Probability Density Functions

Probability Density Functions

Probability Density Functions Definition 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 with a b, P(a X b) = b a f (x)dx That is, the probability that X takes on a value in the interval [a, b] is the area above this interval and under the graph of the density function. The graph of f (x) is often referred to as the density curve.

Probability Density Functions

Probability Density Functions Figure: P(60 X 70)

Probability Density Functions

Probability Density Functions Definition A continuous rv X is said to have a uniform distribution on the interval [A, B], if the pdf of X is f (x; A, B) = { 1 B A A x B 0 otherwise

Probability Density Functions Definition A continuous rv X is said to have a uniform distribution on the interval [A, B], if the pdf of X is f (x; A, B) = { 1 B A A x B 0 otherwise The graph of any uniform pdf looks like the graph in the previous example:

Cumulative Distribution Functions

Cumulative Distribution Functions Definition The cumulative distribution function F (x) for a continuous rv X is defined for every number x by F (x) = P(X x) = x f (y)dy For each x, F (x) is the area under the density curve to the left of x.

Cumulative Distribution Functions

Cumulative Distribution Functions Proposition Let X be a continuous rv with pdf f (x) and cdf F (x). Then for any number a, P(X > a) = 1 F (a) and for any two numbers a and b with a < b, P(a X b) = F (b) F (a).

Cumulative Distribution Functions Proposition Let X be a continuous rv with pdf f (x) and cdf F (x). Then for any number a, P(X > a) = 1 F (a) and for any two numbers a and b with a < b, P(a X b) = F (b) F (a).

Cumulative Distribution Functions

Cumulative Distribution Functions Proposition If X is a continuous rv with pdf f (x) and cdf F (x), then at every x at which the derivative F (x) exists, F (x) = f (x).

Cumulative Distribution Functions

Cumulative Distribution Functions Definition The expected value or mean valued of a continuous rv X with pdf f (x) is µ X = E(X ) = x f (x)dx

Cumulative Distribution Functions Definition The expected value or mean valued of a continuous rv X with pdf f (x) is µ X = E(X ) = x f (x)dx Definition The variance of a continuous random variable X with pdf f (x) and mean value µ is σx 2 = V (X ) = (x µ) 2 f (x)dx = E[(X µ) 2 ] The standard deviation (SD) of X is σ X = V (X ).

Cumulative Distribution Functions

Cumulative Distribution Functions Definition Let p be a number between 0 and 1. The (100p)th percentile of the distribution of a continuous rv X, denoted by η(p), is defined by p = F (η(p)) = η(p) f (y)dy

Normal Distribution

Normal Distribution Definition A continuous rv X is said to have a normal distribution with parameter µ and σ (µ and σ 2 ), where < µ < and σ > 0, if the pdf of X is f (x; µ, σ) = 1 2πσ e (x µ)2 /(2σ 2 ) We use the notation X N(µ, σ 2 ) to denote that X is rormally distributed with parameters µ and σ 2.

Normal Distribution Definition A continuous rv X is said to have a normal distribution with parameter µ and σ (µ and σ 2 ), where < µ < and σ > 0, if the pdf of X is f (x; µ, σ) = 1 2πσ e (x µ)2 /(2σ 2 ) We use the notation X N(µ, σ 2 ) to denote that X is rormally distributed with parameters µ and σ 2. Remark: 1. Obviously, f (x) 0 for all x; 2. It is guaranteed that 1 2πσ e (y µ)2 /(2σ 2) dy = 1.

Normal Distribution

Normal Distribution Proposition For X N(µ, σ 2 ), we have E(X ) = µ and V (X ) = σ 2

Normal Distribution Proposition For X N(µ, σ 2 ), we have E(X ) = µ and V (X ) = σ 2 σ = 1 σ = 2 σ = 0.5

Normal Distribution

Normal Distribution The cdf of a normal random variable X is F (x) = P(X x) = = x x f (y; µ, σ)dy 1 2πσ e (y µ)2 /(2σ 2) dy = 1 x µ e (z)2 /(2σ 2) dz change of variable:z = y µ 2πσ = 1 x µ σ 2πσ = x µ σ e (w)2 /2 σdw 1 2π e (w)2 /2 dw change of variable:w = z σ

Normal Distribution

Normal Distribution Definition The normal distribution with parameter values µ = 0 and σ = 1 is called the standard normal distribution. A random variable having a standard normal distribution is called a standard normal random variable and will be denoted by Z. The pdf of Z is f (z; 0, 1) = 1 2π e z2 /2 < z < The graph of f (z; 0, 1) is called the standard normal (or z) curve. The cdf of Z is P(Z z) = z f (y; 0, 1)dy, which we will denote by Φ(z).

