Mathematical Statistics. Gregg Waterman Oregon Institute of Technology

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1 Mathematical Statistics Gregg Waterman Oregon Institute of Technolog

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3 Contents Joint Probabilit Distributions 87. Discrete Joint Distributions Discrete Marginal Distributions Discrete Conditional Distributions, Independent Random Variables Continuous Joint Probabilit Densit Functions More on Continuous Joint Probabilit Epected Value and Covariance of Joint Distributions Multinomial and Multivariate Hpergeometric Distributions H Solutions to Eercises H. Chapter Solutions i

4 Joint Probabilit Distributions Performance Criteria:. Understand and work with joint distributions. (a) Create a joint distribution table for a discrete probabilit distribution of two random variables. (b) Give probabilities of events for a discrete probabilit distribution of two random variables. (c) Find marginal probabilities for a discrete joint distribution of two random variables. (d) Give the two marginal distributions for a discrete joint distribution of two random variables. (e) Give a conditional probabilit associated with a discrete joint distribution of two random variables. (f) Give a conditional probabilit distribution for a discrete joint distribution of two random variables. (g) Determine whether two random variables are independent. (h) Findthevalueofaconstant forwhichagivenfunctionoftwovariables is a joint probabilit densit function. (i) Find probabilities for a continuous joint distribution. (j) Give the two marginal distributions for a continuous joint distribution of two random variables. (k) Give a conditional probabilit distribution for a continuous joint distribution of two random variables. (l) Determine whether two continuous random variables are independent. (m) Find the epected value and covariance of a discrete joint probabilit distribution. (n) Find the epected value and covariance of a continuous joint probabilit distribution. (o) Calculate probabilities using a multinomial probabilit distribution or multivariate hpergeometric distribution. In this chapter we stud joint probabilit distributions, which arise when we are considering two random variables having the same sample space. 87

5 . Discrete Joint Distributions Performance Criteria:. (a) Create a joint distribution table for a discrete probabilit distribution of two random variables. (b) Give probabilities of events for a discrete probabilit distribution of two random variables. The table below shows political registrations of residents of the state of Montana. PARTY REGISTRATION Dem Rep Ind GENDER Male Female 7. Consider the eperiment of randoml selecting one person out of this group. What is the probabilit that the are (a) male? (b) Republican? (c) male and Republican? (d) male or Republican? (f) Republican, given that the are male? (e) male, given that the are Republican? The above tpe of table is sometimes called a two-wa table. Using terminolog analogous to what we used when we have worked with Ecel, we call each of the spaces occupied b the numbers in the table cells. The numbers in the cells are called cell counts. Note that each cell represents a particular gender/part registration combination. Considering the eperiment of randoml selecting an individual from this group, let us define two random variables X and Y. X assigns to an outcome (individual selected) a number representing their part registration (zero for Democrat, one for Republican, two for Independent). Y assigns to each outcome a zero or one depending on gender (zero for male, one for female). We can make a new table based on the values of the random variables, with the number in each cell being the probabilit that an individual who is randoml selected from the group has the gender and part registration that the cell represents. X Y

6 . Give each of the following probabilities. Take (X =,Y = ) to mean X = and Y =. (a) P(X =,Y = ) (b) P(X = ) (c) P(Y = ) (d) P(X = Y = ) (e) P(X = or Y = ) We will now define a function f of two variables, called a joint probabilit distribution (in this case, a discrete joint probabilit distribution), b f(,) = P(X =,Y = ) for =,,, =, Thus, f(,) =., f(,) =. and so on. At the risk of being a bit redundant, we now show a table of values for the function f: f(,) Eample.(a): Give each of the probabilities from Eercise in terms of the probabilit function f. The onl one that might need some eplanation is P(X = Y = ), where we must divide the probabilit P(X =,Y = ) b P(Y = ): P(X = Y = ) = P(X =,Y = ) P(Y = ) = f(,) f(,)+f(,)+f(,). In the net section we will encounter a concept that will allow us to give a simpler epression for the denominator. Here are the other probabilities in terms of f: P(X =,Y = ) = f(,) P(X = ) = f(,)+f(,) P(Y = ) = f(,)+f(,)+f(,) P(X = or Y = ) = f(,)+f(,)+f(,)+f(,). The concept from the net section will allow us to give a cleaner epression for this last probabilit as well.. Consider the eperiment of rolling a single die (one half of a pair of dice). (a) Give the sample space. (b) Define a random variable X on the sample space b X = if the number rolled is odd, X = if the number rolled is even. Give Ran(X). 89

