C-N M406 Lecture Notes (part 4) Based on Wackerly, Schaffer and Mendenhall s Mathematical Stats with Applications (2002) B. A.

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1 Lecture Next consider the topic that includes both discrete and continuous cases that of the multivariate probability distribution. We now want to consider situations in which two or more r.v.s act in conjunction with each other. It is not hard to see the myriad of applications for this; temperatures on a hot plate, price predictions based on time and place, or any other probability based on a multidimensional vector. Of course, we will begin our study with the 2--3 dimensional function. This is a function whose domain is of dim = 2, range is of dim =, and whose overall graph is of dim = 3. Definition 5. (p. 2) tells us that for discrete random variables Y, and Y 2, the joint probability function (jpf) is given by p(y, y 2 ) = P(Y = y and Y 2 = y 2 ) for y, y 2 R 2. Theorem 5. says that any jpf p(y, y 2 ) has the following properties.... p(y, y 2 ) 0 for all y, y 2 R Σ p(y, y 2 ) =. Lecture Now, this is good for discrete r.v.s, but Definition 5.2 (p. 22) is for discrete or continuous r.v.s. BUT, for any r.v.s Y, and Y 2, the joint cumulative distribution function (jcdf), is given by F(y, y 2 ) = P(Y y and Y 2 y 2 ) for y, y 2 R 2. Definition 5.3 then says that if for continuous r.v.s Y and Y 2 with jcdf F(y, y 2 ), there exists nonnegative function f(y, y 2 ) such that F(y, y 2 ) = y y2 f(x, x 2 ) dx 2 dx for y, y 2 R 2, then Y and Y 2 are said to be jointly continuous r.v.s and f(y, y 2 ) is called the joint probability density function (jpdf). Note that in the continuous case the probability is represented by volume for the 2--3 dimensional jpdf. Remember that in the case of the --2 dimensional pdf probability was represented by area. Is there a pattern here? Absolutely! The last dimension number of the graph gives the dimension for the probability. Theorems 5.2 and 5.3 (p. 24) give some important properties for jcdf s. Using the extended Real numbers (R ext ) for r.v.s Y and Y 2 having jpdf f (in the joint continuous case) and jcdf F (regardless), we see that. F(-,- ) = F(-, y 2 ) = F(y, - ) = F(, ) =. 3. F(y, y 2 ) 0. An interesting probability form is given here. For y b y a, and y 2b y 2a ( R), P(y a < Y < y b and y 2a < Y 2 < y 2b ) = F(y b, y 2b ) F(y b, y 2a ) F(y a, y 2b ) + F(y a, y 2a ). An illustration of the r.v. space is helpful. It is easy to see that the value of F in the upper right corner of the rectangle must be decreased by each of the values at the upper left and lower right corners. Then the value at the lower left corner needs to be added back since it s been discounted twicet. Now, let s do some examples. S

2 Ex. In problem 5.6 (p. 29) we have the jpdf f(y, y 2 ) = ky y 2 x I (y, y2) (0, ) 2. In this instance, the area (0, ) 2 is called the support of the jpdf. This is the positive value location for the function. Integration of f on the support yields 0 0 ky y 2 dy dy 2 = k/4 =. => k = 4. => f(y, y 2 ) = 4y y 2 x I (y,y2) (0, ) 2. To find the jcdf, we really only need find the functional value on the support and utilize the properties of Theorems 5.2 and 5.3. So, for (y, y 2 ) (this is a vector) (0, ) 2, y2 y 0 0 4x x 2 dx dx 2 = 4(y 2 /2) (y 22 /2) = y 2 y 22. So the final result is 2 2 F(y, y 2 ) = y x I (y, y2) ((0, ) X (, ))+ y 2 x I (y, y2) ((, ) X (0, )) 2 + y 2 y 2 x I (y, y2) (0, ) 2 + I (y, y2) (, ) 2. Whew!!! I m glad we have indicator functions! Here's a graph of F, the cumulative distribution function illustrated in light brown (right). So P(Y <.5 and Y 2 <.75) = F(.5,.75) =.5 2 x.75 2 =.406. [] Ex. In problem 5.0 (p. 220) we have the vector (Y, Y 2 ) representing proportions of component chemicals in an insecticide, and their jpdf is given by f(y, y 2 ) = 2 x I (y, y2) (S) where S is the support. f is illustrated (right) with support in light orange. The actual jpdf looks like this when viewed from the third quadrant of R 2 : We are asked to find P(Y and Y 2 <.75) and P(Y and Y 2 <.5). We'll start by locating the vector endpoint (.75,.75) in R 2. Next we'll use the symmetry of the support and integrate accordingly. P(Y and Y 2 <.75) = F(.75,.75) -y = dy 2 dy -y = ( 2y 2 ] 0 dy = ( 2 2y )dy = 2(.0625) =.875. The second probability is found by noting that the vector endpoint (.5,.5) is found on the boundary of S so the volume determined by F(.5,.5) is actually half of the total volume ( cu). Hence, P(Y and Y 2 <.5) =.5. Consider what this means in terms of the problem, and the nature of the jcdf F. [] 2S