Normal Distribution

Normal Distribution Shaded area = Φ(0.5)

Normal Distribution

Normal Distribution Table A.3 Standard Normal Curve Areas z.00.01.02.03.04.09-1.2 0.1151 0.1131 0.1112 0.1094 0.1075 0.0985-1.1 0.1357 0.1335 0.1314 0.1292 0.1271 0.1170 1.6 0.9452 0.9463 0.9474 0.9484 0.9495 0.9545 1.7 0.9554 0.9564 0.9573 0.9582 0.9591 0.9633

Normal Distribution

Normal Distribution Z N(0, 1), calculate (a)p(z 1.61); (b)p(z > 1.12); and (c)p( 1.12 < Z 1.61).

Normal Distribution Z N(0, 1), calculate (a)p(z 1.61); (b)p(z > 1.12); and (c)p( 1.12 < Z 1.61). z.00.01.02.03.04.09-1.2 0.1151 0.1131 0.1112 0.1094 0.1075 0.0985-1.1 0.1357 0.1335 0.1314 0.1292 0.1271 0.1170 1.6 0.9452 0.9463 0.9474 0.9484 0.9495 0.9545 1.7 0.9554 0.9564 0.9573 0.9582 0.9591 0.9633

Normal Distribution Z N(0, 1), calculate (a)p(z 1.61); (b)p(z > 1.12); and (c)p( 1.12 < Z 1.61). z.00.01.02.03.04.09-1.2 0.1151 0.1131 0.1112 0.1094 0.1075 0.0985-1.1 0.1357 0.1335 0.1314 0.1292 0.1271 0.1170 1.6 0.9452 0.9463 0.9474 0.9484 0.9495 0.9545 1.7 0.9554 0.9564 0.9573 0.9582 0.9591 0.9633 P(Z 1.61) = 0.9463;

Normal Distribution Z N(0, 1), calculate (a)p(z 1.61); (b)p(z > 1.12); and (c)p( 1.12 < Z 1.61). z.00.01.02.03.04.09-1.2 0.1151 0.1131 0.1112 0.1094 0.1075 0.0985-1.1 0.1357 0.1335 0.1314 0.1292 0.1271 0.1170 1.6 0.9452 0.9463 0.9474 0.9484 0.9495 0.9545 1.7 0.9554 0.9564 0.9573 0.9582 0.9591 0.9633 P(Z 1.61) = 0.9463; P(Z > 1.12) = 1 P(Z 1.12) = 1 0.1314 = 0.8686;

Normal Distribution Z N(0, 1), calculate (a)p(z 1.61); (b)p(z > 1.12); and (c)p( 1.12 < Z 1.61). z.00.01.02.03.04.09-1.2 0.1151 0.1131 0.1112 0.1094 0.1075 0.0985-1.1 0.1357 0.1335 0.1314 0.1292 0.1271 0.1170 1.6 0.9452 0.9463 0.9474 0.9484 0.9495 0.9545 1.7 0.9554 0.9564 0.9573 0.9582 0.9591 0.9633 P(Z 1.61) = 0.9463; P(Z > 1.12) = 1 P(Z 1.12) = 1 0.1314 = 0.8686; P( 1.12 < Z 1.61) = P(Z 1.61) P(Z 1.12) = 0.9463 0.1314 = 0.8149.

Normal Distribution

Normal Distribution Many tables for the normal distribution contain only the nonnegative part.

Normal Distribution Many tables for the normal distribution contain only the nonnegative part. z.00.01.02.03.04.09 1.6 0.9452 0.9463 0.9474 0.9484 0.9495 0.9545 1.7 0.9554 0.9564 0.9573 0.9582 0.9591 0.9633 What is P(Z < 1.63)?

