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S
tat 110A
,
UCLA
,
Ivo Dinov Slide 1
UCLA STAT 110 A
Applied Probability & Statistics for
Engineers
zInstructor: Ivo Dinov,
Asst. Prof. In Statistics and Neurology
zTeaching Assistant: Neda Farzinnia, UCLA Statistics
University of California, Los Angeles, Spring 2004
http://www.stat.ucla.edu/~dinov/
Stat 110A, UCLA, Ivo DinovSlide 2
Chapter 4
Continuous
Random Variables
and Probability
Distributions
Stat 110A, UCLA, Ivo DinovSlide 3
4.1
Continuous Random
Variables and
Probability
Distributions
Stat 110A, UCLA, Ivo DinovSlide 4
Continuous Random Variables
A random variable Xis continuous if its
set of possible values is an entire
interval of numbers (If A< B, then any
number xbetween Aand B is possible).
Stat 110A, UCLA, Ivo DinovSlide 5
Probability Distribution
Let Xbe a continuous rv. Then a
probability distribution or probability
density function (pdf) of Xis a function
f(x) such that for any two numbers a
and b,
()
()
b
a
P
aXb fxdx≤≤=
The graph of f is the density curve.
Stat 110A, UCLA, Ivo DinovSlide 6
Probability Density Function
For f(x) to be a pdf
1. f (x) > 0 for all values of x.
2.The area of the region between the
graph of fand the x axis is equal to 1.
Area = 1
()yfx=
pf3
pf4
pf5
pf8
pf9
pfa

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Stat 110A, UCLA, Ivo Dinov Slide 1

UCLA STAT 110 A

Applied Probability & Statistics for

Engineers

z Instructor: Ivo Dinov,

Asst. Prof. In Statistics and Neurology

z Teaching Assistant: Neda Farzinnia , UCLA Statistics

University of California, Los Angeles, Spring 2004 http://www.stat.ucla.edu/~dinov/

Slide 2 Stat 110A, UCLA, Ivo Dinov

Chapter 4

Continuous

Random Variables

and Probability

Distributions

Slide 3 Stat 110A, UCLA, Ivo Dinov

Continuous Random

Variables and

Probability

Distributions

Slide 4 Stat 110A, UCLA, Ivo Dinov

Continuous Random Variables

A random variable X is continuous if its set of possible values is an entire interval of numbers (If A < B , then any number x between A and B is possible).

Slide 5 Stat 110A, UCLA, Ivo Dinov

Probability Distribution

Let X be a continuous rv. Then a probability distribution or probability density function (pdf) of X is a function f ( x ) such that for any two numbers a and b , ( ) ( )

b P aXb = (^) ∫ a f x dx

The graph of f is the density curve.

Slide 6 Stat 110A, UCLA, Ivo Dinov

Probability Density Function

For f ( x ) to be a pdf

  1. f ( x ) > 0 for all values of x. 2.The area of the region between the graph of f and the x – axis is equal to 1.

Area = 1

y = f ( ) x

Slide 7 Stat 110A, UCLA, Ivo Dinov

Probability Density Function

is given by the area of the shaded region.

y = f ( ) x

a b

P a ( ≤ Xb )

Slide 8 Stat 110A, UCLA, Ivo Dinov

Continuous RV’s

z A RV is continuous if it can take on any real value in a non-trivial interval (a ; b). z PDF, probability density function, for a cont. RV, Y, is a non-negative function p (^) Y (y), for any real value y, such that for each interval (a; b), the probability that Y takes on a value in (a; b), P(a<Y<b) equals the area under p (^) Y (y) over the interval (a: b). z

p (^) Y (y)

a b

P(a<Y<b)

Slide 9 Stat 110A, UCLA, Ivo Dinov

Convergence of density histograms to the PDF

z For a continuous RV the density histograms converge to the PDF as the size of the bins goes to zero. „ AdditionalInstructorAids\BirthdayDistribution_1978_systat.SYD z

Slide 10 Stat 110A, UCLA, Ivo Dinov

Convergence of density histograms to the PDF

z For a continuous RV the density histograms converge to the PDF as the size of the bins goes to zero. z

Slide 11 Stat 110A, UCLA, Ivo Dinov

Uniform Distribution

A continuous rv X is said to have a uniform distribution on the interval [ A , B ] if the pdf of X is

( )

1 ; , 0 otherwise

A x B f x A B (^) B A

 (^) ≤ ≤ =   − 

Slide 12 Stat 110A, UCLA, Ivo Dinov

Probability for a Continuous rv

If X is a continuous rv, then for any number c , P ( x = c ) = 0. For any two numbers a and b with a < b,

P ( aXb ) = P a ( < Xb )

= P ( aX < b )

= P ( a < X < b )

