Density Curve. Continuous Distributions. Continuous Distribution. Density Curve. Meaning of Area Under Curve. Meaning of Area Under Curve

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1

Continuous Distributions

1 Random Variables of the

Continuous Type

2

Density Curve

Percent

A smooth curve that fit the distribution 10 20 30 40 50 60 70 80 90 100 110 Test scores Density function, f (x) f(x) 3

Density Curve

A smooth curve that fit the distribution

Percent

Use a mathematical model to describe the variable.

10 20 30 40 50 60 70 80 90 100 110

Test scores

Shows the probability density for all values of x.

Probability density is not probability! Probability Density Function, f (x) f(x) 4

Continuous Distribution

1. f (x)  0, xS,

2.S f(x) dx = 1, (Total area under curve is 1.)

3. For a and b in S,

Probability Density Function (p.d.f.)of a random variable X of continuous type with a space S is an integrable function, f(x), that satisfying the following conditions:



   b a f xdx b X a P( ) ( ) f(x) x a b 5

Meaning of Area Under Curve

Example: What percentage of the

distribution is in between 72 and 86?

f(x)

X (Height) 72 86

P(X = 72) =

P(72 X  86)

0, density is not probability. 6

Meaning of Area Under Curve

Example: What percentage of the

distribution is in between 72 and 86?

f(x)

X (Height) 72 86

P(72  X  86) =P(72 < X < 86) =P(72  X <86) =P(72 < X  86)

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7 4 / 1 0 2 2 8x  1/4 0 2 4x        elsewhere , 0 0.5 0 for , 8 ) (x x x f 0.5 f (x) = 8x is a line with a slope 8 and passes through (0,0)

0 x

f (x) = y

Example:

If the density function of a continuous distribution is

Find the proportion of values in this distribution that is less than 1/4.

= 4(1/4)2_{– 4(0)}2_{= 4/16 = 1/4.}



1/4 0 8 xdx The area under the f (x) between 0 and 1/4 =

8

Review of Calculus

c x n dx xn n    



1 1 1 c e a adx du ax u du a e dx e u u ax     

_



1 , 1



e2xdx e2xc 2 1



      c x c x dx x3 31 4 4 1 1 3 1 9 b x b e ₃ 5 / ) ( lim      0 for , 5 1 ) (_x  _e5 _x f x Example:

If the density function of a continuous distribution X, waiting time between arrivals of cars at a intersection, is



 3 5 5 1 e dx x

-The area under the f (x) and x > 3 =

5488 . 0 ) 5488 . 0 ( 0 ) ( ) ( lim_ /5 __ 3/5 _ _ _ _      e e b b

Find the probability that the waiting time (in seconds) till the next arrival of car at this intersection is more than 3 seconds.

3 0 x f (x) = y .2 10 3 0 5 / ₎ (__ex  0 for , 5 1 ) (_x  _e5 _x f x Example:

If the density function of a continuous distribution X, waiting time between arrivals of cars at a intersection, is



3 0 5 5 1 e dx x

-The area under the f (x) below 3 =

4522 . 0 5488 . 0 1 1 ) ( ) (_ 3/5 __ 0/5 _ _ 3/5_ _ _     e e e

Find the probability that the waiting time (in seconds) till the next arrival of car at this intersection is less than 3 seconds.

3 0 x f (x) = y .2 11

Cumulative Distribution Function

The cumulative distribution function (c.d.f. or distribution function, d.f.) of a continuous random variable is defined as











x

dt

t

f

x

X

P

x

F

(

)

(

)

(

)

• F( ) = 0, F() = 1 • P(a < X < b) = F(b) – F(a) • F′(x) = f(x), if derivative exists 12   / 0 / 1 ) ( t x x e e      0 for , 1 ) (x  e x f x   Example:

If the p.d.f. of a continuous distribution X, waiting time between arrivals of cars at a intersection, is, where  is a constant parameter of this distribution,



  x dt t f x F( ) ()

Find the distribution function.



 x -t dt e 0 1               _ x e x x F _x 0 , 1 0 , 0 ) ( _/_

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13

Measure of Center for a

Continuous Distribution

The mean value (expected value) of a continuous random variable (distribution)

X, denoted by _Xor just(or E[X]) is defined as









x

f

(

x

)

dx

X



14

Measure of Spread for a

Continuous Distribution

The variance of a continuous random variable (distribution) X, denoted by _X2 or just2₍_{or Var}_[_X_{]) is defined as}



        2 2 2 ) ( ) ( ] ) [(X x f xdx E   

The standard deviation of X is ] ) [( 2 2     E X 15

Moment Generating Function

for a Continuous Distribution

The moment generating function, if exists, of a continuous random variable (distribution) X, denoted by M(t) is defined as

h

t

h

dx

x

f

e

E

t

M



tX



tx









(

)

,

]

[

)