Normal Distribution Many tables for the normal distribution contain only the nonnegative part. z.00.01.02.03.04.09 1.6 0.9452 0.9463 0.9474 0.9484 0.9495 0.9545 1.7 0.9554 0.9564 0.9573 0.9582 0.9591 0.9633 What is P(Z < 1.63)? By symmetry of the pdf of Z, we know that P(Z < 1.63) = P(Z > 1.63) = 1 P(Z 1.63) = 1 0.9484 = 0.0516

Normal Distribution Many tables for the normal distribution contain only the nonnegative part. z.00.01.02.03.04.09 1.6 0.9452 0.9463 0.9474 0.9484 0.9495 0.9545 1.7 0.9554 0.9564 0.9573 0.9582 0.9591 0.9633 What is P(Z < 1.63)? By symmetry of the pdf of Z, we know that P(Z < 1.63) = P(Z > 1.63) = 1 P(Z 1.63) = 1 0.9484 = 0.0516

Normal Distribution

Normal Distribution Recall: The (100p)th percentile of the distribution of a continuous rv X, η(p), is defined by p = F (η(p)) = η(p) f (y)dy

Normal Distribution Recall: The (100p)th percentile of the distribution of a continuous rv X, η(p), is defined by p = F (η(p)) = η(p) f (y)dy Similarly, the (100p)th percentile of the standard normal rv Z is defined by η(p) 1 p = F (η(p)) = e y 2 /2 dy 2π

Normal Distribution Recall: The (100p)th percentile of the distribution of a continuous rv X, η(p), is defined by p = F (η(p)) = η(p) f (y)dy Similarly, the (100p)th percentile of the standard normal rv Z is defined by η(p) 1 p = F (η(p)) = e y 2 /2 dy 2π We need to use the table for normal distribution to find (100p)th percentile.

Normal Distribution

Normal Distribution e.g. Find the 95th percentile for the standard normal rv Z

Normal Distribution e.g. Find the 95th percentile for the standard normal rv Z z.00.01.02.03.04 0.5.09 1.6 0.9452 0.9463 0.9474 0.9484 0.9495 0.9505 0.9545 1.7 0.9554 0.9564 0.9573 0.9582 0.9591 0.9599 0.9633

Normal Distribution e.g. Find the 95th percentile for the standard normal rv Z z.00.01.02.03.04 0.5.09 1.6 0.9452 0.9463 0.9474 0.9484 0.9495 0.9505 0.9545 1.7 0.9554 0.9564 0.9573 0.9582 0.9591 0.9599 0.9633 η(95) = 1.645, a linear interpolation of 1.64 and 1.65.

Normal Distribution e.g. Find the 95th percentile for the standard normal rv Z z.00.01.02.03.04 0.5.09 1.6 0.9452 0.9463 0.9474 0.9484 0.9495 0.9505 0.9545 1.7 0.9554 0.9564 0.9573 0.9582 0.9591 0.9599 0.9633 η(95) = 1.645, a linear interpolation of 1.64 and 1.65. Remark: If p does not appear in the table, we can either use the number closest to it, or use the linear interpolation of the closest two.

Normal Distribution

Normal Distribution In statistical inference, the percentiles corresponding to right small tails are heavily used. Notation z α will denote the value on the z axis for which α of the area under the z curve lies to the right of z α.

Normal Distribution In statistical inference, the percentiles corresponding to right small tails are heavily used. Notation z α will denote the value on the z axis for which α of the area under the z curve lies to the right of z α. z α

Normal Distribution

Normal Distribution Remark: 1. z α is the 100(1 α)th percentile of the standard normal distribution.

Normal Distribution Remark: 1. z α is the 100(1 α)th percentile of the standard normal distribution. 2. By symmetry the area under the standard normal curve to the left of z α is also α.