Slide 19 Stat 110A, UCLA, Ivo Dinov

Expected Value

The expected or mean value of a continuous rv X with pdf f ( x ) is

μ X E ( X ) x f ( ) x dx

−∞

= = (^) ∫ ⋅

Slide 20 Stat 110A, UCLA, Ivo Dinov

Expected Value of h ( X )

If X is a continuous rv with pdf f ( x ) and h ( x ) is any function of X , then

E h x [ ( ) (^) ] μ h X ( ) h x ( ) f ( ) x dx

−∞

= = (^) ∫ ⋅

Slide 21 Stat 110A, UCLA, Ivo Dinov

Variance and Standard Deviation

The variance of continuous rv X with pdf f ( x ) and mean is

(^2) ( ) ( ) (^2) ( ) σ X V x x μ f x dx

−∞

= = (^) ∫ − ⋅

= E [ (^) ( X − μ)^2 ]

The standard deviation is

μ

σ (^) X = V x ( ). Slide 22 Stat 110A, UCLA, Ivo Dinov

Short-cut Formula for Variance

( ) [^ ]

V ( X ) = E X^2 − E X ( )^2

Slide 23 Stat 110A, UCLA, Ivo Dinov

The Normal

Distribution

Slide 24 Stat 110A, UCLA, Ivo Dinov

Normal Distributions

(^1) ( ) 2 /(2 2 ) ( ) 2

f x e x^ μ^ σ x σ π

= −^ − − ∞ < < ∞

A continuous rv X is said to have a normal distribution with parameters

μ and σ, where − ∞ < μ< ∞and

0 < σ, if the pdf of X is

Slide 25 Stat 110A, UCLA, Ivo Dinov

Standard Normal Distributions

(^12) / 2 ( ;0,1) 2

f z e^ z σ π

= −

The normal distribution with parameter values is called a standard normal distribution. The random variable is denoted by Z. The pdf is

μ= 0 and σ= 1

The cdf is

− ∞ < z < ∞

( ) ( ) ( ;0,1)

z z P Z z f y dy −∞

Φ = ≤ = (^) ∫

Slide 26 Stat 110A, UCLA, Ivo Dinov

Standard Normal Cumulative Areas

0 z

Standard normal curve

Shaded area = Φ( ) z

Slide 27 Stat 110A, UCLA, Ivo Dinov

Standard Normal Distribution

a. Area to the left of 0.85 = 0.

b. P ( Z > 1.32)

Let Z be the standard normal variable. Find (from table)

P Z ( ≤0.85)

1 − P Z ( ≤ 1.32) =0.

Slide 28 Stat 110A, UCLA, Ivo Dinov

Find the area to the left of 1.78 then subtract the area to the left of –2.1.

c. P ( 2.1− ≤ Z ≤1.78)

= P Z ( ≤ 1.78) − P Z ( ≤ −2.1)

Slide 29 Stat 110A, UCLA, Ivo Dinov

z α Notation

will denote the value on the measurement axis for which the area

under the z curve lies to the right of z α.

z α

0 z α

Shaded area = P Z ( ≥ z α )=α

Slide 30 Stat 110A, UCLA, Ivo Dinov

= 2[ P ( z < Z ) – ½]

P ( z < Z < – z ) = 2 P (0 < Z < z )

z = 1.

Ex. Let Z be the standard normal variable. Find z if a. P ( Z < z ) = 0.9278. Look at the table and find an entry = 0.9278 then read back to find z = 1.46. b. P (– z < Z < z ) = 0.

= 2 P ( z < Z ) – 1 = 0. P ( z < Z ) = 0.

Slide 37 Stat 110A, UCLA, Ivo Dinov

Let X be a binomial rv based on n trials, each with probability of success p. If the binomial probability histogram is not too skewed, X may be approximated by a normal distribution with μ= np and σ= npq.

Normal Approximation to the

Binomial Distribution

P X ( x ) x^ 0.5 np npq

 (^) + −  ≤ ≈ Φ (^)     Slide 38 Stat 110A, UCLA, Ivo Dinov

Ex. At a particular small college the pass rate of Intermediate Algebra is 72%. If 500 students enroll in a semester determine the probability that at least 375 students pass. μ = np = 500(.72) = 360 σ = npq = 500(.72)(.28) ≈ 10

( 375) 375.5^360 (1.55) 10

P X ≤ ≈ Φ ^ − = Φ   = 0.

Slide 39 Stat 110A, UCLA, Ivo Dinov

Normal approximation to Binomial

z Suppose Y~Binomial(n, p) z Then Y=Y 1 + Y 2 + Y 3 +…+ Y (^) n, where „ Y (^) k~Bernoulli(p) , E(Y (^) k)=p & Var(Y (^) k)=p(1-p) Î „ E(Y)=np & Var(Y)=np(1-p), SD(Y)= (np(1-p)) 1/ „ Standardize Y: ‰ Z=(Y-np) / (np(1-p))1/ ‰ By CLT Î Z ~ N(0, 1). So, Y ~ N [np, (np(1-p)) 1/2^ ] z Normal Approx to Binomial is reasonable when np >=10 & n(1-p)> (p & (1-p) are NOT too small relative to n).

Slide 40 Stat 110A, UCLA, Ivo Dinov

Normal approximation to Binomial – Example z Roulette wheel investigation: z Compute P(Y>=58), where Y~Binomial(100, 0.47) – „ The proportion of the Binomial(100, 0.47) population having more than 58 reds (successes) out of 100 roulette spins (trials). „ Since np=47>=10 & n(1-p)=53>10 Normal approx is justified. z Z=(Y-np)/Sqrt(np(1-p)) = 58 – 1000.47)/Sqrt(1000.470.53)=2.* z P(Y>=58) Í Î P(Z>=2.2) = 0. z True P(Y>=58) = 0.177, using SOCR (demo!) z Binomial approx useful when no access to SOCR avail.

Roulette has 38 slots 18red 18black 2 neutral

Slide 41 Stat 110A, UCLA, Ivo Dinov

Normal approximation to Poisson

z Let X 1 ~Poisson( λ ) & X 2 ~Poisson( μ ) Æ X 1 + X 2 ~Poisson( λ+μ ) z Let X 1 , X 2 , X 3 , …, X (^) k ~ Poisson( λ ), and independent, z Y (^) k = X 1 + X 2 + ··· + Xk ~ Poisson(k λ ), E( Yk )=Var( Yk )= k λ. z The random variables in the sum on the right are independent and each has the Poisson distribution with parameter λ. z By CLT the distribution of the standardized variable ( Y (^) k k λ ) / ( k λ )1/2^ Î N(0, 1), as k increases to infinity.

z So, for k λ >= 100, Zk = {( Y (^) k k λ ) / ( k λ ) 1/2^ } ~ N(0,1).

z Î Yk ~ N( k λ , ( k λ )1/2^ ).

Slide 42 Stat 110A, UCLA, Ivo Dinov

Normal approximation to Poisson – example z Let X 1 ~Poisson( λ ) & X 2 ~Poisson( μ ) Æ X 1 + X 2 ~Poisson( λ+μ ) z Let X 1 , X 2 , X 3 , …, X 200 ~ Poisson( 2 ), and independent, z Y (^) k = X 1 + X 2 + ··· + Xk ~ Poisson(400), E( Yk )=Var( Yk )= 400****. z By CLT the distribution of the standardized variable ( Y (^) k 400 ) / ( 400 ) 1/2^ Î N(0, 1), as k increases to infinity. z Zk = ( Y (^) k 400 ) / 20 ~ N(0,1) Î Yk ~ N( 400 , 400 ). z P(2 < Yk < 400) = (std’z 2 & 400) = z P( ( 2 400 )/20 < Zk < ( 400 400 )/20 ) = P( -20 < Zk < 0 ) = 0.

Slide 43 Stat 110A, UCLA, Ivo Dinov

Poisson or Normal approximation to Binomial?

z Poisson Approximation ( Binomial(n, p (^) n ) Æ Poisson(λ) ):

„ n>=100 & p<=0.01 & λ =n p <=

z Normal Approximation ( Binomial(n, p) Æ N ( np, (np(1-p))1/2 ) ) „ np >=10 & n(1-p)>

y

p p ye y

n n pn

n

ny n

y n

λ^ λ

λ

 − ^ →

×  →

→∞

WHY?

Slide 44 Stat 110A, UCLA, Ivo Dinov

The Gamma

Distribution and Its

Relatives

Slide 45 Stat 110A, UCLA, Ivo Dinov

The Gamma Function

For α > 0,the^ gamma function

Γ ( α ) is defined by

1 0

( α) x^ α e xdx

∞ Γ = −^ − ∫

Slide 46 Stat 110A, UCLA, Ivo Dinov

Gamma Distribution

A continuous rv X has a gamma distribution if the pdf is

(^1 1) / 0 ( ; , ) (^) ( ) 0 otherwise

x e x x f x

α β α β (^) β α α

 (^) − − ≥ =  Γ  

where the parameters satisfy α > 0, β>0. The standard gamma distribution has β =1.

Slide 47 Stat 110A, UCLA, Ivo Dinov

Mean and Variance

The mean and variance of a random variable X having the gamma distribution f ( ; x α β, ) are

E X ( ) = μ = αβ V X ( )= σ 2 =αβ^2

Slide 48 Stat 110A, UCLA, Ivo Dinov

Probabilities from the Gamma

Distribution

Let X have a gamma distribution with parameters

( ) ( ; , ) ;

x P X x F x α β F α β

  ≤ = = (^)    

Then for any x > 0, the cdf of X is given by

α and β.

where 1

0

x (^) y y e F x dy

α

− −

∫ Γ

Slide 55 Stat 110A, UCLA, Ivo Dinov

Identifying Common Distributions – QQ plots

z Quantile-Quantile plots indicate how well the model distribution agrees with the data. z q -th^ quantile, for 0<q<1, is the (data-space) value, Vq , at or below which lies a proportion q of the data.

