Random Variables
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Transcript Random Variables
Random Variables
Probability Continued
Chapter 7
Part 2
Developing Transformation Rules
Consider the following distribution for
the random variable X:
X
1 2
P( X ) .2 .8
af af
1 18
. faf
.2 a
2 18
. faf
.8 .16
Var(X) = a
E( X) 1 .2 2 .8 18
.
2
2
X+1
What is the probability
distribution for X+1?
X
1
2
P( X )
.2
.8
X 1 2 3
P( X ) .2 .8
af
2 2.8faf
.2 a
3 2.8faf
.8
Var(X +1) = a
E(X +1) = 2 .2 3(.8) 2.8
2
E(X) = 1.8 Var(X) =.16
2
.16
2X
What is the probability
distribution for 2X?
X
1
2
P( X )
.2
.8
2 4
2X
P( X ) .2 .8
E(2X) = 2(.2) 4(.8) 3.6
Var(2X) = 2 3.6 .2 4 3.6 .8 .64
2
2
E(X) = 1.8 Var(X) =.16
Consider
Suppose that E(X) = 2.5, Var(X) = 0.2
What is E(X+5) = ?, Var(X+5) = ?
E(X+5) = 7.5, Var(X+5) = 0.2
What is E(X – 2.2) = ?, Var(X – 2.2) = ?
E(X – 2.2) = 0.3, Var(X – 2.2) = 0.2
What is E(3X) = ?, Var(3X) = ?
E(3X) = 7.5, Var(3X) = 1.8
What is E(2X – 1) = ?, Var(2X – 1) = ?
E(2X – 1) = 4, Var(2X – 1) = 0.8
Rule 1
Rule 1: If X is a random variable and a
and b are fixed numbers, then
ma + bX = a + bmX
Rule 1: If X is a random variable and a
and b are fixed numbers, then
s2a + bX =b2s2X
X
1
2
P( X )
.2
.8
X+X
3
4
X+X 2
What is the probability P( X ) .04 .32 .64
distribution for X+X?
.2
.8
1
.2
1
.8
2
.2
1
2
.8
2
X+X
X
1
2
P( X )
.2
.8
X+X 2
3
4
P( X ) .04 .32 .64
af
E(X + X) = 2 .04 3(.32) 4(.64) 3.6
Var(X + X) =
2 3.6 .04 3 3.6 .32 4 3.6 .64 .32
2
2
2
X
1
2
P( X )
.2
.8
X–X
0
1
X X 1
What is the probability P( X ) .16 .68 .16
distribution for X–X ?
.2
.8
1
.2
1
.8
2
.2
1
2
.8
2
X–X
XX
P( X )
1 0
1
.16 .68 .16
E(X X) = 1(.16) 0(.68) 1(.16) 0
Var(X X) =
a1 0fa.16f a0 0fa.68f a1 0fa.16f .32
2
2
2
Consider
Suppose that E(X) = 2.5, Var(X) = .16,
E(Y) = 1.2, Var(Y) = .36
What is E(X+Y) = ?, Var(X+Y) = ?
E(X+Y) = 3.7, Var(X+Y) = .52
E(X – Y) = ?, Var(X – Y) = ?
E(X – Y) = 1.3, Var(X – Y) = .52
What is s(X) = ?, s(Y) = ?, s(X+Y) = ?
s(X) = .4, s(Y) = .6, s(X+Y) = .7211
Rule 2
Rule 2: If X and Y are random variables,
then
mX + Y = mX + mY
mX – Y = mX – mY
Rule 2: If X and Y are independent
random variables, then
s2X + Y = s2X + s2Y
s2X Y = s2X + s2Y
Note: cannot combine standard deviation
directly.