Normal Distribution Remark: 1. z α is the 100(1 α)th percentile of the standard normal distribution. 2. By symmetry the area under the standard normal curve to the left of z α is also α. 3. The z α s are usually referred to as z critical values. Percentile 90 95 97.5 99.95 α (tail area) 0.1 0.05 0.025 0.0005 z α 1.28 1.645 1.96 3.27

Normal Distribution

Normal Distribution Proposition If X has a normal distribution with mean µ and stadard deviation σ, then Z = X µ σ has a standard normal distribution. Thus P(a X b) = P( a µ Z b µ σ σ ) = Φ( b µ σ ) Φ(a µ σ ) P(X a) = Φ( a µ µ ) P(X b) = 1 Φ(b σ σ )

Normal Distribution

Normal Distribution Example (Problem 38): There are two machines available for cutting corks intended for use in wine bottles. The first produces corks with diameters that are normally distributed with mean 3cm and standard deviation 0.1cm. The second produces corks with diameters that have a normal distribution with mean 3.04cm and standard deviation 0.02cm. Acceptable corks have diameters between 2.9cm and 3.1cm. Which machine is more likely to produce an acceptable cork?

Normal Distribution Example (Problem 38): There are two machines available for cutting corks intended for use in wine bottles. The first produces corks with diameters that are normally distributed with mean 3cm and standard deviation 0.1cm. The second produces corks with diameters that have a normal distribution with mean 3.04cm and standard deviation 0.02cm. Acceptable corks have diameters between 2.9cm and 3.1cm. Which machine is more likely to produce an acceptable cork? P(2.9 X 1 3.1) = P( 2.9 3 Z 3.1 3 0.1 0.1 ) = P( 1 Z 1) = 0.8413 0.1587 = 0.6826 2.9 3.04 3.1 3.04 P(2.9 X 2 3.1) = P( Z ) 0.02 0.02 = P( 7 Z 3) = 0.9987 0 = 0.9987

Normal Distribution

Normal Distribution Example (Problem 44): If bolt thread length is normally distributed, what is the probability that the thread length of a randomly selected bolt is (a)within 1.5 SDs of its mean value? (b)between 1 and 2 SDs from its mean value?

Normal Distribution Example (Problem 44): If bolt thread length is normally distributed, what is the probability that the thread length of a randomly selected bolt is (a)within 1.5 SDs of its mean value? (b)between 1 and 2 SDs from its mean value? P(µ 1.5σ X 1 µ + 1.5σ) = P( µ 1.5σ µ Z µ + 1.5σ µ ) σ σ = P( 1.5 Z 1.5) = 0.9332 0.0668 = 0.8664

Normal Distribution Example (Problem 44): If bolt thread length is normally distributed, what is the probability that the thread length of a randomly selected bolt is (a)within 1.5 SDs of its mean value? (b)between 1 and 2 SDs from its mean value? P(µ 1.5σ X 1 µ + 1.5σ) = P( µ 1.5σ µ Z µ + 1.5σ µ ) σ σ = P( 1.5 Z 1.5) = 0.9332 0.0668 = 0.8664 2 P(µ + σ X 1 µ + 2σ) = 2P( µ + σ µ Z µ + 2σ µ ) σ σ = 2P(1 Z 2) = 2(0.9772 0.8413) = 0.0.2718

Normal Distribution

Normal Distribution Proposition {(100p)th percentile for N(µ, σ 2 )} = µ + {(100p)th percentile for N(0, 1)} σ

Normal Distribution Proposition {(100p)th percentile for N(µ, σ 2 )} = µ + {(100p)th percentile for N(0, 1)} σ Example (Problem 39) The width of a line etched on an integrated circuit chip is normally distributed with mean 3.000 µm and standard deviation 0.140. What width value separates the widest 10% of all such lines from the other 90%?

Normal Distribution Proposition {(100p)th percentile for N(µ, σ 2 )} = µ + {(100p)th percentile for N(0, 1)} σ Example (Problem 39) The width of a line etched on an integrated circuit chip is normally distributed with mean 3.000 µm and standard deviation 0.140. What width value separates the widest 10% of all such lines from the other 90%? η N(3,0.140 2 )(90) = 3.0+0.140 η N(0,1) (90) = 3.0+0.140 1.28 = 3.1792

Normal Distribution

Normal Distribution Proposition Let X be a binomial rv based on n trials with success probability p. Then if the binomial probability histogram is not too skewed, X has approximately a normal distribution with µ = np and σ = npq, where q = 1 p. In particular, for x = a posible value of X, ( ) area under the normal curve P(X x) = B(x; n, p) to the left of x+0.5 x+0.5 np = Φ( ) npq In practice, the approximation is adequate provided that both np 10 and nq 10, since there is then enough symmetry in the underlying binomial distribution.

