MTH 202 : Probability and Statistics
Lecture 12 - 13 :
4 - 5 February, 2014
5. Moments of a Distribution Function
5.1 : Moments of a random variable of discrete type
Definition 5.1.1 : Let X be a random variable of the discrete type
with probability mass function
P (X = xk ) = pk , if k = 1, 2, . . .
P∞
If the sum k=1 |xk |pk < ∞, then we say that the expected value (or
the mean or the mathematical expectation) of X exists and write
∞
X
xk p k
µ = EX =
k=1
P
may be converRecall from the calculus course that a series ∞
k=1 yk P
∞
gent even when it is not absolutely convergent (i.e.
k=1 |yk | is not
convergent). A first such example you learned is the alternating sum
1 1 1 1 1
loge 2 = 1 − + − + − + . . .
2 3 4 5 6
while the absolute sum
1 1 1 1 1
1 + + + + + + ...
2 3 4 5 6
P∞
P∞
diverges. However, if
k=1 |yk | is convergent, then
k=1 yk is
convergent. For this reason, before we compute the expected value
EX, we need to check the convergence of the corresponding absolute
sum.
Example 5.1.2 : Let X have the PMF given by
3j
2
pj = P (X = (−1)j+1 . ) = j , j = 1, 2, 3, . . .
j
3
P∞
P
Then the absolute sum k=1 |xk |pk diverges, while the series ∞
k=1 xk pk
converges. However, according to the definition EX does not exist.
Example 5.1.3 : Find the expected number of throws of a fair die
until a 6 is obtained.
1
2
Solution : In every turn, the probability of getting 6 is 1/6 while
there are five undesired outcomes. We would be continuing until n-th
experiment provided all (n − 1) experiment results in the undesired
outcomes. Hence if pk denote the probability that you stop after n
throws, then we have the PMF
5 n−1 1
P (X = n) =
. , (n = 1, 2, 3, . . . )
6
6
Hence the expected value is
∞
5 n−1 1
X
EX =
n.
.
6
6
n=1
If p = 5/6, using the identity
1 + 2x + 3x2 + 4x3 + . . . =
1
if |x| < 1
(1 − x)2
we have
EX = (1 − p)
∞
X
npn−1 = (1 − p).
n=1
1
=6
(1 − p)2
We would now turn to continuous RVs.
Definition 5.1.4 : Let X be a random variable of the continuous type
with probability density function f . If the absolute improper integral
Z ∞
Z b
|t|f (t)dt = lim lim
|t|f (t)dt
−∞
a→−∞ b→∞
a
exists, we say that the expected value EX of X exists and
Z ∞
EX =
tf (t)dt
−∞
We would recall again that the improper integral
R∞
|t|f (t)dt < ∞.
−∞
R∞
−∞
tf (t)dt exists if
Example 5.1.5 : Consider an RV X with PDF is given by
(
e−x if x ≥ 0,
f (x) =
0
otherwise.
Find EX if it exists.
3
Solution : Since the PDF
R ∞ is non zero while x ≥ 0, the expectation
EX exists if and only if 0 te−t dt exists and
Z ∞
1+x
te−t dt = lim [−te−t − e−t ]xt=0 = − lim
EX =
x→∞
x→∞ ex
0
Using L’Hôpital’s rule we have
1
EX = − lim x = 0
x→∞ e
We will note some important properties of the expectation :
Prop. 5.1.6 : If X is bounded, then EX exists.
Proof : Let X : Ω → R is a bounded RV; i.e. there is a M > 0 such
that |X(ω)| ≤ M for all ω ∈ Ω. But this means :
P (X ≥ M + 1) = 0 = P (X ≤ −M − 1)
Case - I : X is discrete.
Suppose the PMF is given by
P (X = xk ) = pk , if k = 1, 2, . . .
Since the set T = {x1 , x2 , . . . } is discrete we have T ∩ [−M − 1, M + 1]
is finite (Why?)
P∞ which means pk 6= 0 only for finitely many k and hence
the sum k=1 |xk |pk is a finite sum.
Case - II : X is continuous.
Suppose the PDF is f . According to the fact above, F (x) = 0 if
x ≤ −M − 1. Hence differentiating F we have f (x) = 0 if x ≤ −M − 2.
Now
4
TO BE CONTINUED
References :
[RS] An Introduction to Probability and Statistics, V.K. Rohatgi and
A.K. Saleh, Second Edition, Wiley Students Edition.