Relationship between probability set function and random variable - 2 -

Transcription

1 2.0 Random Variables A rat is selected at random from a cage and its sex is determined. The set of possible outcomes is female and male. Thus outcome space is S = {female, male} = {F, M}. If we let X be a function defined on S such that X(F) = 0 and X(M) = 1, X is a real-valued function and S is a domain. X: S = {F, M} {0, 1} ( R) 1) random variable Definition For sample space, a function, which assigns to only one number i.e., : random variable ** Random variable is defined on the element of sample space. The probability set function, P, is defined on the subset of sample space. In our notation, we define that for A R X -1 {A} {c S: X(c) A} : make sure that X -1 {A} indicates the inverse image of A P(X -1 {A}) P(X A) P X (A) For example, {c S: X(c) a} = X -1 {(, a]} - 1 -

2 Relationship between probability set function and random variable - 2 -

3 Example: tossing a coin When we toss a coin, we are usually interested in the total number of spots on the sides that are "up" are counted. If we define X(H) = 1 and X(T) = 0, P X (1) = P(X = 1) = P(X -1 {1}) = P({c S : X(c) = 1}) = P({H}) - 3 -

4 Example: when we cast a die, (i) we define X(1) = 1, X(2) = 2,..., X(6) = 6. Then, P(X 4) = P(X -1 {(, 4]}) = P({c S : X(c) 4}) = P({1, 2, 3, 4}) (ii) we define X(c) = 4 if c is a even number and otherwise X(c) = 5. Then, P(X 4) = P(X -1 {(, 4]}) = P({c S : X(c) 4}) = P({2, 4, 6}) Example (uniform distribution) Let be the number of upfaces when two coins are cast. It is known that P(H) = P(T) = 0.5. Then, find P(X (-, 1/4]). (sol) P(X (-, 1/4]) = P({c S : X(c) 1/4}) = P({TT}) = 1/4 Example (uniform distribution) Let be the number of upfaces when two coins are cast. It is known that P(H) = P(T) = 0.5. Then, find P(X x). (sol) P(X (-, x]) = P({c S : X(c) x}) (i) x < 0, P(X x) = 0 (ii) 0 x < 1, P(X x) = 1/4 (iii) 1 x < 2, P(X x) = 1/4 + 1/2 = 3/4 (iii) 2 x, P(X x) = 1/4 + 1/2 = 1-4 -

5 Example: Point Chosen at Random in the Unit Circle : the distance of the selected point from the origin Find sample space and P(X x). P(X x) = P(X -1 {(, x]}) = P(C) = πx 2 /π = x 2-5 -

6 Example: casting a die,. There are two major difficulties. a) In many practical situations, the probabilities assigned to the events are unknown. (i) We can estimate it by the use of the relative frequency. (ii) Under the certain assumptions, we can estimate it. In general, if we know the distribution of X, we can estimate the parameters and we can calculate P(X<c). b) There are many ways of defining a random variable X on S. (i), (ii) space of the set of real numbers Sp types of random variable a) : discrete r.v. Sp: a countable set b) : continuous r.v. Sp: an interval of real numbers c.f.: countable: one-to-one with positive integers - 6 -

7 2) probability distribution function Roll a four-sided die twice and let X equal the larger of the two outcomes if they are different and the common value if they are same. Then, (1) the outcome space: (2) what are P(X = 1), P(X = 2), P(X = 3) and P(X = 4)? It can be written as for, and it is called probability function. a) probability mass function and probability density function (1) probability mass function (pmf) For : discrete r.v., Sp ow where Sp. Example (uniform distribution) Let be the number of the upfaces on a roll of a dice. Find sample space, space of, and pdf. (sol) S = {1, 2, 3, 4, 5, 6}, Sp(X) = {1, 2,..., 6}, - 7 -

8 for Example (uniform distribution) When we cast a dice, let be 1 for the even number of the upfaces on a roll of a dice and otherwise 0. Find sample space, space of, and pdf. (sol) S = {1, 2, 3, 4, 5, 6}, Sp(X) = {0, 1}, for For a given probability set function, its probability mass(density) function can be different, depending on definition of a random variable. Example (hypergeometric distribution) Consider a collection of N = N 1 + N 2 similar objects, N 1 of them belonging to one of two dichotomous classes (red chips) and N 2 of them belonging to the second class (blue chips). A collection of n objects is selected from these N objects at random and without replacement. Find the probability that exactly x of these n objects are red. (sol) - 8 -

9 for (2) probability density function (pdf) For : continuous r.v., for on where and are any real numbers. : probability density function of Support of, a) : discrete r.v. b) : continuous r.v. Example: Point Chosen at Random in the Unit Circle : the distance of the selected point from the origin - 9 -

10 Find sample space, space of, and pdf. (sol) S = {x: x unit circle}, Sp(X) = [0, 1] If, It should be noted that there are several functions that satisfies this equation. b) cumulative distribution function cumulative distribution function (cdf) of When : discrete r.v. and, When : continuous r.v.,

11 Example: : a real number chosen at random between 0 & 1 Find cdf of. (sol) S = {x: x (0, 1)}, S(X) = (0, 1) if if if Basic properties of cdf : cdf of, (1) If, ( : non-decreasing function) (sol) (, a] (, b]. Thus, X -1 {(, a]} X -1 {(, b]} (2) (sol) F(x) = P(X -1 {(, x]}) 0 and F(x) = P(X -1 {(, x]}) P(S) = 1 (3) lim, lim (sol) lim {c S: X(c) k}: increasing P({c S: X(c) (, )}) P(S) =

12 lim {c S: X(c) k}: increasing P(Ø) 0 (4) lim ( : right continuous) (sol) lim lim lim D n {c S: X(c) x + 1/n}: decreasing

13 Theorems w.r.t. cdf (1) ; however, the converse does not hold. (2) (sol) Let A {c S: X(c) (, a]} and B {c S: X(c) (, b]} Then, A and B A c : mutually exclusive We have P(B A c ) P(B) P(A) F X (b) F X (a) and P(B A c ) P({c S: X(c) (a, b]}) P(a < X b). (3) where lim. lim lim Thus, (4) : continous r.v. where is differentiable Example : the life time in years of a mechanical part Find pdf

14 (sol) f X (x) e -x for x 0 and otherwise f X (x) 0. Example Find and where has the discontinous cdf. (sol) lim

15 2.1 Discrete Random Variables 1) Properties of pmf a) b) c) Example: Suppose r.v. has the pmf ow (sol). Example: (Geometric distribution) Consider a sequence of independent flips of a coin : the # of flips needed to obtain the first head. Find the space, pmf of, and the prob. that the first heads appears on an odd number. (sol)

16 Sp(X) = {1, 2,... } Example: (Hypergeometric distribution) From 100 fuses where there are 20 defective fuses, inspect five of them at random Then let X be the number of defective fuses among the 5 that are inspected. What are the pmf and space of X (sol) for x = 0, 1,..., 5 and otherwise px (x) = 0. Sp(X) = { 0, 1,..., 5} Definition of Expectation : discrete If has a pmf and,

17 Properties of Expectation (1) If c is a constant, then E(c) = c. (2) If c is a constant and u is a function, then E[cu(X)] = c E[u(X)] (3) If c 1 and c 2 are constants and u 1 and u 2 are functions, then E[c 1 u 1 (X) + c 2 u 2 (X)] = c 1 E[u 1 (X)] + c 2 E[u 2 (X)] Example: Let r.v. have the pmf, Find (sol)

18 Example: Let u(x) = (x b) 2, where b is a constant. Find b that minimizes E[u(X)]. (sol) Thus it is minimized at b = E(X) Mean, Variance, Standard deviation a) Definition (i) mean : (ii) variance : (iii) standard deviation : (iv) rth moment around b : b) Properties (1) (2) (3)

19 transformation (a) Space of : Sp Sp (b) pdf of : : 's pdf, transformation: case 1 When is one-to-one from Sp to Sp and pmf of X is p X (x), the pmf of Y is. (Proof) Example: (Geometric distribution) : flip number on which the first head appears : flip number before the first head : (sol). Thus, we have if {1, 2,...} if {1, 2,...}

20 Example: Suppose has the pmf Find pmf of r.v.. (sol) Thus, we have. ow if transformation: case 2 When is not one-to-one, the pmf of Y is (Proof) P({c S: Y(c) = y}) = P({c S: g(x(c)) = y}) = P({d 1, d 2, d 3,... }) if we denote {c S: g(x(c)) = y} = {d 1, d 2, d 3,... }

21 Example: Let When r.v., find pmf of. (Sol) Sp(Z) = {0, 1, 4, 9,... } and if or if if

22 if if if Example: Let and where X i are independent. Then if, find pmf of. (Sol) Bernoulli distribution

23 Bernoulli experiment Random experiment of which the outcome can be classified in one of two mutually exclusive and exhaustive ways. That is, S = {success, failure} Bernoulli trial to perform a Bernoulli experiment several independent times Bernoulli random variable a r.v. associated with a Bernoulli trial by defining it as follows: {success}, {failure} definition (Bernoulli distribution) ; P(X=1) = pmf:, mean : variance : var Binomial distribution In a sequence of Bernoulli trials, we are often interested in the total number of successes and not in the order of their occurrence. binomial theorem

24 definition mean :, where, variance : (Proof 1) (i) (ii)