7 (c) Define the random variable Y on the same sample space to be the number of letters in the spelling of the number rolled. For eample, Y() = because the word five has four letters. Give Ran(Y). (d) Create a table for the joint probabilit distribution f(,). It should have the form of the table on the previous page, but give all probabilities as fractions. We now make a formal definition, based on things that should be intuitivel clear. First, though, we recall that for two sets A and B, we define the Cartesian product of A and B, denoted b A B, to be the set of all ordered pairs (,) for which is in A and is in B. Recall also that Ran(X) is the range of X, and similarl for Y. Discrete Joint Probabilit Distribution Let X and Y be random variables defined on the same discrete sample space. A function f, defined on Ran(X) Ran(Y), is a joint probabilit distribution if ) f(,) for all (,) Ran(X) Ran(Y), ) f(,) =, Ran(X) Ran(Y) ) P(X =,Y = ) = f(,) for all (,) Ran(X) Ran(Y). Although the following theorem ma be obvious, we state it because of its importance when we introduce continuous joint distributions. Theorem.: Let f be a discrete joint probabilit distribution for the random variables X and Y, and let A be an subset of Ran(X) Ran(Y). Then P((X,Y) A) = f(,). (,) A. The following table gives the joint probabilit distribution function f for two discrete random variables X and Y. f(,) 9

8 Give each of the following probabilities of events, first as an epression in terms of f evaluated at specific values of and, then give a numerical answer, as a fraction. Take (X = a,y = b) to mean X = a and Y = b, and take P(X = a Y = b) to mean the conditional probabilit of X = a given that Y = b. You might consider giving one of the probabilities in terms of its complementar event. (a) P(X =,Y = ) (b) P(X = ) (c) P(X +Y ) (d) P(X = or Y = ) (e) P(X +Y ) (f) P(Y = X = ). (a) Suppose that three marbles are to be drawn without replacement from an urn containing three blue marbles, two red marbles and five ellow marbles. Let X be the number of blue marbles drawn and Y be the number of red marbles drawn. Give a table like the one above for the joint probabilit distribution, giving all probabilities in fraction form. It is suggested that ou make all fractions have denominators of 7; check our answers b making sure that the sum of all the probabilities is one. (b) Repeat part (a) under the conditions that the marbles are drawn with replacement. Here I would suggest that ou make the denominators of all the fractions. 9

9 . Discrete Marginal Distributions Performance Criteria:. (c) Find marginal probabilities for a discrete joint distribution of two random variables. (d) Give the two marginal distributions for a discrete joint distribution of two random variables. Consider again the table from the last section, but with probabilities in place of counts: PARTY REGISTRATION Dem Rep Ind GENDER Male.7.. Female...8 This table gives us probabilities of specific part/gender combinations, but does not directl give probabilities such as whether a randoml selected individual will be Republican, or be male, and so on. In terms of the random variables X and Y, we are asking for P(X = ) and P(Y = ). It should be obvious that to compute probabilities involving onl one of the random variables X or Y one needs to add up the values in each each column and row of the distribution table: X Y These additional probabilities are called marginal probabilities. The give the values of two additional distributions g and h defined b g() = P(X = ) for =,, and h() = P(Y = ) for =,. For this particular eample, g and h have the distributions g()..9. h()..8 9