Normal Distribution

Normal Distribution A graphical explanation for ( ) area under the normal curve P(X x) = B(x; n, p) to the left of x+0.5 x+0.5 np = Φ( ) npq

Normal Distribution A graphical explanation for ( ) area under the normal curve P(X x) = B(x; n, p) to the left of x+0.5 x+0.5 np = Φ( ) npq

Normal Distribution

Normal Distribution Example (Problem 54) Suppose that 10% of all steel shafts produced by a certain process are nonconforming but can be reworked (rather than having to be scrapped). Consider a random sample of 200 shafts, and let X denote the number among these that are nonconforming and can be reworked. What is the (approximate) probability that X is between 15 and 25 (inclusive)?

Normal Distribution Example (Problem 54) Suppose that 10% of all steel shafts produced by a certain process are nonconforming but can be reworked (rather than having to be scrapped). Consider a random sample of 200 shafts, and let X denote the number among these that are nonconforming and can be reworked. What is the (approximate) probability that X is between 15 and 25 (inclusive)? In this problem n = 200, p = 0.1 and q = 1 p = 0.9. Thus np = 20 > 10 and nq = 180 > 10

Normal Distribution Example (Problem 54) Suppose that 10% of all steel shafts produced by a certain process are nonconforming but can be reworked (rather than having to be scrapped). Consider a random sample of 200 shafts, and let X denote the number among these that are nonconforming and can be reworked. What is the (approximate) probability that X is between 15 and 25 (inclusive)? In this problem n = 200, p = 0.1 and q = 1 p = 0.9. Thus np = 20 > 10 and nq = 180 > 10 P(15 X 25) = Bin(25; 200, 0.1) Bin(14; 200, 0.1) 25 + 0.5 20 15 + 0.5 20 Φ( ) Φ( ) 200 0.1 0.9 200 0.1 0.9 = Φ(0.3056) Φ( 0.2500) = 0.6217 0.4013 = 0.2204

Exponential Distribution

Exponential Distribution Definition X is said to have an exponential distribution with parameter λ(λ > 0) if the pdf of X is { λe λx x 0 f (x; λ) = 0 otherwise

Exponential Distribution Definition X is said to have an exponential distribution with parameter λ(λ > 0) if the pdf of X is { λe λx x 0 f (x; λ) = 0 otherwise Remark: 1. Usually we use X EXP(λ) to denote that the random variable X has an exponential distribution with parameter λ.

Exponential Distribution Definition X is said to have an exponential distribution with parameter λ(λ > 0) if the pdf of X is { λe λx x 0 f (x; λ) = 0 otherwise Remark: 1. Usually we use X EXP(λ) to denote that the random variable X has an exponential distribution with parameter λ. 2. In some sources, the pdf of exponential distribution is given by { 1 f (x; θ) = θ e x θ x 0 0 otherwise The difference is that λ 1 θ.

Exponential Distribution

Exponential Distribution

Exponential Distribution

Exponential Distribution Proposition If X EXP(λ), then And the cdf for X is E(X ) = 1 λ and V (X ) = 1 λ 2 F (x; λ) = { 1 e λx x 0 0 x < 0

Exponential Distribution

Exponential Distribution Proof: E(X ) = = 1 λ = 1 λ 0 0 0 xλe λx dx (λx)e λx d(λx) ye y dy = 1 λ [ ye y 0 + = 1 λ [0 + ( e y 0 )] 0 e y dy] y = λx integration by parts: u = y, v = e y = 1 λ

Exponential Distribution

Exponential Distribution Proof (continued): E(X 2 ) = 0 x 2 λe λx dx = 1 λ 2 (λx) 2 e λx d(λx) 0 = 1 λ 2 y 2 e y dy 0 = 1 λ 2 [ y 2 e y 0 + = 1 λ 2 [0 + 2( ye y 0 + = 1 λ 2 2[0 + ( ye y 0 )] 0 2ye y dy] 0 e y dy)] y = λx integration by parts integration by parts = 2 λ 2

