1. 1 Continuous Distributions ch3

2. 2   A random variable X of the continuous type has a support or space S that is an interval(possibly unbounded) or a union of intervals, instead of a set of real numbers (discrete points).   The probability density function(p.d.f.) of X is an integrable function f(x) satisfying:   f(x)>0, x ∈ S.   ∫ S f(x)dx= 1,   P(X ∈ A) = ∫ A f(x)dx, where A ⊂ S.   Ex.3.2-1: Assume X has the p.d.f.   P(X>20)=?   The distribution function of X is (cumulative distribution function) (c.d.f.) If the derivative F’(x) exists,

3. 3 Zero Probability of Points   For the random variable of the continuous type, the probability of any point is zero.   Namely, P(X=b)=0.   Thus, P(a≤X≤b) = P(a<X<b) = P(a≤X<b) = P(a<X≤b) = F(b)-F(a).   For instance,   Let X be the times between calls to 911. 105 observations aremade to construct a relative frequent histogram h(x).   It is compared with the exponential model   in Ex3.2-1. 30 17 65 8 38 35 4 19 7 14 12 4 5 4 2 7 5 12 50 33 10 15 2 10 1 5 30 41 21 31 1 18 12 5 24 7 6 31 1 3 2 22 1 30 2 13 12 129 28 6 5063 5 17 11 23 2 46 90 13 21 55 43 5 19 47 24 4 6 27 4 6 37 16 41 68 9 5 28 42 3 42 8 52 2 11 41 4 35 21 3 17 10 16 1 68 105 45 23 5 10 12 17

4. 4 (Dis)continuity, Differentiable, Integration   Ex3.2-4: Let Y be a continuous random variable with p.d.f. g(y)=2y, 0<y<1.   The distribution function of Y is   P(½≤Y≤¾)=G(¾)-G(½)=5/16.   P(¼≤Y<2)=G(2)-G(¼)=15/16.   Properties of a continuous random variable:   The area between the p.d.f. f(x) and x-axis must equal 1.   f(x) is possibly unbounded (say, >1).   f(x) can be discontinuous function (defined over a set of intervals),   However, its c.d.f. F(x) is always continuous since integration.   It is possible that F’(x) does not exist at x=x 0.

5. 5 Mean, Variance, Moment-generating Fn   Suppose X is a continuous random variable.   The expected value of X, the mean of X is   The variance of X is   The standard deviation of X is   The moment-generating function of X, if it exists, is   The r th moment E(X r ) exists and is finite ⇒ E(X r-1 ), …, E(X 1 ) do.   The converse is not true.   E(e tX ) exists and is finite –h<t<h ⇒ all the moments do.   The converse is not necessarily true.   Ex3.2-5: Random variable Y with p.d.f. g(y)=2y, 0<y<1. (Ex3.2-4)

6. 6 Percentile (π) and Quartiles   Ex.3.2-6: X has p.d.f. f(x)=xe -x, 0≤x<∞.   The (100p) th percentiles a number π p s.t. the area under f(x) to the left of π p is p.   The 50 th percentile is called the median: m = π.5.   The 25 th and 75 th percentiles are called the first and third quartiles.   Namely, π.25 = q 1, π.75 = q 3, and m = π.5 = q 2 the second quartile.   Ex: X with p.d.f. f(x)=1-|x-1|, 0≤x<2. To find 32 rd percentile π.32 is to solve F(π.32 )=.32 ∵ F(1)=.5>.32 To find 92 rd percentile π.92 is to solve F(π.92 )=.92

7. 7 More Example   Ex3.2-8: X has the p.d.f. f(x)=e -x-1, -1<x<∞.   The median m = π.5 is

8. 8 Uniform Distribution   Random variable X has a uniform distribution if its p.d.f. equals a constant on its support.   If the support is the interval [a, b], then p.d.f. ⇒   This distribution, denoted as U(a, b), is also referred to as rectangular due to the shape of f(x).   The mean, variance, distribution function and moment-generating function are   Pseudo-random number generator: a program applies simple arithmetical operations on the seed (starting number) to deterministically generate a sequence of numbers, whose distribution follows U(0, 1).   Table IX on p.695 shows an example of these (random) numbers*104.   Ex.3.3-1: X has p.d.f. f(x)=1/100, 0<x<100, namely U(0,100).   The mean and variance are μ=(0+100)/2=50, σ 2 =10000/12.   The standard deviation is, 100 times of U(0,1).

9. 9 Exponential Distribution   The waiting (inter-change) times W between successive changes whose number X in a given interval is a Poisson distribution is indeed an exponential distribution.   Such time is nonnegative ⇒ the distribution function F(w)=0 for w<0.   For w ≥0, F(w) = P(W≤w) = 1 -P(W>w) = 1 -P(no changes in [0, w]) = 1 –e –λw,   For w>0, the p.d.f. f(w) = F’(w) = λe –λw ⇒   Suppose λ=7, the mean number of changes per minute; ⇒ θ= 1/7, the mean waiting time for the first (next) change.   Ex3.3-2: X has an exponential distribution with a mean of θ=20.

