with Probability Applications--Spring 2008
Problem Set 2 for ISYE 2027, Section B (February 26, 2008)
Purpose. The second exam will focus on material largely covered in Chapters 3 and 4 of Walpole et al. However, this should be considered a cumulative exam in that prior material will be presupposed and may be directly treated. In its final version, this document will contain two practice exams and a list of recommended exercises from the textbook. In the lecture periods prior to the exam, I will solve these practice exams at the board, along with additional exercises as time permits. Please see also the document "Study Guide 2," which provides a theoretical overview of the new material you are responsible for. Also important are your lecture notes and the document "Homework 2 Solutions."
Notes, books, and electronic devices may not be used during the exam. First practice exam. The following formulas are provided without context, as a memory aid: F (x) = P (X x) for all x R g(x) = h(y) =
y
f (x, y) or g(x) = or h(y) =
+ - f (x, y) dy + - f (x, y) dx
x f (x, y)
f (y|x) = f (x, y)/g(x) and f (x|y) = f (x, y)/h(y) E(X) =
x xf (x)
or E(X) =
+ - xf (x) dx
Var(X) = E[(X - E[X])2 ] = E(X 2 ) - (E(X))2 Cov(X, Y ) = E[(X - E[X])(Y - E[Y ])] = E(XY ) - E(X)E(Y ) E(aX + b) = aE(X) + b and Var(aX + b) = a2 Var(X) Var(X + Y ) = Var(X) + Var(Y ) + 2 Cov(X, Y ) 1. Suppose that the set of possible values for the random variable X is the interval of real numbers x satisfying a x b, for some particular a, b R with a < b. (a) What type of random variable is X: discrete or continuous? (b) Suppose that f (x) = c for x in the set of possible values for X. If X is discrete, interpret f as a probability mass function. If X is continuous, interpret f as a probability density function. What value must c take? Your final answer should be a simplified expression involving a and b, and it should not involve a sum or an integral. 2. Suppose we are to toss three fair coins. Assume that any given toss is independent of any collection of other tosses. Define the random variable X to be the number of heads we see. Find F (x), the cumulative distribution function of X. You must specify a numerical value for F (x) in the form of a simplified fraction (where necessary) for all real numbers x R.
3. Suppose we are to toss two loaded, six-sided dice; one die is red and the other is green. For each die, the probability of seeing any given even number is twice the probability of seeing any given odd number. Assume that the two tosses are independent. Let X be the number that comes up on the red die, and let Y be the number that comes up on the green die. (a) Identify the following statement as either true or false. For all possible values y of Y , the conditional probability mass function of Y given that X = 5 is equal to the marginal pmf of Y . (b) What is the conditional probability that Y is less than or equal to 3, given that X = 5? A simplified fraction is required.
1 4. Suppose that X and Y have joint probability density function f (x, y) = 16 over the region described by 0 x 4 and 0 y 4. Let f (x, y) = 0 outside of this region of possible values for X and Y .
(a) What is P (Y X 2 )? Your final answer should be a simplified fraction. (b) Is it true that P (Y X 2 ) = P (Y > X 2 )? 5. Consider the situation in Question 2. Give exact numerical values (simplified fractions where necessary) for the following: (a) E(X) (b) Var(X) 6. Consider the situation in Question 3. Give exact numerical values (simplified fractions where necessary) for the following: (a) E(2X - 3Y + 1) (b) Var(2X - 3Y + 1) 7. Suppose we toss two fair coins, and assume the two tosses are independent. Define the random variable X to be the number of heads we see. Define the random variable Y to be 1 if the two tosses yield different results (i.e., one sees H and the other sees T) and 0 otherwise (i.e., if both see H or both see T). (a) Are X and Y independent? (b) What is Cov(X, Y )? Second practice exam. The following formulas are provided without context, as a memory aid:
f (y|x) = f (x, y)/g(x) and f (x|y) = f (x, y)/h(y) Var(X) = E[(X - E[X])2 ] = E(X 2 ) - (E(X))2 Cov(X, Y ) = E[(X - E[X])(Y - E[Y ])] = E(XY ) - E(X)E(Y ) E(aX + b) = aE(X) + b and Var(aX + b) = a2 Var(X) Var(X + Y ) = Var(X) + Var(Y ) + 2 Cov(X, Y ) 1. Suppose that the set of possible values for the random variable X consists of the integers x satisfying a x b, for some particular a, b Z with a < b.
(a) What type of random variable is X: discrete or continuous? (b) Suppose that f (x) = c for x in the set of possible values for X. If X is discrete, interpret f as a probability mass function. If X is continuous, interpret f as a probability density function. What value must c take? Your final answer should be a simplified expression involving a and b, and it should not involve a sum or an integral. 2. Suppose that the cumulative distribution function of X is given for all x R by xa 0 x-a a<xb F (x) = b-a 1 x>b for some a, b R with a < b. (a) What type of random variable is X: discrete or continuous? (b) What is P (X > b+a )? Give a simplified fraction. 2
1 3. Suppose that X and Y have joint probability mass function f (x, y) = 25 over the region described by 0 x 4 and 0 y 4, where x and y are restricted to be integers.