(

16       elsewhere , 0 0.5 0 for , 8 ) (x x x f Example:

Find the mean and variance of this distribution

3 1 ) 0 125 (. 3 8 3 8 8 5 . 0 0 3 0.5 0      

_

x x dx x The mean is



        0.5 0 ( ) ) ( ) (X x f x dx x f x dx E  17       elsewhere , 0 0.5 0 for , 8 ) (x x x f Example:

Find the mean and variance of this distribution

The variance is 2 2 2 ]) [ ( ] [ ) (X E X E X Var     8 1 ) 0 0625 (. 4 8 4 8 8 ) ( ) ( ] [ 5 . 0 0 4 0.5 0 2 0.5 0 2 2 2          



   x x dx x dx x f x dx x f x X E 72 1 3 1 8 1 ]) [ ( ] [ 2 2 2 2 _ _         EX EX  M ″(0) M ′(0)2 18

The Percentile

The (100p)th percentile (quantile of order p) is the number p_psuch that the area under f(x) to the left of p_pis p.

)

(

)

(

x

dx

F

_p

f

p





pp



p

  pp p

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Median of a distribution Example:

Find the median of this distribution.       elsewhere , 0 0.5 0 for , 8 ) (x x x f 0.5 x 0 f (x) = y 20 Median is c such that

What is c?

5

8

0

x

dx

.

c





354 . 8 1 4 / 5 . 5 . 0 4 5 . 2 8 8 2 2 0 2 0          



c c c x dx x c c 21 Percentile Example:

Find the 25th_{percentile of this distribution.}

      elsewhere , 0 0.5 0 for , 8 ) (x x x f 0.5 0 x f (x) = y 22 25th_{percentile is}_c_{such that}

What is c?

25 .

8

0





c

dx

x

25 . 16 1 4 / 25 . 25 . 0 4 25 . 2 8 8 2 2 0 2 0          



c c c x dx x c c 23

Sample Quantile

Let y₁ y₂...y_n be the order statistics associated with the sample x₁, x₂, ···, x_n, then

y_r is called the sample quantile of order

as well as percentile. 1  n r 1 100  n r Example: 1, 3, 7, 8, 9, 13

n = 6, and the value 8, y4 ,

is the [4/(6+1)]th quantile of the distribution of the sample, i.e. 0.571 sample quantile or 57.1 percentile.

n r.5 25 . 375 .   n r Other options: 24

Examine Distribution with

Quantile-Quantile Plot

Good model for the observations

Not a good model for the observations Theoretical Distribution, pp Sample, yr Sample, yr Theoretical Distribution, pp

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Continuous Distributions

2 The Uniform and Exponential

Distributions

26

Uniform Distribution

The continuous random variable X has a uniform distribution if its p.d.f. is equal to a constant on its support. If the support is the interval [a, b], then its p.d.f. is

.

,

1 )

(

a

x

b

a

b

x

f







It is usually denoted as U(a, b). a b a b 1 f(x) 0 a b F(x) 1 0 27

Uniform Distribution

The mean, variance, and m.g.f. of a continuous random variable X that has a uniform distribution are:

             . 0 , 1 , 0 , ) ( ) ( , 12 ) ( , 2 2 2 t t a b t e e t M a b b a ta tb  

Pseudo-Random Number Generator on most computers U(0, 1)

28

Uniform Distribution

Example: Let X be U(0,10), find the mean and the variance, and the m.g.f. of X.

              . 0 , 1 , 0 , 10 1 ) ( , 3 25 12 ) 0 10 ( , 5 2 10 0 10 2 2 t t t e t M t   29

Exponential Distribution

The continuous random variable X has an exponential distribution if its p.d.f. is

    _   elsewhere , 0 0 for , 1 ) ( / _x x x f x 

where  is the mean of the distribution. * X can be the waiting time until next success in a Poisson process.

30

Exponential Distribution

The mean, variance, and m.g.f. of a continuous random variable X that has an exponential distribution are:

      1 , 1 1 ) ( , , 2 2      t t t M             _ x e x x F _x 0 , 1 0 , 0 ) ( _/_ c.d.f. 

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Exponential Distribution

Example: Let X have an exponential distribution with a mean of 30, what is the first quartile of this distribution?             _ x e x x F _x 0 , 1 0 , 0 ) ( _/₃₀ p_.25/ = ln(.75) p_.25 = ·ln(.75) p.25 =  30·ln(.75) {= 30} p.25 =8.63 75 . 25 . 1 ) ( .25 / / 2 5 . 2 5 .      p  p  p e e F pp =  · ln(1 – p) 32

Exponential Distribution

Let W be the waiting time until next success in a Poisson process in which the average number of success in unit interval is l, then, for w 0

F(w) = P(Ww) = 1 – P(W > w) = 1 – P(W > w) = 1 – P(nosuccess in[0,w]) = 1 – e-lw_{d.f. of exponential distribution.} F (w) = f(w) = le-lw _{p.d.f. of exponential distribution.}   / 1 x e 33

