Download Lecture 6, Sep 15.

Math/Stat 352
Lecture 6
Section 2.5
Linear Functions of Random Variables
Linear Functions of Random Variables
If X is a random variable, and a and b are constants, then
Y= aX+b is a linear function of X. Y is also a random variable, and
aµ X + b
µaX=
+b
2
2 2
a
σ aX
=
σX
+b
σ aX +b = a σ X
2
More Linear Functions
If X and Y are random variables, and a and b are constants, then
µaX +bY = µaX + µbY = aµ X + bµY .
More generally, if X1, …, Xn are random variables and c1, …, cn are
constants, then the mean of the linear combination c1 X1+…+cn Xn is given
by
µc X + c X
1
3
1
2
= c1µ X1 + c2 µ X 2 + ... + cn µ X n .
2 +...+ cn X n
Two Independent Random Variables
If X and Y are independent random variables, and S and T are sets of
numbers, then
P ( X ∈ S and Y ∈ T )= P ( X ∈ S ) P (Y ∈ T ).
More generally, if X1, …, Xn are independent random variables, and S1, …,
Sn are sets, then
𝑷 𝑿𝟏 ∈ 𝑺𝟏 , 𝑿𝟐 ∈ 𝑺𝟐 , … , 𝑿𝒏 ∈ 𝑺𝒏 = 𝑷(𝑿𝟏 ∈ 𝑺𝟏 )P(𝑿𝟐 ∈ 𝑺𝟐 ) ⋯ 𝑷(𝑿𝒏 ∈ 𝑺𝒏 ).
4
Variance Properties
If X1, …, Xn are independent random variables, then the variance of the
sum X 1 +  + X n is given by
σ X2 + X
1
2 +...+ X n
= σ X2 1 + σ X2 2 + .... + σ X2 n .
If X1, …, Xn are independent random variables and c1, …, cn are constants,
then the variance of the linear combination c1 X 1 +  + cn X n is given
by
2
2 2
2 2
2 2
σ c X +c X
1 1
2
= c1 σ X1 + c2σ X 2 + .... + cnσ X n .
2 +...+ cn X n
If X and Y are independent random variables with variances σ X and σ Y ,
2
2
then the variance of the sum X + Y is σ X2 =
σ
+
σ
+Y
X
Y.
2
The variance of the difference X – Y is
5
2
2
σ X2 =
σ
+
σ
−Y
X
Y.
2
Example
A piston is placed inside a cylinder. The clearance is the distance between
the edge of the piston and the wall of the cylinder and is equal to one-half the
difference between the cylinder diameter and the piston diameter. Assume
the piston diameter has a mean of 80.85 cm with a standard deviation of 0.02
cm. Assume the cylinder diameter has a mean of 80.95 cm with a standard
deviation of 0.03 cm. Find the mean clearance. Assuming that the piston
and cylinder are chosen independently, find the standard deviation of the
clearance.
Sln: Let X=piston diameter, EX=80.85 cm, St. DevX=0.02 cm.
Y=cylinder diameter, EY=80.95, St. dev. Y= 0.03cm. X and Y are independent.
C=clearance= 0.5(Y – X).
EC= 0.5(EY – EX) = 0.5(80.95 – 80.85)=0.05cm = average clearance.
Var(C)=Var(0.5(Y – X))= (0.5)2(VarX + VarY)= (0.25)(0.022 + 0.032)= 0.000325.
St. Dev(C)= 𝟎. 𝟎𝟎𝟎𝟎𝟎𝟎 = 𝟎. 𝟎𝟎𝟎 cm.
6
Independence and Simple Random Samples
Definition: If X1, …, Xn is a simple random sample, then X1, …, Xn may be
treated as independent random variables, all with the same
distribution.
When X1, …, Xn are independent random variables, all with the same
distribution, we sometimes say that X1, …, Xn are independent and
identically distributed (i.i.d).
7
Properties of
X
If X1, …, Xn is a simple random sample from a population with mean µ and
variance σ2, then the sample mean
and variance
8
σ =
2
X
σ2
n
.
X is a random variable with µ X = µ
The standard deviation of
X
is
σX =
σ
n
.
Example
A process that fills plastic bottles with a beverage has a mean fill volume
of 2.013 L and a standard deviation of 0.005 L. A case contains 24
bottles. Assuming that the bottles in a case are a simple random sample
of bottles filled by this method, find the mean and standard deviation of
the average volume per bottle in a case.
Soln: X1, …, X24 volume of beverage in the 24 bottles. Xi’s have the same
distribution and are independent.
EXi =2.013 L, and st.dev Xi=0.005L.
Average volume per bottle =
st.dev
9
X = (st.devXi)/
X
,E
X =E Xi = 2.013L,
𝒏 = 0.005/ 𝟐𝟐 = 𝟎. 𝟎𝟎𝟎 𝑳