(a) What is P (Y X 2 )? Your final answer should be a simplified fraction. (b) Is it true that P (Y X 2 ) = P (Y > X 2 )? 4. Suppose that X and Y have joint probability density function f (x, y) = 2 over the region in the plane described by 0 < x < 1, 0 < y < 1, and 0 < x + y < 1, with f (x, y) = 0 outside of this region. (a) What is the marginal density function of X, g(x)? Simplify your answer, and specify an appropriate value of g(x) for all x R. (b) What is the marginal density function of Y , h(y)? Simplify your answer, and specify an appropriate value of h(y) for all y R. 5. Consider the situation in the preceding question. (a) What is the conditional pdf of X, given that Y = 0.2? Simplify your answer, and specify an appropriate value of f (x|0.2) for all x R. (b) What is the conditional pdf of Y , given that X = x, assuming that 0 < x < 1? Simplify your answer, and specify an appropriate value of f (y|x) for all y R.
1 6. Suppose that we are given a, b R with a < b. Let the pdf of X be f (x) = b-a for a < x < b, with f (x) = 0 for all other x R. Give simplified expressions (involving a or b as necessary) for the following:
(a) E(X + X) (b) Var(-X - X) 7. Let X have pdf fX (x) = defined for all x R. Let Y have pdf fY (y) =
2 1 e-(y-5) /2 2 2 1 e-(x+5) /2 2
defined for all y R. Assuming X and Y are independent random variables:
(a) What is the joint pdf of X and Y ? You may use the notation established above. Be sure to specify this function over the entire plane R2 . (b) What is Cov(Y, X)? Recommended textbook exercises. In these exercises, do not worry too much about needing a calculator to simplify complicated numerical expressions. Note that solutions to odd-numbered exercises are available in the student solutions manual, and answers to these exercises are also in the back of the textbook itself. Any exercises marked with `' below are ones I know to have dubious or incorrect published answers. Contact me with any doubts about published solutions; I can verify answers on request (usually by working out problems during office hours), but I am not doing all the problems on my own by default. Exercises beginning on p.88: 2 (assume the agency does not distinguish the order in which the three cars are received), 3, 4, 5, 6, 7, 8 (assume each toss is independent of any collection of other tosses), 9, 10, 11, 12, 13, 14, 15, 17, 18, 19, 20 Exercises beginning on p.101: 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50, 56, 57, 60, 61 (There are review exercises beginning on p.103; I have not examined these.) Exercises beginning on p.113: 1 (use symmetry rather than getting lost in the details of integration), 2, 3, 4, 5, 6, 7, 10, 12, 15, 17, 18, 23 Exercises beginning on p.122: 34, 35, 36, 39, 40, 43, 45, 47, 48 Exercises beginning on p.134: 51, 54, 55, 56, 58, 69, 70, 72 (There are review exercises beginning on p.136; I have not examined these.) Solutions for the first practice exam. 1. (a) The set of possible values is an interval of real numbers, so X is continuous. (b) Since f is a pdf, we have
b a c dx + - f (x) dx
= 1. By the definition of f (x) this implies that
= 1. Since a and b are finite (being real numbers), we conclude after integrating 1 that c = b-a . 2. The appropriate sample space here is {H, T }3 . Given our independence assumption and the fairness of the we coins, may reason as in Homeworks 1 and 2 that the probability of any given sample space element is ( 1 )3 = 1 . To calculate P (X = x) = f (x), we simply need to 2 8 multiply this fraction by the number of sample points corresponding to seeing x heads. Thus the pmf is 1/8 x = 0 3/8 x = 1 f (x) = 3/8 x = 2 1/8 x = 3
We want the cdf. In the discrete case, the cdf F (x) = P (X x) may be calculated from the pmf as tx f (t); here, we obtain: 0 1/8 1/2 F (x) = 7/8 1 x<0 0x<1 1x<2 2x<3 x3