Exponential Distribution

Let W be the waiting time until next success in a Poisson process in which the average number of success in unit interval is l, then, l = 1/, where  is the average waiting time until next success, for w 0 F(w) = 1 – e-lw₌_{1 –}_e-w/ f(w) = p.d.f. of exponential distribution.  / 1_ew d.f. of exponential distribution. 34

Exponential Distribution

Suppose that number of arrivals of

customers follows a Poisson process with a mean of 10 per hour. What is the probability that the next customer will arrive within 15 minutes? (15 min. = 0.25 hour)

l= 10  = 0.1 P(X  x) = F (x) = 1 – e-x/ P(X 0.25) = F (0.25) = 1 – e-0.25/ F (0.25) = 1 – e-0.25/0.1₌_.918 35

Continuous Distributions

3 The Gamma and Chi-square

Distributions

36

Gamma Distribution

The continuous random variable X has a Gamma distribution,Gamma(a, ). if its p.d.f. is     _     elsewhere. , 0 0 for , ) ( 1 ) ( / 1_e _x x x f x a a  a

* X can be the waiting time until ath success in a Poisson process.

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37

Graphs of Gamma Distributions

a = 1/4 a = 1 a = 2 a = 3 _a = 4  = 5/6  = 2  = 4  = 4 a = 4  = 1  = 3 0 10 20 30 0 10 20 30 38

Gamma Distribution

The mean, variance, and m.g.f. of a continuous random variable X that has an Gamma distribution are:

  a  a  a 1 , ) 1 ( 1 ) ( , , 2 2      t t t M

•Gamma(1,  )  Exponential Distribution •Gamma(2, r/2 )  Ch-square with d.f. = r.

39 40

Special Notation

c

2 a

(

r

)

c2 a(r) Probability of X greater than or equal to c2_a₍_r₎_isa.

a

c

_a





_

(

)]

[

2 1

r

X

P

Example: Findc2 .05(3) = ? Tail area

a

c

a





(

)]

[

X

2

r

P

Let X be a random that has Chi-square distribution with degrees of freedom r, 7.815 41

Try This!

• Findc2 .01(7) = ? • Findc2 .025(4) = ?

• Find the 10th_{percentile from a}c2 distribution with degrees of freedom 6.

42

Continuous Distributions

4 Distributions of Functions of

Random Variable

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43

A Good Generator

• Simple

• Widespread use

• Long cycle length 231_{– 2 (all number} besides 1 and 231_{– 1 can be generated.)} • U_i= X_i/(231_{– 1),}_U

i ~ U(0,1)

X_i = (397,204,094X_i-₁) mod (231_{– 1)}

Starts with a Seed number

44

Methods for Generating

Non-Uniform RN’s

• CDF Inversion • Transformations • Accept/Reject Methods • … 45

CDF Inversion

Theorem: Let U have a distribution that is

U(0,1). Let F(x) have the properties of a distribution function of the continuous type with

F(a) = 0 and F(b) = 1, and suppose that F(x) is strictly increasing on the support a < x < b, where a and b could be  and ,

respectively. Then the random variable X

defined by X = F 1₍_U_{) is a continuous} random variable with distribution function F(x).

46

Proof:

Let X = F 1₍_U_{) , the distribution function of}_X_is

P (Xx ) = P [ F1₍_U₎_x_],_a_<_x_<_{b .}

Since F (x) is strictly increasing, { F1₍_U₎_x_{} is}

equivalent to { UF (x) } and hence

P ( X x ) = P [ U F (x) ], a < x < b .

But U is U(0, 1); so P ( U u ) = u for 0 < u < 1, and accordingly,

P ( X x ) = P [ UF (x) ] = F (x) , 0  F(x) < 1.

That is the distribution function of X is F(x).

47

Generate random numbers from

Exponential distribution,



= 10

• F(x) = 1 ex/10_and_{x =F}1₍_y_{) =}_10·_ln₍₁_y₎

• Use uniform U(0, 1) random number generator to generate random numbers y1, y2, ..., yn .

• The exponentially distributed random numbers xi's

would be xi= 10·ln(1yi) , i = 1, ..., n.

• Therefore, if the uniform random number generator generates a number 0.1514, then 1.6417 = 10·ln(1 0.1514) would be a random observation from exponential distribution with  = 10.

48

Try this!!!

1. Use the CDF Inversion method to convert this

U(0,1) random number to simulate an observation from an exponential distribution with  = 12.

xi= F1(yi) = 12·ln(1yi)

2. Use CDF Inversion method to generate a random number from the continuous random variable that has the following p.d.f.       elsewhere , 0 0.5 0 for , 8 ) (x x x f

An U(0,1) random number generator has generated a value of 0.26.