statistical methods in transport analysis

8/2/2019 Statistical Methods in Transport Analysis

1/20

Click to edit Master subtitle style

3/5/12

Statistical Methods inTransport Analysis

Chapter-0 4

MATHAMATICALEXPECTATION


2/20

3/5/12

04.1 _Expected value

Expected Value is another word formean or average.

It gives a general impression of thebehavior of some random variablewithout giving full details of its

probability distribution or probabilitydensity function

Two random variables with the same

expected value can have very different


3/20

3/5/12

Expected value cont

Mean of expectation

Expected value of X

= E(x) = x f(x) - discrete

-

x


4/20

3/5/12

Expected value of X cont

Example_01: Tossing a die (discretevariable)

= E(x) = x f(x)

= 1 x 1/6 + 2 x 1/6 + 3 x /6 + 4 x 1/6 + 5 x 1/6 +6 x 1/6

= 3.5

x 1 2 3 4 5 6

P(X=x)

1/6 1/6 1/6 1/6 1/6 1/6


5/20

3/5/12

Example_02: A carnival game

consists of tossing a dart, which lands

at a random spot within the square

target to the right. The red circle wins

$5, the blue circle wins Rs 7 and either

black triangle wins Rs 10. If it costs Rs 5 to playthis game, do you expect to make money or losemoney?

Let R be the payoff:

Value of R Probability Product

5 /16 0.98175

7 /16 1.3744510 1/4 2.50

Total 4.8562


6/20

3/5/12

Example_03:(Infinite Sum Expected Value)Toss a coin until heads appears, and let R be thenumber of tosses required. What is the expected

value of R?

Let S = 1/2 + 2/4 + 3/8 + 4/16 + 5/32 + ...

Value of R Probability Product

1 1/2 1/2

2 1/4 2/4

3 1/8 3/8

4 1/16 4/165 1/32 5/32

n 1/2n n/2n


7/203/5/12

Mean of expectation

cont Expected value of a function, g(x)

g(x) = E[g(x)] = g(x)f(x) -discrete

g(x) = E[g(x)] = g(x)f(x) -continuous

-

x


8/203/5/12

Expected value of a

function,g(x) cont

Example_02: The total distance measured in unitsof 100 km, that private bus operates during a week isa continuous random variable X that has the density

function,x 0


9/203/5/12

Mean of expectation

cont Expected value of joint

distribution, g(x,y)

= E[g(x,y)] = g(x,y) f(x,y)

= E[g(x,y)] = g(x,y)

f(x,y)dxdy

x y

-

-


10/203/5/12

Expected value of jointdistribution, g(x,y) cont

F(x,y) = 1/50 (x2+y2) 0


11/203/5/12

Properties of Expectations

E(ax+b) = a E(x) + b

= (ax+b)f(x)dx

= a xf(x)dx + b f(x)dx

= a E(x) + b

-

-

-


12/203/5/12

Properties of Expectations

E(ax+b) = a E(x) + b

Expectation of two functions

E[g(x) +h(x)] = E[g(x)] + E[h(x)]

E[g(x) - h(x)] = E[g(x)] - E[h(x)]

For two random variables

E[g(x,y) +h(x,y)] = E[g(x,y)] + E[h(x,y)]

E[g(x,y) - h(x,y)] = E[g(x,y)] - E[h(x,y)]

If x and y are independent,


13/203/5/12

04.2 _ Variance andCovariance

Although the mean or expected

value describes where the

probability distribution is centered,the variance; non-negative number

which gives an idea of how widely

spread the values of the random

variable are likely to be; the larger

the variance, the more scattered the


14/203/5/12

Variance and Covariancecont

Notes

The larger the variance, the further that

individual values of the random variable(observations) tend to be from the mean.

The smaller the variance, the closer thatindividual values of the random variable

(observations) tend to be to the mean. Taking the square root of the variance

gives the standard deviation, S.

The variance and standard deviation of arandom variable are alwa s non-ne ative


15/203/5/12

Variance and Covariancecont

2 =E[(x- ) 2]

= (x- ) 2f(x) - discrete

= (x- ) 2f(x) dx - continuous

Variance of g(x) = E[(g(x) - g(x))2]

= [g(x) - g(x)] 2f(x) -discrete

= [g(x) - g(x)] 2f(x)dx -

continuous

-

-


16/203/5/12

Properties of Variance

2 = E[x2] 2

= E[(x- ) 2]

= E[(x2 2x + 2 )]

= E[x2] 2 E[x] + 2

= E[x2] 2 2+ 2 = E[x2] 2

2ax = a2 2x

2ax+b = 2ax


17/203/5/12

Coefficient of Variance

Coefficient of variance =

, have equal units

So the coefficient of variance is adimensionless value

Higher coefficient of variance largespread

Lower coefficient of variancenegligible spread


18/20

3/5/12

Example_03: The probabilitydistribution of the number of cars

arriving at a toll booth during anyminute is as follows. Calculate the,

Expected value? variance? coefficient of variance?


19/20

3/5/12

= x P(x) = 0.99

E(x2) = x2P(x) = 1.95

2 = E[x2] 2= 1.95 0.992

= 0.97 = 0.984

/ = 0.984/0.99 = 0.994 (Large

X P(x) xP(x) x2P(x) (x-)2P(x)

0 0.37 0 0 0.3621 0.37 0.37 0.37 0

2 0.18 0.36 0.72 0.183

3 0.06 0.18 0.54 0.242

4 0.02 0.08 0.32 0.181

sum 1.00 0.99 1.95 0.968


20/20

3/5/12

THANK YOU.. . . !

statistical methods in transport analysis

Documents