Lecture (10)Lecture (10)

Mathematical Expectation Mathematical Expectation

Mathematical Expectation Mathematical Expectation

{ } . ( )E z z p z dz

2 2{ } . ( )E p z dz z z

22

2

{ }

- { } ( )

z E z E z

z E z p z dz

1

{ } . ( )n

i ii

E z z p z

22

1

{ } . ( )n

i ii

E z z p z

2 2

1

( { }) ( )n

z i ii

z E z p z

The expected value of a variable is the value of a descriptor when averaged over a large number theoretically infinite.

Mathematical Expectation Mathematical Expectation (cont.) (cont.)

2 2

1

22 2

1

22 2

1 1 1

22 2

1 1 1

2 2 2

( { }) ( )

( { } 2[ . { }]) ( )

( ) { } ( ) 2 [ . { }] ( )

( ) { } ( ) 2 { } ( )

{ } { } ( )

n

z i ii

n

z i i ii

n n n

z i i i i ii i i

n n n

z i i i i ii i i

z i

z E z p z

z E z z E z p z

z p z E z p z z E z p z

z p z E z p z E z z p z

E z E z p z

1

2 2 2

2 { }. { }

{ } { }

n

i

z

E z E z

E z E z

Another way to compute the variance

Example 1Example 1

TTTT

THTH

HTHT

HHHH

00

11

22

Sample Space Number of Heads

Example 1 (cont.)Example 1 (cont.)

OUTCOME X

PROBABILITY P(X)

0 1/4

1 2/4=1/2

2 1/4

Example 1 (cont.)Example 1 (cont.)

3210

1

0.5

.25

NUMBER OF HEADS

PR

OB

AB

ILIT

Y

Experiment: Toss Two Coins

Example 1 (cont)Example 1 (cont)

• E.G. Toss 2 coins, count heads, compute expected value:

• = 0 .25 + 1 .50 + 2 .25 = 1

E.G. Toss 2 coins, count heads, compute variance: variance = (0 - 1)2 (.25) + (1 - 1)2 (.50) + (2 - 1)2(.25) = .50

Example 2Example 2

1

2 2 2

1

2 2

2 2 2 2

1, 1, 2,3,.....,

( )0

{ } . ( )

1 1 11. 2. ...

1 ( 1) 1(1 2 ... )

2 2

. ( ) { }

1 1 1 1(1. 4. ... ) ( )

21 1 1

(1 2 ... ) ( 1 2 ) ( 1)4 12

n

k

n

xk

k np k n

E x k p k

nn n n

n n nn

n n

k p k E x

nn

n n n

n n n nn

• Find the mean of the number of spots that appear when a die is tossed. The probability distribution is given below.

Discrete Uniform Distribution Example

X 1 2 3 4 5 6

P(X) 1/6 1/6 1/6 1/6 1/6 1/6

X 1 2 3 4 5 6

P(X) 1/6 1/6 1/6 1/6 1/6 1/6

X P X( )

( / ) ( / ) ( / ) ( / )

( / ) ( / )

/ .

1 1 6 2 1 6 3 1 6 4 1 6

5 1 6 6 1 6

21 6 3 5

That is, when a die is tossed many times, the theoretical mean will be 3.5.

That is, when a die is tossed many times, the theoretical mean will be 3.5.

Discrete Uniform Distribution Example (cont.)

Binomial Distribution -Binomial Distribution - Example

• A coin is tossed four times. Find the mean, variance and standard deviation of the number of heads that will be obtained.

• Solution:Solution: n = 4, p = 1/2 and q = 1/2. = np = (4)(1/2) = 2. 2 = npq = (4)(1/2)(1/2) = 1. = = 1.1

Poisson Distribution

-

0

-

1

1-

1

-

0

-

e

!

e

.( 1)!

e ( 1)!

e !

e .

i

i

i

i

i

i

k

k

E x ii

ii i

i

k

e

Uniform Distribution Example Uniform Distribution Example

f(x) or F(x)

xa b

1.0

0

F(x)

f(x)

(a-b)

1

2 2

2 2

23 3

2 2 22

1( )

{ } . ( )

1( )

2( ) 2

{ } . ( )

1( )

3( )

1{ } { } ( )

12

b

a

b

a

x

f xb a

E x x f x dx

xdx

b a

a bb a

b a

E f x dxx x

x dx b a

b a b a

E x E x b a

If the probability density function has the formf(x) = ax for a random variable X between 0 and 2.(a)Find the value of a.(b) Find the median of X(c)Find P(1.0 < X < 2.0)

Solution: (a) From the area under the whole density curve is 1,then we have

21

1212221 aaa

7071.022

21

21

221

so 0.5, isleft its toarea the, uemedian val At the (b)

2 mmmm

m

xxxx

x

75.00.221

0.121

21

2 to1 from curvedensity under the area the)0.20.1( (c)

XP

Example

QuizQuiz

22

exp( ), 0( )

0 otherwise

1{ }

1

x xf x

E x

0

1

0

F(x

) o

r f(

x)

x

F(x)

f(x)

Exponential Distribution

0

0

0

0

0

1/

x

x x

x

E x x e dx

xe e dx

e

Distribution Normal

x

LogN

Y =logx

Gamma

x

Exp

t

Mean x y nk 1/k

Variance x y

nk 1/k2

Skewness zero zero 2/n0.5 2

Comparison of Parameters of Dist’n