25 var (Proof 2) If we let Z i ~ Bernoulli, then X = Z Z n and E(Z i ) = p & var(z i ) = p(1 p). varvar var Example Suppose that an urn contains N 1 success balls and N 2 failure balls. Let X equal the number of success balls in a random sample of size n that is taken from this urn. If the sampling is done with replacement, then what is the pmf of X? If the sampling is done without replacement, then what is the pmf of X? *** When N 1 + N 2 is large and n is relatively small, it makes little difference if the sampleing is done with or without replacement. That is, binomial distribution and hypergeometric distribution are similar

26 N_1=8, N_2=32, n=8 x y N_1=16, N_2=64, n=8 x y N_1=16, N_2=64, n=16 x y N_1=32, N_2=128, n=16 x y Blue: hypergeometric, green: binomial

27 Geometric distribution In a sequence of Bernoulli trials, we are often interested in the total number of Bernoulli trials needed to get the first success. definition X ~ Geo(p) pmf: ; P(X=1) =, mean : variance : (1) (Proof) If we let Then, lim lim

28 Therefore, lim lim lim 1 When c > 1, ln ln ln ln for x > 0 lim lim lim ln lim ln ln lim 2 When c < 1, then 1/c > 1 and lim lim

29 (2) If we let, lim Thus, lim lim

30 because we showed that lim As a result, in (1). By Taylor expansion, for p < 1 Negative Binomial distribution In a sequence of Bernoulli trials, we are often interested in the total number of Bernoulli trials needed to observe the rth success. definition X ~ Negbin(r, p) pmf:, mean : variance :

31 By Taylor's theorem,, for p <

32 moment generating function (mgf) th moment : Definition of mgf of : Let be a discrete r.v. such that for some and. Then, mgf of : Example: Let have the pdf. Find the mgf of (Use the fact that (sol) For any t > 0, ) lim lim and lim lim If t > 0, this series diverges. However M X (t) should be constant for -h < t < h

33 for some h>0. Therefore, M X (t) does not exist. Relationship between mgf and moment ( ) For any type of random variable X, where exists.,, for some Example: Let have the pdf. Find the mgf of, and calculate E(X) and E(X 2 ). (Sol)

34 Example: negative binomial distribution Consider the situation in which we observe a sequence of Bernouli trials until exactly r successes occur, where r is a fixed positive integer. Let the random variable, X, denote the number of trials needed to observe the rth success. Find its pmf and calculate mgf. (Sol) log ** By Taylor's theorem,, p <

35 Uniqueness of mgf (it is also true for continuous random variable) for some and Example: Let the moments of X be defined by Then what are P(X = 0) and P(X = 1)? (sol) Thus, P(X = 0) = 0.2 and P(X = 1) = 0.8 because the equivalence of cdf for the discrete random variable indicates the equivalence of the pmf. Poisson Distribution Example of Poisson Distribution Some experiments result in counting the number of times particular events occur at given times or with given physical objects

36 - The number of phone calls arriving at a switchboard between 9 and 10 AM. - The number of flaws in 100 feet of wire - The number of customers that arrive at a ticket window between 12 noon and 2PM. We can have an approximate Poisson process with parameter λ if the following conditions are satisfied. (a) The numbers of changes occurring in nonoverlapping intervals are independent. (b) The probability of exactly one change occurring in a sufficiently short interval of length h is approximately λh. (c) The probability of two or more changes occurring in a sufficiently short interval is essentially zero. Under these conditions, we assume that the probability of exactly one change in unit interval is λ. Then, by the condition (c), the one change in a sufficiently short interval of length 1/n follows Bernoulli distribution and its probability is λ/n. Thus the probability of the one change in a sufficiently short interval of length 1/n is λ/n. lim P(we observe x changes from n units with 1/n length) lim (sol) lim lim

37 lim (i) lim (ii) lim (iii) lim Thus, lim - Approximation: if n is large and p is small, definition (Poisson distribution),, **** (by Taylor expansion) mgf : exp,

38 (sol) exp mean : (sol) variance : (sol) Example For Poisson, compute. (sol) Example We assume that the number of people who comes to bank is 4 per an hour. What is the probability that a new person comes to the bank after one person came to the bank in 30 minutes? (sol) X: the number of people who comes to bank per hour Y: the number of people who comes to bank per half an hour

39 Then, X ~ Poisson(4) and Y ~ Poisson(2). P(Y>0) = 1 P( Y = 0 ) = Example A collection of 1000 parts is shipped to a company. A sampling plan dictates that n = 100 parts are to be taken at random and without replacement and the collection is accepted if no more than 2 of these 100 parts are defective. Denote the proportion of defective parts by p. Then what is the P(X 2) by using both approximation of hypergeometric distribution to binomial distribution and approximation of binomial distribution to Poisson distribution. (sol) If we let p be the fraction defective in the collection, By the approximation,. Thus, if p = 0.01,