10 The functions g and h are called the marginal distributions associated with the joint distribution f defined in the last section. The value of. in the lower right is the sum of the probabilities for both g and h - as it should be, the total probabilit for each of those marginal distributions is one. For an joint distribution f, we define g and h more formall as follows. Marginal Distributions Let f be a discrete joint probabilit distribution for the random variables X and Y. We define the marginal distribution functions g and h on Ran(X) and Ran(Y), respectivel, b g() = f(,) and h() = f(,). Ran(Y) Ran(X) Note carefull what these two formulas sa. To compute the value of g for a particular, we fi the value of and add up all the values of f(,) as ranges over all possible values of the random variable Y. A similar process is carried out for finding values of h. For the eample we have been looking at, this means that g() = f(,)+f(,) and h() = f(,)+f(,)+f(,). We can perhaps see this more readil b adding the marginal probabilities to our table for the distribution function f(, ): f(, ) h() g()..9. Eample.(a): Three marbles are drawn, without replacement, from an urn containing blue, ellow and red marbles. Let X be the random variable that assigns to an outcome the number of red marbles drawn, and let Y assign to an outcome the number of ellow marbles drawn. The table for the joint probabilit distribution f is shown below. Determine the marginal distributions g() and h(). f(,) 8 9

11 To get the marginal distributions, we simpl sum up the rows and columns to obtain the following table: f(, ) h() g() From this we can see that the marginal distributions are g() 7 h() 7 Eample.(b): For the distribution from the previous eample, give each of the following probabilities in terms as efficientl as possible in terms of f, g and h. P(X = ) P(X = Y = ) P(X = or Y = ) Using the definitions of f, g, h and conditional probabilit, along with the addition rule, we have P(X = ) = g() P(X = Y = ) = f(,) h() P(X = or Y = ) = g()+h() f(,). Add the marginal probabilities to our table from Eercise (d) of the previous section, then give the marginal distribution g of X and the marginal distribution h of Y.. Consider again the joint distribution for gender and political affiliation. Use to to give each of the following probabilities. (a) P(X = ) in terms of f (c) P(X =,Y = ) in terms of f (d) P(Y = X = ) in terms of f and g or h (e) P(X = or Y = ) in terms of f alone (f) P(X = or Y = ) in terms of f, g and h (b) P(X = ) in terms of g or h 9

12 . The following table gives the joint probabilit distribution function f for two discrete random variables X and Y. f(,) Give the two marginal distributions g() and h(). : : g() : h() :. Continue using the joint distribution given b the table in the previous eercise. Give each of the following probabilities in terms of the marginal distributions g and h, and use the joint distribution f as well, when necessar. Then give a numerical value for each, as a fraction. (a) P(X = ) (b) (b) P(Y = ) (c) P(Y ) (d) P(X = or Y = ) (e) P(X = Y = ) (f) P(Y = X = ). Add the marginal probabilities to our table from Eercise (a) of the previous section, then give the marginal distribution g of X and the marginal distribution h of Y. 9

13 . Discrete Conditional Distributions, Independent Random Variables Performance Criteria:. (e) Give a conditional probabilit associated with a discrete joint distribution of two random variables. (f) Give a conditional probabilit distribution for a discrete joint distribution of two random variables. (g) Determine whether two random variables are independent. Consider one last time the table for the Montanans: PARTY REGISTRATION Dem Rep Ind GENDER Male Female 7 Suppose that we wish to determine the probabilit that a randoml selected male voter from this group would be a Democrat. Because there are + + = males, the probabilit of selecting a Democrat from among them is =.7. You ll recall that what we have computed hereis aconditional probabilit, inthis casetheprobabilit that oneof thevoters isademocrat, given that the are male. Suppose that we wished to determine the same probabilit from the table of values for the joint distribution function f(, ): f(, ) h() g()..9. In this case we would compute the desired probabilit b taking.7 =.7. You can see here. that the quotient that gives us the desired probabilit is f(,) h() We would find the probabilit of a voter being Republican, given that the are male, to be f(,) and the probabilit that a voter is Independent, given that the are male, to be f(,) h() h(). These three values are the probabilities of a probabilit distribution function that we will denote b f( ). This is defined more formall on the net page. 9

