Download 4. MOMENTS 4.1 Expected values of discrete random variables

4. MOMENTS
4.1 Expected values of discrete random variables
The expected value (or the mean) of a discrete random
variable X is the sum of all its possible values multiplied by
their probabilities, i.e.,
EX = ∑ x ⋅ P(X = x) .
x∈X ( Ω )
Experiment: A coin is tossed twice.
Random variables: X … number of heads
Y … number of heads - number of tails
Expected value of X, Y, U= 13 X, V=7, and Z=X+Y:
EX=0.P(X=0)+1.P(X=1)+2.P(X=2)=0. 14 +1. 12 +2. 14 =1
EY=-2.P(Y=-2)+0.P(Y=0)+2.P(Y=2)=(-2). 14 +0. 12 +2. 14 =0
EU= 03 .P(U= 03 )+ 13 .P(U= 13 )+ 32 .P(U= 23 )
= 03 . 14 + 13 . 12 + 32 . 14 = 13 = 13 EX
EV=7⋅P(V=7)=7⋅1=7
EZ=(-2). 14 +1. 12 +4. 14 =1=EX+EY
Expected value rules: E(λX)=λEX
Eλ=λ
E(X+Y)=EX+EY
4.1
4.2 Expected values of continuous random variables
The expected value (or the mean) of a continuous random
variable X with probability density function f(x) is given by
∞
EX = ∫ xf(x)dx .
−∞
Since the expected value of X depends only on its density
function f, we will call it also the expected value of f.
• Exercise 4.a: Let X be a continuous random variable with
a standard uniform distribution, i.e., its density function is
given by
1 if x ∈ [0,1],
f(x)= 
0 if x ∉ [0,1].
Find the expected value of X.
∞
0
1
∞
−∞
−∞
0
0
1
∞
1
−∞
0
1
EX = ∫ xf(x)dx = ∫ xf(x)dx + ∫ xf(x)dx + ∫ xf(x)dx
= ∫ x ⋅ 0 dx + ∫ x ⋅ 1 dx + ∫ x ⋅ 0 dx
=
x2
0+ 2
1
0
+0 =
12
2
−
02
2
= 12
4.2
4.3 Variance and standard deviation
Let X be a random variable with expected value µ. The
variance of X is defined by
var(X)=E(X-µ)2
and is usually denoted by σ2. The variance is the expected
squared deviation from the mean and is a measure of
variability. Its square root is called the standard deviation
of X.
• Exercise 4.b: Let X be the number of heads when a coin
is tossed twice. Calculate the variance and the standard
deviation of X.
µ=EX=0.P(X=0)+1.P(X=1)+2.P(X=2)=0. 14 +1. 12 +2. 14 =1
Y=(X-µ)2=(X-1)2 ⇒ P(Y=0)=P(X=1)= 12
P(Y=1)=P(X=0 ∨ X=2)= 14 + 14 = 12
σ2=var(X)=E(X-µ)2=EY=0. 12 +1. 12 = 12
σ=
1
2
4.3
4.4 Alternative representation of the variance
• Exercise 4.c: Use the rules
(i) Eλ=λ
(ii) E(λX)=λEX
(iii) E(X+Y)=EX+EY
E(X-Y)=EX-EY
to show that
var(X)=EX2-µ2=E(X-µ)X.
var(X) =E(X-µ)2=E(X2-2µX+µ2)
=EX2-E2µX+Eµ2
using (iii)
=EX2-2µEX+Eµ2
using (ii)
=EX2-2µ2+Eµ2
=EX2-2µ2+µ2
using (i)
=EX2-µ2
var(X) =E(X-µ)2=E(X-µ)(X-µ)
=E[(X-µ)X-(X-µ)µ]
=E(X-µ)X-E(X-µ)µ
using (iii)
=E(X-µ)X-µE(X-µ)
using (ii)
=E(X-µ)X-µ(EX-Eµ) using (iii)
=E(X-µ)X-µ(EX-µ)
using (i)
=E(X-µ)X
4.4
4.5 Central and noncentral moments
Noncentral moments of the random variable X:
1st (noncentral) moment: EX=µ
2nd (noncentral) moment: EX2
(mean)
M
kth (noncentral) moment:
EXk
Central moments of the random variable X:
1st central moment:
2nd central moment:
E(X-µ)=EX-µ=0
E(X-µ)2=σ2
(variance)
M
kth central moment:
E(X-µ)k
4.5
4.6 Further exercises
• Exercise 4.d: Let X be a discrete random variable with
P(X=-1)=0.2, P(X=0)=0.3, P(X=1)=0.2, and P(X=3)=0.3.
Calculate the mean and the variance of X.
EX=-1⋅0.2+0⋅0.3+1⋅0.2+3⋅0.3=0.9
EX2=(-1)2⋅0.2+02⋅0.3+12⋅0.2+32⋅0.3=3.1
var(X)=EX2-(EX)2=3.1-0.81=2.29
4.6