∞
k=1
q
k−1
X∞
j=0
q
j
(j = y − k)
= p
X∞
k=1
q
k−1
1
1 − q
= ✁
p
1
✁
p
X∞
k=1
q
k−1
=
X∞
j=0
q
j
(j = k − 1)
=
1
1 − q
=
1
p
.
Remark. The same trick can actually be used to show that if Y is a nonnegative integer-valued
random variable, then
E(Y ) = X∞
y=0
P(Y > y).
The derivation above for the case of a geometric random variable is just a special case of this. In
fact, if Y ∼ Geom(p), then P(Y > y) = q
y
for y = 0, 1, 2, . . . (see Exercise 3.71(a) in the course
textbook), and thus
E(Y ) = X∞
y=0
q
y =
1
1 − q
=
1
p
.
1
For the variance of Y , we proceed similarly. Here the trick is to notice that
Xn
i=1
i =
n(n + 1)
2
=⇒
X
y−1
k=1
k =
(y − 1)y
2
=⇒ y(y − 1) = 2X
y−1
k=1
k.
Using this we have
E[Y (Y − 1)] = X
y y(y − 1)p(y) = X∞
y=1
y(y − 1)q
y−1
p
= p
X∞
y=2
y(y − 1)q
y−1
(term w/ y = 1 equals 0)
= p
X∞
y=1 " 2
X
y−1
k=1
k
!
q
y−1
#
= 2p
X∞
y=1 X
y−1
k=1
kqy−1
!
(the trick)
= 2p
XX
1≤k<y<∞
kqy−1 = 2p
X∞
k=1
X∞
y=k+1
kqy−1
(exchange order of summation)
= 2p
X∞
k=1
kqk
X∞
y=k+1
q
y−k−1
= 2p
X∞
k=1
kqk
X∞
j=0
q
j
(j = y-k-1)
= 2p
X∞
k=1
kqk
1
1 − q
= 2✁
p
1
✁
p
X∞
k=1
kqk = 2X∞
k=1
kqk
=
2q
p
X∞
k=1
kqk−1
p =
2q
p
· E(Y ) = 2q
p
·
1
p
=
2q
p
2
.
Thus,
E(Y
2
) = E[Y (Y − 1)] + E(Y ) = 2q
p
2
+
1
p
=
2q + p
p
2
,
and hence
Var(Y ) = E(Y
2
) − [E(Y )]2 =
2q + p
p
2
−
1
p
2
=
2q + p
p
2
−
1
p
2
=
2q − (1 − p)
p
2
=
2q − q
p
2
=
q
p
2
.
Step-by-step explanation: