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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 a ≤ X ≤ b = (^) ∫ 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
- 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 ( ≤ X ≤ b )
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 ( a ≤ X ≤ b ) = P a ( < X ≤ b )
= P ( a ≤ X < 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
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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
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= 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.
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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