3 Student Problem set H H. Given here is the joint probability function associated with data obtained in a study of automobile accidents in which a child (under age 5) was in the car and at least one fatality occurred. Probabilities are given in gold. Specifically, the study focused on whether or not the child survived and what type of seat belt (if any) he or she used. Y = {0, if the child survived Y 2 = {, if not {0, if no belt used {, if adult belt used {2, if car-seat belt used Note that Y is the number of fatalities per child and, since children's car seats use two belts, Y 2 is the number of seat belts in use at the time of the accident. a) Verify that the preceding probability function satisfies theorem 5. (p ). b) Find F(,2). What is the interpretation of this value? c) Find F(0, 2). What is the interpretation of this value? d) Find P(Y = 0 Y 2 = 2) and interpret this result. H2. Suppose that the random variables Y and Y 2 have joint probability density function f(y,y 2 ) given by f(y,y 2 )= {6y 2 y 2, 0 y y 2, y + y 2 2 {0, elsewhere a) Verify that this is a valid joint density function. b) What is P(Y +Y 2 ). H3. Use the joint probability function from H regarding the correlation between car accident fatalities and the use of seat belts. a) Give the marginal probability functions for Y and Y 2. b) Give the conditional probability function for (Y 2 Y = 0). c) What is the probability that a child survived given that she wore a car-seat belt? H4. Let Y and Y 2 ~ iid U(0, ), and let 3 random observations be taken from each variable. Denote by Y (max), and Y 2(max) represent the maximum observation for each variable. Find the jpdf and jcdf for the vector <Y (max), Y 2(max) >. Student Answer Set H H. a) Notice that all of the probabilities are least 0 and sum to. b) Note F(,2) = P(Y, Y 2 2) =. The interpretation of this value is that every child in the experiment either survived or didn't and used either 0,, or 2 seat belts. H2. a) Clearly, f(y,y 2 ) 0. Then we take the double integral of our joint density function from limits 0 to and y to 2-y. Doing so gives us, meaning that the joint density function is valid. b) P(Y + Y 2 < ) is the double integration of our joint density function from 0 to ½ and y to -y = /8 6/64 = /32 = H3. a) P(Y =0)=.76, P(Y =)=.24 P(Y 2 =0)=.55, P(Y 2 =) =.6, P(Y 2 =2)=.29 b) P(Y 2 =0 Y =0)=P(Y 2 = 0, Y = 0)/P(Y = 0)=.38/.76=.5, P(Y 2 = Y = 0) =.8, P(Y 2 =2 Y =0)=.32 3S