14 Discrete Conditional Distributions Let f be a discrete joint probabilit distribution for the random variables X and Y with marginal distributions g and h. For a fied Ran(Y) we define the conditional distribution function v( ) on Ran(X) b v( ) = f(,) h() if h() >. Similarl, for a fied Ran(X), if g() > the conditional distribution function w( ) is defined b for all Ran(Y) w( ) = f(,) g() The purpose of writing v( ) instead of v( ) is to indicate that is a variable, whereas is fied (for the purposes of the conditional distribution). Eample.(a): Find the conditional distributions v( ) and w( ) for the probabilit distribution from Eample.(a), shown below. f(, ) h() g() Here we have v( ) = / 7 =, v( ) = 8 / 7 v( ) = / 7 = 8, so the function v( ) is given b = v( ) 8 8, v( ) = / 7 = 8 and Similar computations give us w( ) 97

15 You will recall that if two events A and B are independent, the occurrence of B makes it no more or less likel that A will occur. That is, P(A B) = P(A). This leads us to the definition that A and B are independent if, and onl if, P(A B) = P(A)P(B). We define independent random variables in a similar manner. Independent Random Variables Let X and Y be discrete random variables with joint distribution function f and marginal distribution functions g and h, respectivel. If f(,) = g()h() for all Ran(X) and Ran(Y), then we sa that the random variables X and Y are independent. Eample.(b): Determine whether the random variables X and Y from the distribution given in Eample.(a) (the table for the distribution is reproduced below) are independent. f(, ) h() g() We see that g()h() = 7 dependent. = 7 8 = f(,), so the random variables X and Y are. The following table gives the joint probabilit distribution function f for two discrete random variables X and Y. Give the conditional distributions v( ) and w( ) in the manner shown in Eample.(a). Are the random variables X and Y independent? f(,) 98

16 . Continuous Joint Probabilit Densit Functions Performance Criteria:. (h) Findthevalueofaconstant forwhichagivenfunctionoftwovariables is a joint probabilit densit function. (i) Find probabilities for a continuous joint densit function. As with the single random variable situation, when we can have two random variables that are continuous rather than discrete. In that case we again have a probabilit densit function, but of two variables. And as in the single random variable case, evaluating such a function at a single point has no real meaning - it is onl when we integrate values over some region in the R plane that we get a probabilit. That is, the probabilit has densit, but no mass at an single point. Here is the definition of such a function: Continuous Joint Probabilit Densit Function Let X and Y be continuous random variables defined on the same sample space. A function f : R R is a joint probabilit densit function if ) f(,) for all (,) R, ) f(,) dd =, ) For an subset A of R, P[(X,Y) A] = A f(,) dd Eample.(a): Find the value of c for which { c for (,) [,] [,] f() = otherwise is a continuous joint probabilit densit function. It should be clear that f meets condition () above. We must choose c so that condition () is met. We see that f(,) dd = c dd = so c must be. c [ ] d = c d = c [ ] = c, 99

17 Eample.(b): For the continuous joint probabilit densit function { f() = for (,) [,] [,] otherwise of the previous eample, find P( X, Y ). Here we appl () from the definition, where A is the set [,] [,]: P( X, ) = [ ] d = dd = [ ] d = 7 [ d = 7 ] = 7 Although the integrals for both of the previous eamples were computed b hand, I would encourage ou to use some technolog like the Wolfram Alpha R Double Integral Calculator or our handheld calculator to make such computations. Eample.(c): Find P(X + Y ) for the continuous joint probabilit densit function { e for, f() =. otherwise Setting up the iterated integral for the desired probabilit will be easier if we first determine the region in R where we are integrating. We first consider the equation + =, which is a line with -intercept (found b setting = ) two and -intercept four. Because the point (,) satisfies the inequalit + the region of interest is the triangle shown to the right (because we onl have nonzero probabilit in the first quadrant). We then see that the probabilit is given b either of the iterated integrals e dd = resulting in P(X +Y ) = e.98. e dd, Eample.(d): For the continuous joint probabilit densit function f() = { for (,) [,] [,] otherwise, find P(Y X +).

