The Distribution of a Linear Combination

5.5 The Distribution of a Linear Combination

  The sample mean X and sample total T o are special cases of a type of random vari- able that arises very frequently in statistical applications.

  DEFINITION Given a collection of n random variables X 1 ,...,X n and n numerical constants

  a 1 ,...,a n , the rv

  n

  Y5a 1 X 1 1 …1 a n X n 5 o a i X i (5.7)

  i5 1

  is called a linear combination of the X i ’s.

  For example, consider someone who owns 100 shares of stock A, 200 shares of stock

  B, and 500 shares of stock C. Denote the share prices of these three stocks at some

  particular time by X 1 ,X 2 , and X 3 , respectively. Then the value of this individual’s stock holdings is the linear combination Y 5 100X 1 1 200X 2 1 500X 3 . Taking a 1 5a 2 5...5a n 5 1 gives Y 5 X 1 1...1X n 5T o , and a 1 5

  a 2 5…5 a n 5 1 yn yields

  Notice that we are not requiring the X i ’s to be independent or identically distrib- uted. All the X i ’s could have different distributions and therefore different mean values and variances. Our first result concerns the expected value and variance of

  a linear combination.

  prOpOSITION Let X 1 ,X 2 ,...,X n have mean values m 1 ,...,m n , respectively, and variances

  s 2 1 2 , …, s n , respectively.

  1. Whether or not the X i ’s are independent,

  E (a 1 X 1 1a 2 X 2 1...1a n X n )5a 1 E (X 1 )1a 2 E (X 2 )1...1a n E (X n )

  5 a 1 m 1 1...1a n m n (5.8)

  5.5 the Distribution of a Linear Combination 239

  2. If X 1 ,...,X n are independent,

  a 1 X 1 1…1 a n X n 5 Ïa 1 s 1 1…1 a n s n (5.10)

  3. For any X 1 ,...,X n ,

  i5 o 1 j5 o 1

  a i a j Cov(X i ,X j ) (5.11)

  Proofs are sketched out at the end of the section. A paraphrase of (5.8) is that the expected value of a linear combination is the same as the linear combination of

  the expected values—for example, E(2X 1 1 5X 2 ) 5 2m 1 1 5m 2 . The result (5.9) in

  Statement 2 is a special case of (5.11) in Statement 3; when the X i ’s are independent, Cov(X i , X j ) 5 0 for i ? j and 5 V(X i ) for i 5 j (this simplification actually occurs when the X i ’s are uncorrelated, a weaker condition than independence). Specializing to the case of a random sample (X i ’s iid) with a i 5 1n for every i gives E(X) 5 m and

  V (X) 5 s 2 n, as discussed in Section 5.4. A similar comment applies to the rules for T o . ExamplE 5.30 A gas station sells three grades of gasoline: regular, extra, and super. These are

  priced at 3.00, 3.20, and 3.40 per gallon, respectively. Let X 1 ,X 2 , and X 3 denote

  the amounts of these grades purchased (gallons) on a particular day. Suppose the

  X i ’s are independent with m 1 5 1000, m 2 5 500, m 3 5 300, s 1 5 100, s 2 5 80, and s 3 5 50. The revenue from sales is Y 5 3.0X 1 1 3.2X 2 1 3.4X 3 , and

  E (Y) 5 3.0m 1 1 3.2m 2 1 3.4m 3 5 5620

  V (Y) 5 (3.0) 2 s 2 1 1 (3.2) 2 s 2 1 (3.4) 2 s 2 3 5 184,436

  s Y 5 Ï184,436 5 429.46

  n

  the difference Between two random Variables

  An important special case of a linear combination results from taking n 5 2, a 1 5 1,

  and a 2 5 21: Y 5a 1 X 1 1a 2 X 2 5X 1 2X 2 We then have the following corollary to the proposition.

  COrOllarY E (X 1 2X 2 ) 5 E(X 1 ) 2 E(X 2 ) for any two rv’s X 1 and X 2 .

  V (X 1 2X 2 ) 5 V(X 1 ) 1 V(X 2 ) if X 1 and X 2 are independent rv’s.

  The expected value of a difference is the difference of the two expected values. However, the variance of a difference between two independent variables is the sum,

  not the difference, of the two variances. There is just as much variability in X 1 2X 2 as in X 1 1X 2 [writing X 1 2X 2 5X 1 1 (21)X 2 , (21)X 2 has the same amount of

  variability as X 2 itself].