4 c) The desired probability is P(Y =0 Y 2 =2)=.24/.29 = [] Lecture Now that we have mastered the basics of multidimensional probability we are ready for some extensions into conditional probability. First, some definitions which we shall present in the continuous cases (the discrete cases can be defined in the appropriate way). Definition 5.4 (p. 223) tells us that if Y, and Y 2 are jointly continuous random variables with jpdf f(y, y 2 ), then the marginal density functions for Y and Y 2 respectively are f (y ) = f(y, y 2 )dy 2 and f 2 (y 2 ) = f(y, y 2 )dy. It is interesting to note again that the dimensionality is consistent. In the case of the --2 dimensional pdf, probability = area, and the height of the pdf is meaningless because it represents a line segment which has 0 area. Here, in the 2--3 dimensional case, we see that probability = volume, and the value of the marginal pdf is given by area (having 0 volume), but serving the same purpose. In the discrete case, this can be seen quite well in a contingency table. Recall that in that case using the jpmf we had p(y y 2 ) = p(y, y 2 )/p(y 2 ) for p(y 2 ) > 0. Well, the same type of thing occurs for the jointly continuous r.v.s. Definition 5.6 (p. 227) says that if Y, and Y 2 have jpdf f(y, y 2 ), then the conditional distribution function of Y given Y 2 is F(y y 2 ) = P(Y y Y 2 = y 2 ) (note that y 2 is a constant). Similarly, definition 5.7 (p. 228) gives the conditional density function for Y given Y 2 by f(y y 2 ) = f(y, y 2 )/ f 2 (y 2 ) and is only defined for y 2 such that f 2 (y 2 ) > 0. Clearly, we must now pay close attention to our indicator function. Ex Problem 5.22 (p. 230) gives us an excellent starting point. Here, recall from problem 5.6 that f(y, y 2 ) = 4y y 2 x I (y, y2) (0, ) 2. We will proceed to find the marginal and conditional densities for f. f (y ) = 4y y 2 dy 2 = 4y (y 22 /2] 0 = 2y x I y (0, ), and f 2 (y 2 ) = 4y y 2 dy = 4y 2 (y 2 /2] 0 = 2y 2 x I y2 (0, ). From this we can obtain P(Y <.5 Y 2 >.75) = P(Y <.5, and Y 2 >.75)/P(Y 2 >.75) = ( y y 2 dy dy 2 )/.75 2y 2 dy 2 =.094/.4375 =.25. The conditional densities are now easy to obtain. f(y y 2 ) = f(y, y 2 )/ f 2 (y 2 ) = (4y y 2 x I (y, y2) (0, ) 2 )/( 2y 2 x I y2 (0, )) = 2y x (I y (0, ) x I y2 (0, ))/( I y2 (0, )) = 2y x I (y, y2) (0, ) 2 (why?). Similarly, f(y 2 y ) = f(y, y 2 )/ f (y ) = 2y 2 x (I y (0, ) x I y2 (0, ))/( I y2 (0, )) = 2y 2 x I (y, y2) (0, ) 2. We are now asked to find P(Y <.75 Y 2 =.5). This is given by -.75 f(y y 2 )dy = y x I (y, y2) (0, ) 2 dy = y dy = The moral of the story, again, is that indicator functions are terrific! [] 4S

5 Ex. In prob we have that Y and Y 2 are jointly continuous and the jpdf is given by f(y, y 2 ) = 6y 2 y 2 x I (y, y2) (S), where S is given by The marginal densities are found by 2-y f (y ) = f(y, y 2 )dy 2 = y 6y 2 y 2 dy 2 x I y (0, ) = 2y 2 ( y ) x I y (0, ) (we see here that Y ~ β(3, 2)), and f 2 (y 2 ) = f(y, y 2 )dy y2 2-y2 = 0 6y 2 y 2 dy 2 x I y2 (0, ) + 0 6y 2 y 2 dy x I y2 (, 2) 4 = 2y 2 x I y2 (0, ) + 2y 2 (2 - y 2 ) 3 x I y2 (, 2). Note, again, that the marginals must be functions of the r.v. in question only. The conditional density of Y 2 given Y = y is given by f(y 2 y ) = f(y, y 2 )/ f (y ) = 6y 2 y 2 x I (y, y2) (S)/(2y 2 ( y ) x I y (0, )) = (y 2 /(2( - y ))) x I (y, y2) (S). We can now find P(Y 2 <. Y =.6) which is given by.. - f(y 2 y )dy 2 = - (y 2 /(2( - y ))) x I (y, y2) (S)dy 2. =.25 x.6 y 2 dy 2 = [] Lecture Recall from earlier days in statistics that two events were deemed independent iff P(A B) = P(A) x P(B). Well, the same sort of thing applies here in a larger sense. Definition 5.8 (p. 233) tells us that if Y and Y 2 are jointly continuous r.v.s having cdfs F (y ) and F 2 (y 2 ) respectively, then they are independent iff F(y, y 2 ) = F (y ) x F 2 (y 2 ) for all vectors (y, y 2 ) in R 2. Now, the astute observer will note that the units have a hard time matching up here, and would be correct in noting that. However, we must state that this relationship is purely numerical. If this is not the case, then they are said to be dependent. Note that our previous definition dealt with events being independent, whereas now we are concerned with r.v.s. Clearly, the same type of definition can be developed for discrete r.v.s. Theorem 5.4 (p. 234) tells us that we can take the issue down to the jpdfs specifically, for the previously defined r.v.s, Y and Y 2 are independent iff f(y, y 2 ) = f (y ) x f 2 (y 2 ) for all vectors (y, y 2 ) R 2. Theorem 5.5 (p. 236) gives an even easier way to determine independence. It says that if Y and Y 2 are jointly continuous r.v.s with jpdf f(y, y 2 ) then they are independent iff f can be written as a product of nonnegative functions of y alone and y 2 alone. So to show independence we can use either of theorems 5.4 and 5.5. But to show dependence, we need only demonstrate that f(y, y 2 ) f (y ) x f 2 (y 2 ) for some vector (y, y 2 ) R 2. Ex. We now reconsider our favorite jpdf (from problem 5.6) in problem 5.44 f (y, y 2 ) = 4y y 2 x I (y, y2) (0, ) 2. Note from problem 5.22 that f(y, y 2 ) = 2y I y (0, ) x 2y 2 I y2 (0, ) = f (y ) x f 2 (y 2 ). 5S