Exponential Distribution

Exponential Distribution Proof (continued): V (X ) = E(X 2 ) [E(X )] 2 = 2 λ 2 ( 1 λ )2 = 1 λ 2 F (x) = = = x 0 x 0 λx 0 λe λy dy e λy d(λy) e z dz = e z λx 0 = 1 e λx z = λy

Exponential Distribution

Exponential Distribution Example (Problem 108) The article Determination of the MTF of Positive Photoresists Using the Monte Carlo method (Photographic Sci. and Engr., 1983: 254-260) proposes the exponential distribution with parameter λ = 0.93 as a model for the distribution of a photon s free path length (µm) under certain circumstances. Suppose this is the correct model.

Exponential Distribution Example (Problem 108) The article Determination of the MTF of Positive Photoresists Using the Monte Carlo method (Photographic Sci. and Engr., 1983: 254-260) proposes the exponential distribution with parameter λ = 0.93 as a model for the distribution of a photon s free path length (µm) under certain circumstances. Suppose this is the correct model. a. What is the expected path length, and what is the standard deviation of path length?

Exponential Distribution Example (Problem 108) The article Determination of the MTF of Positive Photoresists Using the Monte Carlo method (Photographic Sci. and Engr., 1983: 254-260) proposes the exponential distribution with parameter λ = 0.93 as a model for the distribution of a photon s free path length (µm) under certain circumstances. Suppose this is the correct model. a. What is the expected path length, and what is the standard deviation of path length? b. What is the probability that path length exceeds 3.0?

Exponential Distribution Example (Problem 108) The article Determination of the MTF of Positive Photoresists Using the Monte Carlo method (Photographic Sci. and Engr., 1983: 254-260) proposes the exponential distribution with parameter λ = 0.93 as a model for the distribution of a photon s free path length (µm) under certain circumstances. Suppose this is the correct model. a. What is the expected path length, and what is the standard deviation of path length? b. What is the probability that path length exceeds 3.0? c. What value is exceeded by only 10% of all path lengths?

Exponential Distribution

Exponential Distribution Proposition Suppose that the number of events occurring in any time interval of length t has a Poisson distribution with parameter αt (where α, the rate of the event process, is the expected number of events occurring in 1 unit of time) and that numbers of occurrences in nonoverlappong intervals are independent of one another. Then the distribution of elapsed time between the occurrence of two successive events is exponential with parameter λ = α.

Exponential Distribution Proposition Suppose that the number of events occurring in any time interval of length t has a Poisson distribution with parameter αt (where α, the rate of the event process, is the expected number of events occurring in 1 unit of time) and that numbers of occurrences in nonoverlappong intervals are independent of one another. Then the distribution of elapsed time between the occurrence of two successive events is exponential with parameter λ = α. e.g. the number of customers visiting Costco in each hour = Poisson distribution;

Exponential Distribution Proposition Suppose that the number of events occurring in any time interval of length t has a Poisson distribution with parameter αt (where α, the rate of the event process, is the expected number of events occurring in 1 unit of time) and that numbers of occurrences in nonoverlappong intervals are independent of one another. Then the distribution of elapsed time between the occurrence of two successive events is exponential with parameter λ = α. e.g. the number of customers visiting Costco in each hour = Poisson distribution; the time between every two successive customers visiting Costco = Exponential distribution.

Exponential Distribution

Exponential Distribution Example (Example 4.22) Suppose that calls are received at a 24-hour hotline according to a Poisson process with rate α = 0.5 call per day.

Exponential Distribution Example (Example 4.22) Suppose that calls are received at a 24-hour hotline according to a Poisson process with rate α = 0.5 call per day. Then the number of days X between successive calls has an exponential distribution with parameter value 0.5.

Exponential Distribution Example (Example 4.22) Suppose that calls are received at a 24-hour hotline according to a Poisson process with rate α = 0.5 call per day. Then the number of days X between successive calls has an exponential distribution with parameter value 0.5. The probability that more than 3 days elapse between calls is P(X > 3) = 1 P(X 3) = 1 F (3; 0.5) = e 0.5 3 = 0.223.