3. (a) Using our standard notation, we are to determine whether f (y|5) = h(y). It is equivalent to ask whether P (Y = y | X = 5) = P (Y = y). Since the two tosses are independent by assumption, this is true. The event referred to by the statement Y = y is independent of the event referred to by the statement X = 5; the former specifies only the outcome for the green die, and the latter specifies only the outcome for the red die. We could also obtain this answer by finding the marginal pmfs g(x) and h(y) and the joint pmf f (x, y) (using the independence assumption), and verifying that f (x, y)/g(x) = h(y). Refer to the textbook example 2.24 (discussed in class) to calculate the marginals. (b) P (Y 3 | X = 5) = P (Y 3) by independence (by our assumptions, the event referred to by the statement Y 3 is independent of the event referred to by the statement 4 X = 5). Our answer is therefore P (Y 3) = 3 h(y) = 9 . We could also obtain this y=1 answer by writing the sum 3 f (y|5). y=1 4. (a) Since we have a joint pdf, P (Y X 2 ) = yx2 f (x, y) dx dy. By the definition of f (x, y), we may concentrate on the region where x and y are between 0 and 4. The probability y 1 4 4 1 we want may be calculated as the iterated integral 0 0 16 dx dy = 16 0 y 1/2 dy = 1 2 3/2 = 1 . 16 3 (4) 3 (b) Since we are given a joint pdf, it is true that P (Y X 2 ) = P (Y > X 2 ). Our iterated integral would be the same either way; this is to say that P (Y = X 2 ) = 0. 5. (a) Since X is discrete (having a finite number of possible values), we may use the definition E(X) = x xf (x). Continuing with the usual notation, this means E(X) = 3 1 3 1 xf (x) = 0( 8 ) + 1( 3 ) + 2( 8 ) + 3( 8 ) = 3 in our case. x=0 8 2 (b) We could use the definition of variance directly, but it is probably easier to use the alternate formula Var(X) = E(X 2 )-(E(X))2 . In our discrete case, E(X 2 ) = 3 x2 f (x) = x=0 1 0( 8 ) + 1( 3 ) + 4( 3 ) + 9( 1 ) = 3. Therefore, Var(X) = 3 - ( 3 )2 = 3 . 8 8 8 2 4 6. (a) By the properties of sums (and integrals) reflected in our mean and variance formulas, E(2X - 3Y + 1) = 2E(X) - 3E(Y ) + 1. Since in this case X and Y have equal probability mass functions, we will find that E(X) = E(Y ). Our problem reduces to finding 1 -E(X) + 1. Since E(X) = 6 xg(x) = 1( 1 ) + 2( 2 ) + 3( 1 ) + 4( 2 ) + 5( 9 ) + 6( 2 ) = 11 , x=1 9 9 9 9 9 3 8 our answer is - 3 . (b) Var(2X - 3Y + 1) = Var(2X - 3Y ) since adding a constant to a random variable (in this case, the random variable 2X - 3Y constructed from X and Y ) does not affect the variance. Now we may use the formula Var(aX + bY ) = a2 Var(X) + b2 Var(Y ) + 2ab Cov(X, Y). Here a = 2 and b = -3, so we have Var(2X-3Y ) = 4 Var(X)+9 Var(Y )- 12 Cov(X, Y ). Since in this case X and Y are independent, we have Cov(X, Y ) = 0. Since in this case X and Y have equal probability mass functions, we will find that
Var(X) = Var(Y ), so our task reduces to calculating 13 Var(X). We know already that E(X) = 11 . To calculate the variance of X we may use the formula involving 3 1 E(X 2 ) = 1( 1 )+4( 2 )+9( 9 )+16( 2 )+25( 1 )+36( 2 ) = 147 . Now our answer is 13 Var(X) = 9 9 9 9 9 9 147 11 2 26 13( 9 - ( 3 ) ) = 13( 9 ) = 338 . 9 7. (a) Given (for example) that Y = 1, we must have X = 1. So we know that information about Y is relevant to determining X and vice versa, and X and Y cannot be independent. Mathematically, this is revealed by observing that (for example) f (x|1) = g(x) for (for example) x = 0. Equivalently, we may verify that f (x, y) = g(x)h(y) for (to use the same example) x = 0 and y = 1, where f and g and h are interpreted as the joint probability mass function and the marginal mass functions respectively. (b) Let us find the joint pmf f (x, y). Since X and Y take finitely many values, we may do this by means of a table: Y =1 Y =0 1/4 1/2 1/4 0 1/2 0 1/2 1/4 0 1/4 1/2 X=0 X=1 X=2
For each entry corresponding to X = x and Y = y for some given x and y, the probability of seeing that combination of values has been recorded; this probability has been calculated using the standard reasoning to the effect that each element of the sample space {H, T }2 should be equally likely. Now we may use the alternate formula for covariance, Cov(X, Y ) = E(XY ) - E(X)E(Y ). Since this is a discrete case, E(XY ) = x y xyf (x, y). In our present situation, the only nonzero element of this 1 sum is 1 1 2 . Meanwhile, we have written the marginal probability mass functions of X and Y "in the margins" of the table above, by summing up columns and rows respectively. 1 According to this, we have E(X) = 0( 1 ) + 1( 1 ) + 2( 4 ) = 1 and E(Y ) = 0( 1 ) + 1( 1 ) = 1 . 4 2 2 2 2 1 1 In conclusion, Cov(X, Y ) = 1 1 2 - 1 2 = 0. Observe that we have ...