6 Therefore, Y, and Y 2 are independent by Theorem 5.4. [] Lecture Now here are some problems. Student Problem set I I. In exercise 5.8, we proved that f(y, y 2 ) = I (<Y, Y2>) (0, ) 2 is a valid joint probability density function for Y, the amount of pollutant per sample collected above the stack without the cleaning device, and Y 2, the amount collected above the stack with the cleaner. Are the amounts of pollutants per sample collected with and without the cleaning device independent? I2. Suppose that that random variables Y and Y 2 have joint probability density function, given by f(y, y 2 )={6y 2 y 2, 0 y 2, y +y 2 2 {0, elsewhere Show that Y and Y 2 are dependent random variables. I3. Three balanced coins are tossed independently. Let Y = # of heads, and Y 2 = amount of money won on a side bet in which Y 2 = I Y (head on coin ) + 2I Y (first head on coin 2) + 3I Y (first head on coin 3) I Y (Y = 0). a) derive f 2 (y 2 ). b) find P(Y = 3 Y 2 = ). I4. A radioactive particle is uniformly and randomly located on the unit square so that f(y, y 2 ) = I (<Y, Y2>) (0, ) 2 a) Find the marginal density functions f (Y ) and f 2 (Y 2 ). b) Find the conditional density function f(y 2 Y ). c) Find P(Y 2 <.5 y =.7). I5. The following 2--3 function, f(y,y 2 )={, 0 y 2, 0 y 2, 2y 2 y {0, elsewhere is a valid joint probability density function for Y, the amount of pollutant per sample collected above the stack without the cleaning device, and for Y 2, the amount collected above the stack with the cleaner. a) If we consider the stack with a cleaner installed, find the probability that the amount of pollutant in a given sample will exceed.5. b) Given that the amount of pollutant in a sample taken above the stack with the cleaner is observed to be.5, find the probability that the amount of pollutant exceeds.5 above the other stack (without the cleaner). I6. Suppose Y and Y 2 denoted the proportions of time during which employees I and II actually performed their assigned tasks during a workday. The joint density of Y and Y 2 is given by f(y,y 2 ) = (y +y 2 ) I <y,y2> ((0, ) x (0, )). a) Find the marginal density functions for Y and Y 2. b) Find P(Y >.5 Y 2 >.5). c) If employee II spends exactly 50% of the day working on assigned duties, find the probability that employee I spends more than 75% of the day working on similar duties. 6S

7 I7. The number of defects per yard, Y, for a certain fabric is known to have a Poisson distribution with parameter lambda. However, lambda itself is a random variable with probability density function given by f( ) = e - I [0, ) Find the unconditional probability function for Y. Student Solution Set H I. The variables Y and Y 2 are dependent, as the range of y values on which f(y, y 2 ) is defined depends on Y 2. I2. Dependent as the range of y values on which f(y, y 2 ) is defined depends on y 2. More rigorously, one could verify from the solution to exercise 5.28 that f(y 2 Y =y ) f(y 2 ). I5. a) We want to integrate f 2 (y 2 ) = 2(-y 2 ) from.5 to, which gives us.25. b) First, we determine that f(y y 2 )=/(2(-y 2 )) for 2(y 2 ) y 2. Then, since y 2 =/2, f(y y 2 =/2)= for y 2. Therefore, P(Y >.5 Y 2 =/2)=2-.5=.5. I6. a) f (y ) = y +/2 and f 2 (y 2 ) = y 2 +/2 b) P(Y 2 > /2, Y > ½) = 3/8 and P(Y 2 > ½) = 5/8 Thus, P(Y > /2 Y 2 > (/2))=(6/6)/(5/8)=3/5 c) First, we consider that f(y y 2 )=(y +y 2 )/(y 2 +/2). P(Y >.75 Y 2 =.5) = (the integral of f(y y 2 ) from.75 to )= = I7. We have, if > 0, P(Y=y )=(( y )(e - )/(y!)) and f(λ)=e (- ) I ). Thus, P(y)= 0 P(Y = y, ) d = (/2) (y+). [] 7S

8 Lecture Finally, it is worth noting that the role played by indicator functions is huge. If the support for a jpdf lends itself to a factorization of the type seen in the previous example, then the r.v.s have a good chance of being independent. If, on the other hand, the support looks like that of problem 5.28, then it is unlikely that the r.v.s are independent. 8S