Exponential Distribution Example (Example 4.22) Suppose that calls are received at a 24-hour hotline according to a Poisson process with rate α = 0.5 call per day. Then the number of days X between successive calls has an exponential distribution with parameter value 0.5. The probability that more than 3 days elapse between calls is P(X > 3) = 1 P(X 3) = 1 F (3; 0.5) = e 0.5 3 = 0.223. The expected time between successive calls is 1/0.5 = 2 days.

Exponential Distribution

Exponential Distribution Memoryless Property Let X = the time certain component lasts (in hours) and we assume the component lifetime is exponentially distributed with parameter λ. Then what is the probability that the component can last at least an additional t hours after working for t 0 hours, i.e. what is P(X t + t 0 X t 0 )?

Exponential Distribution Memoryless Property Let X = the time certain component lasts (in hours) and we assume the component lifetime is exponentially distributed with parameter λ. Then what is the probability that the component can last at least an additional t hours after working for t 0 hours, i.e. what is P(X t + t 0 X t 0 )? P(X t + t 0 X t 0 ) = P({X t + t 0} {X t 0 }) P(X t 0 ) = P(X t + t 0) P(X t 0 ) = 1 F (t + t 0; λ) F (t 0 ; λ) = e λt

Exponential Distribution

Exponential Distribution Memoryless Property However, we have P(X t) = 1 F (t; λ) = e λt

Exponential Distribution Memoryless Property However, we have Therefore, we have for any positive t and t 0. P(X t) = 1 F (t; λ) = e λt P(X t) = P(X t + t 0 X t 0 )

Exponential Distribution Memoryless Property However, we have Therefore, we have P(X t) = 1 F (t; λ) = e λt P(X t) = P(X t + t 0 X t 0 ) for any positive t and t 0. In words, the distribution of additional lifetime is exactly the same as the original distribution of lifetime, so at each point in time the component shows no effect of wear. In other words, the distribution of remaining lifetime is independent of current age.

Gamma Distribution

Gamma Distribution Definition For α > 0, the gamma function Γ(α) is defined by Γ(α) = 0 x α 1 e x dx

Gamma Distribution Definition For α > 0, the gamma function Γ(α) is defined by Γ(α) = 0 x α 1 e x dx Properties for gamma function: 1. For any α > 1, Γ(α) = (α 1) Γ(α 1) [via integration by parts];

Gamma Distribution Definition For α > 0, the gamma function Γ(α) is defined by Γ(α) = 0 x α 1 e x dx Properties for gamma function: 1. For any α > 1, Γ(α) = (α 1) Γ(α 1) [via integration by parts]; 2. For any positive integer, n, Γ(n) = (n 1)!;

Gamma Distribution Definition For α > 0, the gamma function Γ(α) is defined by Γ(α) = 0 x α 1 e x dx Properties for gamma function: 1. For any α > 1, Γ(α) = (α 1) Γ(α 1) [via integration by parts]; 2. For any positive integer, n, Γ(n) = (n 1)!; 3. Γ( 1 2 ) = π.

Gamma Distribution Definition For α > 0, the gamma function Γ(α) is defined by Γ(α) = 0 x α 1 e x dx Properties for gamma function: 1. For any α > 1, Γ(α) = (α 1) Γ(α 1) [via integration by parts]; 2. For any positive integer, n, Γ(n) = (n 1)!; 3. Γ( 1 2 ) = π. e.g. Γ(4) = (4 1)! = 6 and Γ( 5 2 ) = 3 2 Γ( 3 2 ) = 3 2 [ 1 2 Γ( 1 2 )] = 3 4 π

Gamma Distribution

Gamma Distribution Definition A continuous random variable X is said to have a gamma distribution if the pdf of X is { 1 β f (x; α, β) = α Γ(α) x α 1 e x/β x 0 0 otherwise where the parameters α and β satisfy α > 0, β > 0. The standard gamma distribution has β = 1, so the pdf of a standard gamma rv is { 1 Γ(α) f (x; α) = x α 1 e x x 0 0 otherwise

Gamma Distribution

Gamma Distribution Remark: 1. We use X GAM(α, β) to denote that the rv X has a gamma distribution with parameter α and β.