18 Because the densit function is nonzero onl on the rectangular region [,] [,], we need onl integrate over the portion of that region that satisfies the inequalit +. The graph of the line = + has slope one and -intercept one and we are considering the points in the rectangle that are below that line, as shown in the graph to the right. Thus the desired probabilit is obtained b integrating over that region: P(Y X +) = + dd = We should make two observations regarding the previous eercise: If we were to change the order of integration, two iterated integrals would be required. This is because when is a fied value between zero and one we enter the region at = and leave at =, but if is a fied value between one and two we enter at = and leave at =. Therefore we would have P(Y X +) = dd + dd Because the densit function is nonzero onl on the rectangular region [,] [,], We could have instead obtained the desired probabilit b integrating over the triangle shown to the right and subtracting from one: P(Y X +) = + dd = 9 =

19 . More on Continuous Joint Probabilit Performance Criteria:. (j) Give the two marginal distributions for a continuous joint densit function of two random variables. (k) Give a conditional probabilit distribution for a continuous joint densit function of two random variables. (l) Determine whether two continuous random variables are independent. Marginal Distributions Let f be a continuous joint probabilit densit function for the random variables X and Y. We define the marginal distribution functions g : R R and h : R R, respectivel, b g() = f(,) d and h() = f(,) d Eample.(a): Find the marginal distributions g and h for the continuous joint probabilit densit function { for (,) [,] [,] f() = otherwise From the above definitions we have and g() = h() = d = [ ] = for [,], otherwise d = [ ] = for [,], otherwise

20 Conditional Distributions Let f be a continuous joint probabilit densit function for the random variables X and Y. For a fied R with h() > we define the conditional distribution function v( ) : R R b v( ) = f(,) h() The notation is meant to indicate that is fied and the is a variable. Similarl, for a fied R with g() > the conditional distribution function w( ) : R R is defined b w( ) = f(,) g() Eample.(b): Find the conditional distributions v( ) and w( ) for the continuous joint probabilit densit function { for (,) [,] [,] f() = otherwise Appling the above definitions and using the results of Eample.(a) we have and v( ) = f(,) h() w( ) = f(,) g() = = for [,], otherwise = = for [,], otherwise. We should make particular note of the results of this last eample. Looking at the first part, we see that v( ) = f(,) = g(). h() This tells us that computing probabilities for the random variable Y for an fied value of the random variable Y is the same as computing probabilities of X regardless of the value of Y; that is, X seems to be independent of Y. Multipling both sides of the second equalit b h() results in f(,) = g()h() which, as in the case of two discrete random variables, we take to be the definition of independence:

21 Independent Random Variables Let X and Y be continuous random variables with joint distribution function f and marginal distribution functions g and h, respectivel. If f(,) = g()h() for all (,) R, then we sa that the random variables X and Y are statisticall independent. Eample.(c): The continuous joint probabilit densit function for two continuous random variables X and Y is { f() = ( +) for (,) [,] [,] otherwise Are X and Y independent? First we see that and g() = h() = ( +) d = [ + ] = ( + ) for [,], otherwise ] ( +) d = [ + = ( + ) for [,], otherwise It looks doubtful that f(,) = g()h(), but I suppose we should check to be sure: g()h() = ( + ) ( + ) = ( ) ( +) Therefore the random variables X and Y are not independent. Consider the following joint probabilit densit function for two continuous random variables X and Y: { (+) for (,) [,] [,] f(,) = otherwise. Find P(X, Y ). Give the integral used, and evaluate it b hand, giving its value in fraction form. Check our answer using a TI-89 or the Wolfram online double integral calculator (do a search for online double integral calculator.. Find P(X + Y ) as follows: (a) Sketch the region over which the probabilit densit function must be integrated to find the desired probabilit. (b) Set up the integral. (c) Evaluate the integral using a TI-89 or the Wolfram double integral calculator. You should 7 get 8 ; if ou don t, tr to correct our integral so that ou do.