  240 ChapteR 5 Joint probability Distributions and Random Samples

  ExamplE 5.31 A certain automobile manufacturer equips a particular model with either a six-cylinder

  engine or a four-cylinder engine. Let X 1 and X 2 be fuel efficiencies for independently and randomly selected six-cylinder and four-cylinder cars, respectively. With m 1 5 22,

  m 2 5 26, s 1 5 1.2, and s 2 5 1.5,

  E (X 1 2X 2 )5m 1 2m 2 5 22 2 26 5 24

  V (X 1 2 X 2 )5s 2 1 1s 2 5 (1.2) 2 1 (1.5) 2 5 3.69

  s X 1 2 X 2 5 Ï3.69 5 1.92

  If we relabel so that X 1 refers to the four-cylinder car, then E(X 1 2X 2 ) 5 4, but the

  variance of the difference is still 3.69.

  n

  the case of normal random Variables

  When the X i ’s form a random sample from a normal distribution, X and T o are both normally distributed. Here is a more general result concerning linear combinations.

  prOpOSITION If X 1 ,X 2 ,…, X n are independent, normally distributed rv’s (with possibly dif- ferent means andor variances), then any linear combination of the X i ’s also

  has a normal distribution. In particular, the difference X 1 2X 2 between two

  independent, normally distributed variables is itself normally distributed.

  ExamplE 5.32 The total revenue from the sale of the three grades of gasoline on a particular day

  (Example 5.30

  was Y 5 3.0X 1 1 3.2X 2 1 3.4X 3 , and we calculated m Y 5 5620 and (assuming

  continued) independence) s Y 5 429.46. If the X i ’s are normally distributed, the probability that

  revenue exceeds 4500 is

  P(Y . 4500) 5 P Z.

  5 P(Z . 22.61) 5 1 2 F(22.61) 5 .9955

  n

  The CLT can also be generalized so it applies to certain linear combinations. Roughly speaking, if n is large and no individual term is likely to contribute too much to the overall value, then Y has approximately a normal distribution.

  Proofs for the Case n52

  For the result concerning expected values, suppose that X 1 and X 2 are continuous

  with joint pdf f(x 1 ,x 2 ). Then

  sa 1

  1 X 1 sx 1 d dx 1

  5a 1 x f

  a 2 x

  f X 2 sx 2 d dx 2

  5 a 1 E (X 1 )1a 2 E (X 2 )

  5.5 the Distribution of a Linear Combination 241

  Summation replaces integration in the discrete case. The argument for the variance result does not require specifying whether either variable is discrete or continuous.

  Recalling that V(Y) 5 E[(Y 2 m Y ) 2 ],

  V (a 1 X 1 1a 2 X 2 ) 5 E{[a 1 X 1 1a 2 X 2 2 (a 1 m 1 1a 2 m 2 )] 2 }

  5 E {a 1 2 sX 1 2m 1 d 2 1 a 2 sX 2 2m 2 d 2 1 2a 1 a 2 sX 1 2m 1 dsX 2 2m 2 d} The expression inside the braces is a linear combination of the variables Y 1 5

  (X 1 2m 1 ) 2 , Y 2 5 (X 2 2m 2 ) 2 , and Y 3 5 (X 1 2m 1 )(X 2 2m 2 ), so carrying the E operation through to the three terms gives a 1 2 V sX 1 d1a 2 V sX 2 d 1 2a 1 a 2 Cov sX 1 ,X 2 d

  as required.

  n

  EXERCISES Section 5.5 (58–74)

  58. A shipping company handles containers in three different

  61. Exercise 26 introduced random variables X and Y, the

  sizes: (1) 27 ft 3 (3 3 3 3 3), (2) 125 ft 3 , and (3) 512 ft 3 .

  number of cars and buses, respectively, carried by a

  Let X i (i 5 1, 2, 3) denote the number of type i containers

  ferry on a single trip. The joint pmf of X and Y is given

  shipped during a given week. With m i 5 E(X i ) and

  in the table in Exercise 7. It is readily verified that X

  s i 2 5 V (X i ), suppose that the mean values and standard

  and Y are independent.

  deviations are as follows:

  a. Compute the expected value, variance, and standard

  m 1 5 200 m 2 5 250 m 3 5 100

  deviation of the total number of vehicles on a single trip.

  s 1 5 10 s 2 5 12 s 3 58

  b. If each car is charged 3 and each bus 10, compute

  a. Assuming that X 1 ,X 2 ,X 3 are independent, calculate

  the expected value, variance, and standard deviation

  the expected value and variance of the total volume

  of the revenue resulting from a single trip.

  shipped. [Hint: Volume 5 27X 1 1 125X 2 1 512X 3 .]

  b. Would your calculations necessarily be correct if the

  62. Manufacture of a certain component requires three

  X i ’ s were not independent? Explain.

  different machining operations. Machining time for each operation has a normal distribution, and the three

  59. Let X 1 ,X 2 , and X 3 represent the times necessary to per-

  times are independent of one another. The mean values

  form three successive repair tasks at a certain service

  are 15, 30, and 20 min, respectively, and the standard

  facility. Suppose they are independent, normal rv’s with

  deviations are 1, 2, and 1.5 min, respectively. What is

  expected values m 1 ,m 2 , and m 3 and variances s 1 2 ,s 2 , and

  the probability that it takes at most 1 hour of machining

  s 3 2 , respectively.

  time to produce a randomly selected component?

  a. If m 5m 2 5m 3 5 60 and s 1 2 5s 2 5s 3 2 5 15, calcu-

  late P( T o 200) and P(150 T o 200)?