Gamma Distribution Remark: 1. We use X GAM(α, β) to denote that the rv X has a gamma distribution with parameter α and β. 2. If we let α = 1 and β = 1/λ, then we get the exponential distribution: f (x; 1, 1 λ ) = { 1 1 λ Γ(1)x 1 1 e x/ 1 λ = λe λx x 0 0 otherwise

Gamma Distribution Remark: 1. We use X GAM(α, β) to denote that the rv X has a gamma distribution with parameter α and β. 2. If we let α = 1 and β = 1/λ, then we get the exponential distribution: { f (x; 1, 1 1 λ ) = 1 Γ(1)x 1 1 e x/ 1 λ = λe λx x 0 λ 0 otherwise 3. When X is a standard gamma rv (β = 1), the cdf of X, F (x; α) = x 0 y α 1 e y Γ(α) dy is called the incomplete gamma function.

Gamma Distribution Remark: 1. We use X GAM(α, β) to denote that the rv X has a gamma distribution with parameter α and β. 2. If we let α = 1 and β = 1/λ, then we get the exponential distribution: { f (x; 1, 1 1 λ ) = 1 Γ(1)x 1 1 e x/ 1 λ = λe λx x 0 λ 0 otherwise 3. When X is a standard gamma rv (β = 1), the cdf of X, F (x; α) = x 0 y α 1 e y Γ(α) dy is called the incomplete gamma function. There are extensive tables of F (x; α) available (Appendix Table A.4).

Gamma Distribution

Gamma Distribution

Gamma Distribution

Gamma Distribution Proposition If X GAM(α, β), then E(X ) = αβ and V (X ) = αβ 2 Furthermore, for any x > 0, the cdf of X is given by ( ) x P(X x) = F (x; α, β) = F β ; α where F ( ; α) is the incomplete gamma function.

Gamma Distribution

Gamma Distribution Example: The survival time (in days) of a white rat that was subjected to a certain level of X-ray radiation is a random variable X GAM(5, 4). Then what is a. the probability that the survival time is at most 16 days;

Gamma Distribution Example: The survival time (in days) of a white rat that was subjected to a certain level of X-ray radiation is a random variable X GAM(5, 4). Then what is a. the probability that the survival time is at most 16 days; b. the probability that the survival time is between 16 days and 20 days (not inclusive);

Gamma Distribution Example: The survival time (in days) of a white rat that was subjected to a certain level of X-ray radiation is a random variable X GAM(5, 4). Then what is a. the probability that the survival time is at most 16 days; b. the probability that the survival time is between 16 days and 20 days (not inclusive); c. the expected survival time.

Gamma Distribution Example: The survival time (in days) of a white rat that was subjected to a certain level of X-ray radiation is a random variable X GAM(5, 4). Then what is a. the probability that the survival time is at most 16 days; b. the probability that the survival time is between 16 days and 20 days (not inclusive); c. the expected survival time.

Chi-Squared Distribution

Chi-Squared Distribution Definition Let ν be a positive integer. Then a random variable X is said to have a chi-squared distribution with parameter ν if the pdf of X is the gamma density with α = ν/2 and β = 2. The pdf of a chi-squared rv is thus f (x; ν) = { 1 2 ν/2 Γ(ν/2) x (ν/2) 1 e x/2 x 0 0 x < 0 The parameter ν is called the number of degrees of freedom (df) of X. The symbol χ 2 is often used in place of chi-squared.

Chi-Squared Distribution

Chi-Squared Distribution Remark: 1. Usually, we use X χ 2 (ν) to denote that X is a chi-squared rv with parameter ν;

Chi-Squared Distribution Remark: 1. Usually, we use X χ 2 (ν) to denote that X is a chi-squared rv with parameter ν; 2. If X 1, X 2,..., X n is n independent standard normal rv s, then X 2 1 + X 2 2 + + X 2 n has the same distribution as χ 2 (n).

Chi-Squared Distribution

Chi-Squared Distribution