22 . Find and simplif the marginal distribution h(), then use it to find P( ).. Give the conditional distribution f X Y ( ) in simplified form.. Use our answer to (d) to find P(X over the desired range of values. Y = ). Do this b setting = and integrating

23 . Epected Value and Covariance of Joint Distributions Performance Criteria:. (m) Find the epected value and covariance of a discrete joint probabilit distribution. (n) Find the epected value and covariance of a continuous joint probabilit distribution. Epected Value of Joint Random Variables Let X and Y be discrete random variables with joint distribution function f. The epected value of the distribution, denoted µ XY = E(XY) is E(XY) = Ran(X) Ran(Y) f(,). If X and Y are continuous random variables with joint distribution function f, the epected value µ XY = E(XY) of the distribution is E(XY) = f(,) dd. Eample.(a): Find the epected value of the joint probabilit distribution, for discrete random variables X and Y, given in the table below. f(,) E(XY) = ()()(.7) +()()(.) +()()(.) + ()()(.) +()()(.) +()()(.8) =.

24 Eample.(b): Find the epected value of the joint probabilit distribution, for continuous random variables X and Y, given b { f() = ( +) for (,) [,] [,] otherwise E(XY) = f(,) dd = ( + ) dd = ( +) dd = [ + ] [ ( + ) d = 8 + ] = d = Covariance of Joint Random Variables Let X and Y be discrete random variables with joint distribution function f. Let µ X and µ Y be the epected values of the marginal distributions g() and h(). The covariance σ XY of X and Y is σ XY = E[(X µ X )(Y µ Y )] = ( µ X )( µ Y )f(,). Ran(X) Ran(Y) If X and Y are continuous random variables, then the covariance is given b σ XY = E[(X µ X )(Y µ Y )] = ( µ X )( µ Y )f(,) dd. As with the variance of a single random variable, the definition of the covariance of a joint distribution does not give the most efficient wa to compute the covariance. Forthat we turn to the following result. Theorem. Let X and Y be random variables (discrete or continuous) with joint distribution function f. The covariance of X and Y is given b σ XY = E(XY) µ X µ Y. 7

25 Eample.(c): The joint probabilit distribution from Eample.(a) is given below, with the marginal distributions added. Find the covariance of the random variables X and Y. f(, ) h() g()..9. We can see that the epected values for X and Y are µ X = ()(.) +()(.9) +()(.) =.79 and µ Y = ()(.) +()(.8) =.8 We computed the epected value in Eample.(a), so we can now compute σ XY = E(XY) µ X µ Y =. (.79)(.8) =.. The table below and to the right gives the joint probabilit distribution for two random variables X and Y. Use it for the following. (a) Give g() and h() in table form, with all values reduced. (b) Compute g(). Reduce our answer. (c) Compute = = = f(,); reduce again. (d) Your answers to (b) and (c) should be the same, and both are µ X. µ Y can be computed analogousl, using either method. Find µ Y. (e) Find the epected value E(XY). (f) Find the covariance σ XY using the formula σ XY = E(XY) µ X µ. (g) Are X and Y independent? Show/eplain. This requires four mathematical statements!. Give the epected value and covariance for the following joint probabilit densit function for two continuous random variables X and Y: { (+) for (,) [,] [,] f(,) = otherwise f(,) 9 8

26 .7 Multinomial and Multivariate Hpergeometric Distributions Performance Criteria:. (o) Calculate probabilities using a multinomial probabilit distribution or multivariate hpergeometric distribution. Consider an eperiment with the following characteristics: It consists of n repeated trials. Each trial can result in an of k outcomes E,E,E,...,E k The probabilities of the outcomes of the outcomes, which we will denote p,p,p,...,p k, respectivel, remain the same for each trial. The outcome of an single trial is independent of the outcomes of the previous trials. We will refer to such an eperiment as a multinomial eperiment. Note that a Bernoulli eperiment is simpl a multinomial eperiment for which k =. Multinomial Probabilit For a multinomial eperiment with outcomes and probabilities as defined above, let X,X,X,...,X k be random variables that assign the number of times the outcomes E,E,E,...,E k occur, respectivel. Then the joint probabilit distribution for these random variables is ( ) n f(,,,..., k ;p,p,p,...,p k ;n) = p,,..., p p k k. k where ( n,,..., k ) = n!!! k! It is instructive to compare this with the formula for the binomial distribution. The multinomial distribution would appl to an eperiment like drawing ten marbles, with replacement, from an urn containing red marbles, blue marbles, ellow marbles and green marbles. In that case, the probabilit of getting two red, two blue, five ellow and one green marble is f(,,,;,,, ( )! ( ) ( ) ( ) ;) =.!!!! Suppose that we had the same urn, same number of draws, but we were drawing without replacement. You might guess that some variation of the hpergeometric distribution would be used, and ou are correct. Note that the hpergeometric situation can be interpreted as follows: We are drawing from N objects with onl two possible outcomes, with respective numbers k and 9