1 Graph of the CDF, FY (y)=P(Y<=Vq )=q

0

q

Vq

Slide 56 Stat 110A, UCLA, Ivo Dinov

Constructing QQ plots z Start off with data {y 1 , y 2 , y 3 , …, y (^) n} z Order statistics y (^) (1) <= y (^) (2) <= y (^) (3) <=…<= y (^) (n) z Compute quantile rank , q(k) , for each observation, y (^) (k) , P(Y<= q(k) ) = (k-0.375) / (n+0.250), where Y is a RV from the (target) model distribution. z Finally, plot the points ( y (^) (k) , q(k) ) in 2D plane, 1<=k<=n. z Note: Different statistical packages use slightly different formulas for the computation of q(k). However, the results are quite similar. This is the formulas employed in SAS. z Basic idea : Probability that: P( (model)Y<=(data)y (^) (1))~ 1/n; P(Y<=y (^) (2)) ~ 2/n; P(Y<=y (^) (3)) ~ 3/n; …

Slide 57 Stat 110A, UCLA, Ivo Dinov

Example - Constructing QQ plots

z Start off with data {y 1 , y 2 , y 3 , …, y (^) n}. z Plot the points ( y(k) , q(k) ) in 2D plane, 1<=k<=n.

0

1

2

3

Expected Value forNormal Distribution C:\Ivo.dir\UCLA_Classes\Winter2002\AdditionalInstructorAidsBirthdayDistribution_1978_systat.SYDSYSTAT, Graph

Æ^ Probability Plot, Var4, Normal Distribution

Slide 58 Stat 110A, UCLA, Ivo Dinov

Other Continuous

Distributions

Slide 59 Stat 110A, UCLA, Ivo Dinov

The Weibull Distribution

A continuous rv X has a Weibull distribution if the pdf is

1 ( / ) (^0) ( ; , ) 0 0

x e x x f x x

α β^ α α

α α β (^) β

 (^) − − ≥ =    (^) <

where the parameters satisfy α > 0, β>0.

Slide 60 Stat 110A, UCLA, Ivo Dinov

Mean and Variance

The mean and variance of a random variable X having the Weibull distribution are 2 μ β 1 1 σ 2 β^21 2 α α α

  ^      = Γ (^)  + (^)  = (^) Γ (^)  + (^)  − (^)  Γ (^)  +    (^)    (^)   

Slide 61 Stat 110A, UCLA, Ivo Dinov

Weibull Distribution

The cdf of a Weibull rv having parameters

1 (^ /^ ) 0 ( ; , ) 0 < 0

e x x F x x

β^ α α β

 − − ≥ = (^)  

α and βis

Slide 62 Stat 110A, UCLA, Ivo Dinov

Lognormal Distribution

A nonnegative rv X has a lognormal distribution if the rv Y = ln( X ) has a normal distribution the resulting pdf has parameters

(^1) [ln( ) ] /(2^2 2 ) 0 ( ; , ) (^2) 0 0

e x x f x (^) x x

μ σ μ σ πα

 (^) − − ≥ = (^)   (^) <

μ and σand is

Slide 63 Stat 110A, UCLA, Ivo Dinov

Mean and Variance

The mean and variance of a variable X having the lognormal distribution are

( )

(^2) / 2 2 2 2 E X ( ) = e^ μ^ +σ^ V X ( ) = e μ^ +σ^ e σ − 1

Slide 64 Stat 110A, UCLA, Ivo Dinov

Lognormal Distribution

F ( ; x μ , α ) = P X ( ≤ x ) = P [ln( X ) ≤ln( )] x

The cdf of the lognormal distribution is given by

ln( ) x ln( ) x P Z

μ μ σ σ

 −^   −  = (^)  ≤ (^)  = Φ     

Slide 65 Stat 110A, UCLA, Ivo Dinov

Beta Distribution

A rv X is said to have a beta distribution with parameters A , B, (^) α> 0, and β> 0 if the pdf of X is

f ( ; x α β, , A B , )=

1 ( ) 1 1 0 ( ) ( ) 0 otherwise

x A B x (^) x B A B A B A

α β^ α β α β

 (^) Γ +  −  −^  − −  ⋅ (^)     ≥  − Γ ⋅Γ (^)  − (^)   −   

Slide 66 Stat 110A, UCLA, Ivo Dinov

Mean and Variance

The mean and variance of a variable X having the beta distribution are

A ( B A )

α μ α β

2 2 2

B A αβ σ α β α β