  63. Refer to Exercise 3.

  b. Using the m i ’ s and s

  i ’ s given in part (a), calculate

  a. Calculate the covariance between X

  1 5 the number

  both P(55 X) and P(58 X 62).

  of customers in the express checkout and X 2 5 the

  c. Using the m i ’ s and s i ’ s given in part (a), calculate

  number of customers in the superexpress checkout.

  and interpret P(210 X 1 2 .5X 2 2 .5X

  b. Calculate V(X

  1 1X 2 ). How does this compare to

  and s 3 2 5 14, calculate P(X 1 1X 2 1X 3 160) and

  64. Suppose your waiting time for a bus in the morning is

  also P(X 1 1X 2 2X 3 ).

  uniformly distributed on [0, 8], whereas waiting time in

  60. Refer back to Example 5.31. Two cars with six-cylinder

  the evening is uniformly distributed on [0, 10] indepen-

  engines and three with four-cylinder engines are to be driv-

  dent of morning waiting time.

  en over a 300-mile course. Let X 1 ,...X 5 denote the resulting

  a. If you take the bus each morning and evening for a

  fuel efficiencies (mpg). Consider the linear combination

  week, what is your total expected waiting time? [Hint: Define rv’s X , ..., X and use a rule of

  Y 5 (X 1 1X 2 )

  y2 2 (X 10

  3 1X 4 1 1 X 5 ) y3 expected value.]

  which is a measure of the difference between four-

  b. What is the variance of your total waiting time?

  cylinder and six-cylinder vehicles. Compute P(0 Y)

  c. What are the expected value and variance of the dif-

  and P(Y . 22). [Hint: Y 5 a 1 X 1 1 ... 1a 5 X 5 , with

  ference between morning and evening waiting times

  a 1 5 1 y2 ,…, a 5 52 1 y3.]

  on a given day?

  242 ChapteR 5 Joint probability Distributions and Random Samples

  d. What are the expected value and variance of the dif-

  value 1 in. and standard deviation .1 in. Assuming that

  ference between total morning waiting time and total

  the lengths and amount of overlap are independent of one

  evening waiting time for a particular week?

  another, what is the probability that the total length after

  65. Suppose that when the pH of a certain chemical

  insertion is between 34.5 in. and 35 in.?

  compound is 5.00, the pH measured by a randomly

  68. Two airplanes are flying in the same direction in adjacent

  selected beginning chemistry student is a random vari-

  parallel corridors. At time t 5 0, the first airplane is

  able with mean 5.00 and standard deviation .2. A large

  10 km ahead of the second one. Suppose the speed of the

  batch of the compound is subdivided and a sample

  first plane (kmhr) is normally distributed with mean 520

  given to each student in a morning lab and each student

  and standard deviation 10 and the second plane’s speed

  in an afternoon lab. Let X 5 the average pH as deter-

  is also normally distributed with mean and standard

  mined by the morning students and Y 5 the average pH

  deviation 500 and 10, respectively.

  as determined by the afternoon students.

  a. What is the probability that after 2 hr of flying, the

  a. If pH is a normal variable and there are 25 students

  second plane has not caught up to the first plane?

  in each lab, compute P(2.1 X 2 Y .1). [Hint:

  b. Determine the probability that the planes are sepa-

  X 2 Y is a linear combination of normal variables,

  rated by at most 10 km after 2 hr.

  so is normally distributed. Compute m X2Y and s X2Y .]

  69. Three different roads feed into a particular freeway

  b. If there are 36 students in each lab, but pH determi-

  entrance. Suppose that during a fixed time period, the

  nations are not assumed normal, calculate (approxi-

  number of cars coming from each road onto the freeway

  mately) P(2.1 X 2 Y .1).

  is a random variable, with expected value and standard

  66. If two loads are applied to a cantilever beam as shown in

  deviation as given in the table.

  the accompanying drawing, the bending moment at 0 due

  Road 1

  Road 2 Road 3

  to the loads is a 1 X 1 1a 2 X 2 .