27 N k in the original urn. We draw n objects and want the probabilit of getting of the first outcome and n of the second outcome. The probabilit is given b ( )( ) k N k n ( ) N n Lets change notation a bit b letting a be the number of the N total objects that are of the first kind, and let a be the number of objects of the second kind. Also, let and be the numbers of each of those kinds of objects that we ant the probabilit of obtaining when we draw n objects. Then the above epression becomes ) ( a )( a ( N n Now suppose we have N objects of k tpes, a of tpe A, a of tpe A, and so on through a k of tpe A k, and we are selecting n objects, without replacement. ) Multivariate Hpergeometric Distribution For aneperiment as justdescribed, let X,X,X,...,X k berandomvariables that assign the number of objects of tpes A,A,A,...,A k selected, respectivel. Then the joint probabilit distribution for these random variables is ( )( ) ( ) a a ak k f(,,,..., k ;a,a,a,...,a k,n,n) = ( ). N n

28 H Solutions to Eercises H. Chapter Solutions Section.. (a) ++ = = (b) + = 78 = 9 (c) = 9 (d) +++ = 7 (e) + = 78 = (f) ++ =. (a) P(X =,Y = ) =. (b) P(X = ) =.+. =.9 (c) P(Y = ) = =.8 (d) P(X = Y = ) = =.. =. (e) P(X = or Y = ) = =.. (a) S = {,,,,,} (b) Ran(X) = {,} (c) Ran(Y) = {,,} (d) See to the right. f(,). (a) P(X =,Y = ) = f(,) = = (b) P(X = ) = f(,)+f(,)+f(,) = + + = 7 (c) P(X +Y ) = f(,+f(,)+f(,) = + + = = (d) P(X = or Y = ) = f(,)+f(,)+f(,)+f(,)+f(,)+f(,) = = (e) P(X +Y ) = [f(,)+f(,)+f(,)] = [+ +] = f(,) (f) P(Y = X = ) = f(,)+f(,)+f(,) = + + = =. (a) f(,)

29 (b) f(,) Section.. (a) (b) g() h(). (a) P(X = ) = f(,)+f(,) (b) P(X = ) = g() (c) P(X =,Y = ) = f(,) (d) P(Y = X = ) = f(,) g() (e) P(X = or Y = ) = f(,)+f(,)+f(,)+f(,) (f) P(X = or Y = ) = g()+h() f(,). g() 7 h() 7. (a) P(X = ) = g() = = (b) (b) P(Y = ) = h() = = 8 (c) P(Y ) = h()+h() = + 7 = = 7 8 (d) P(X = or Y = ) = g()+h() f(,) = + 7 = = (e) P(X = Y = ) = f(,) h() = 7 = 7 (f) P(Y = X = ) = f(,) g(). (a) (b) g() h() = = Section.. v( ) w( ) Because f(,) = = = g()h() X and Y are not independent.

30 Section.. P(X, Y ) =. (a) See graph to right. (b) P(X +Y ) =. h() = P( Y ) = f(,)d =. (a) v( ) = f(,) h() h()d = 8 = + + (b) P(X Y = ) = (+)dd = 7 (+)dd (+)d = (+), v( )d = 8 Section.. (a) (b) g() g() = = h() 8 (c) = = (e) E(XY) = 9 (f) σ XY = 8 f(,) = (d) µ Y = (g) f(,) = = 8 = g()h(), f(,) = = 8 = g()h(), f(,) = = 8 = g()h(), f(,) = 9 = 8 = g()h() h() = 8 =. g() = + for [,], elsewhere, h() = + for [,], elsewhere µ X = E(X) = 7, µ Y = E(Y) = 7, σ XY = E(XY) µ µ = 7 7