10. 10 Examples   Ex.3.3-3: Customers arrivals follow a Poisson process of 20 per hr.   What is the probability that the shopkeeper will have to wait more than 5 minutes for the arrival of the first (next) customer?   Let X be the waiting time in minutes until the next customer arrives.   Having awaited for 6 min., what is the probability that the shopkeeper will have to wait more than 3 min. additionally for the new arrival?   Memory-less, forgetfulness property!   Percentiles:   To exam how close an empirical collection of data is to the exponential distribution, the q-qplot (y r,π p ) from the ordered statistics can be constructed, where p=r/(n+1), r=1,2,…,n.   If θis unknown, π p =-ln(1-p) can be used in the plot, instead.   As the curve ≈a straight line, it matches well. (Slope: an estimate of 1/θ)

11. 11 Gamma Distribution  Generalizing the exponential distribution, the Gamma distribution considers the waiting time W until the α th change occurs, α≥1.  The distribution function F(w) of W is given by  Leibnitz's rule:  The Gamma function is defined by F(w) F’(w) generalized factorial

12. 12 Fundamental Calculus  There are some formula from Calculus:  Generalized integration by parts (also ref. p.666):  Formula used in the Gamma distribution:  Pf:  Suppose  Then,  m   By the induction hypothesis, the equation holds! α=1:

13. 13 Chi-square Distribution  A Gamma distribution with θ=2, α=r/2, (r ∈ N) is a Chi- square distribution with r degrees of freedom, denoted as χ 2 (r).  The mean μ=r, and variance σ 2 =2r.  The mode, the point for the maximal p.d.f., is x=r-2  Ex.3.4-3: X has a chi-square distribution with r=5.  Using Table IV on p.685 P(X>15.09)1-F(15.09)=1-0.99=0.01. P(1.145 ≦ X ≦ 12.83)=F(12.83)-F(1.145)=0.975-0.05=0.925.  Ex.3.4-4: X is χ 2 (7). Suppose there are two constants a & b s.t. P(a<X<b)=0.95. ⇒ One of many possible is a=1.69 & b=16.01  Percentiles:  The 100(1-α) percentile is  The 100αpercentile is f(x)

14. 14 Distributions of Functions of Random Variable  From a known random variable X, we may consider another Y = u(X), a function of X, and want to know Y’s distribution function.  Distribution function technique:  We directly find G(y) = P(Y≤y) = P[u(X)≤y], and g(y)=G’(y).  E.g., finding the gamma distribution from the Poisson distribution.  Also, N(μ,σ 2 ) ⇒ N(0,1), and N(μ,σ 2 ) ⇒ χ 2 (1).  It requires the knowledge of the related probability models.  Change-of-variable technique:  Find the inverse function X=v(Y) from Y=u(X), and  Find the mapping: boundaries, one-to-one, two-to-one, etc.  It requires the knowledge of calculus and the like.

15. 15 Distribution Function Technique  Ex: [Lognormal] X is N(μ,σ 2 ). If W=e X, G(w)=?, g(w)=?  Ex3.5-2: Let w be the smallest angle between the y-axis and the spinner, and have a uniform distribution on (-π/2, π/2). G(w)= g(w)= <=Cauchy p.d.f. Both limits ⇒ E(x) does not exist g(x) (0,1) y w x

16. 16 Change-of of-variable Technique  Suppose X is a R.V. with p.d.f. f(x) with support c 1 < x < c 2.  Y=u(X) ⇔ the inverse X=v(Y) with support d 1 < y < d 2.  Both u and v are conti. increasing functions, and d 1 =u(c 1 ), d 2 =u(c 2 ).  Suppose both u and v are conti. decreasing functions:  The mapping of c 1 y > d 2  Generally, the support mapped from. G(y)= g(y)= Ex4.5-1 is an example. G(y)=…= g(y)=G’(y)

17. 17 Conversions: Any ⇔ U(0,1)  Thm.3.5-2: X has F(x) that is strictly increasing on Sx={x: a<x<b}.Then R.V. Y, defined by Y=F(X), has a distribution U(0,1).  Pf: The distribution function of Y is  The requirement that F(x) is strictly increasing can be dropped.  It will take tedious derivations to exclude the set of intervals as f(x)=0.  The change-of-variable technique can be applied to the discrete type R.V.  Y=u(X), X=v(Y): there exists one-to-one mapping.  The p.m.f. of Y is g(y)=P(Y=y)=P[u(X)=y]=P[X=v(y)]=f[v(y)], y ∈ S y.  There is no term “|v’(y)|”, since f(x) presents the probability.

18. 18 Examples  Ex.3.5-6: X is a Poisson with λ=4.  If Y=X 1/2, X=Y 2, then  When the transformation Y=u(X) is not one-to-one, say V=Z 2.  For instance, Z is N(0,1): -∞<z<∞, 0≤v<∞. [2-to-1 mapping]  Each interval (case) is individually considered.  Ex.3.5-7: X has p.d.f. f(x)=x 2 /3, -1<x<2.  If X=Y 1/2, Y=X 2, then 0≤y<4:  -1 < x 1 < 0 ⇔ 0 ≤y 1 < 1  0 ≤x 2 < 1 ⇔ 0 ≤y 2 < 1  1 ≤x 3 < 2 ⇔ 1 ≤y 3 < 4 G(v)= g(v)=

19. 19 How to find E(X) & Var(X) now  Ex3.6-6: Find the mean and variance of X in previous example.  Ex3.6-7: Reinsurance companies may agree to cover the wind damages that ranges between $2 and $10 million.  X is the loss in million and has a distribution function:  If losses beyond $10 is set as $10, then  The cases (x>10) will all be attributed to x=10: P(X=10)=1/8. F(x)= μ=E(X) σ 2 =E(X 2 )-μ 2