  Expected value

  X 1 X 2 Standard deviation

  16 25 18 a. What is the expected total number of cars entering

  a 1 a 2 the freeway at this point during the period? [Hint: 0 Let X i 5 the number from road i.]

  b. What is the variance of the total number of entering

  a. Suppose that X 1 and X 2 are independent rv’s with

  cars? Have you made any assumptions about the

  means 2 and 4 kips, respectively, and standard devia-

  relationship between the numbers of cars on the

  tions .5 and 1.0 kip, respectively. If a 1 5 5 ft and

  different roads?

  a 2 5 10 ft, what is the expected bending moment and

  c. With X i denoting the number of cars entering from

  what is the standard deviation of the bending

  road i during the period, suppose that Cov(X 1 ,X 2 ) 5 80,

  moment?

  Cov(X 1 ,X 3 ) 5 90, and Cov(X 2 ,X 3 ) 5 100 (so that the

  b. If X 1 and X 2 are normally distributed, what is the

  three streams of traffic are not independent). Compute

  probability that the bending moment will exceed

  the expected total number of entering cars and the

  75 kip-ft?

  standard deviation of the total.

  c. Suppose the positions of the two loads are random

  70. Consider a random sample of size n from a continuous

  variables. Denoting them by A 1 and A 2 , assume that

  distribution having median 0 so that the probability of

  these variables have means of 5 and 10 ft, respec-

  any one observation being positive is .5. Disregarding the

  tively, that each has a standard deviation of .5, and

  signs of the observations, rank them from smallest to

  that all A i ’s and X i ’s are independent of one another.

  largest in absolute value, and let W 5 the sum of the

  What is the expected moment now?

  ranks of the observations having positive signs. For

  d. For the situation of part (c), what is the variance of

  example, if the observations are 2.3, 1.7, 12.1, and

  the bending moment?

  2 2.5, then the ranks of positive observations are 2 and 3,

  e. If the situation is as described in part (a) except that

  so W 5 5. In Chapter 15, W will be called Wilcoxon’s

  Corr(X 1 ,X 2 ) 5 .5 (so that the two loads are not inde-

  signed-rank statistic. W can be represented as follows:

  pendent), what is the variance of the bending moment?

  W 51?Y 1 12?Y 2 13?Y 3 1...1n?Y n

  67. One piece of PVC pipe is to be inserted inside another n

  o i?Y i i5 1

  piece. The length of the first piece is normally distributed

  with mean value 20 in. and standard deviation .5 in. The length of the second piece is a normal rv with mean and

  where the Y i ’ s are independent Bernoulli rv’s, each with

  standard deviation 15 in. and .4 in., respectively. The

  p 5 .5 (Y i 5 1 corresponds to the observation with rank

  amount of overlap is normally distributed with mean

  i being positive).

  Supplementary exercises 243

  a. Determine E(Y i ) and then E(W) using the equation

  s 3 5 2, m 4 5 12, s 4 5 3. I plan to leave my office at

  for W. [Hint: The first n positive integers sum to

  precisely 10:00 a.m. and wish to post a note on my door

  n (n 1 1) y 2.]

  that reads, “I will return by t a.m.” What time t should I

  b. Determine V(Y i ) and then V(W). [Hint: The sum of

  write down if I want the probability of my arriving after

  the squares of the first n positive integers can be

  t to be .01?

  expressed as n(n 1 1)(2n 1 1) y 6.]

  73. Suppose the expected tensile strength of type-A steel is

  71. In Exercise 66, the weight of the beam itself contributes

  105 ksi and the standard deviation of tensile strength is

  to the bending moment. Assume that the beam is of uni-

  8 ksi. For type-B steel, suppose the expected tensile

  form thickness and density so that the resulting load is

  strength and standard deviation of tensile strength are

  uniformly distributed on the beam. If the weight of the

  100 ksi and 6 ksi, respectively. Let X 5 the sample aver-

  beam is random, the resulting load from the weight is

  age tensile strength of a random sample of 40 type-A

  also random; denote this load by W (kip-ft).

  specimens, and let Y 5 the sample average tensile

  a. If the beam is 12 ft long, W has mean 1.5 and stan-

  strength of a random sample of 35 type-B specimens.

  dard deviation .25, and the fixed loads are as described

  a. What is the approximate distribution of X? Of Y?

  in part (a) of Exercise 66, what are the expected value

  b. What is the approximate distribution of X 2 Y?

  and variance of the bending moment? [Hint: If the

  Justify your answer.

  load due to the beam were w kip-ft, the contribution

  c. Calculate (approximately) P(21 X 2 Y 1).

  to the bending moment would be w 0 12 x dx.]

  d. Calculate P(X 2 Y 10). If you actually observed

  b. If all three variables (X 1 , X 2 , and W) are normally

  X 2 Y 10, would you doubt that m 1 2m 2 5 5?

  distributed, what is the probability that the bending

  74. In an area having sandy soil, 50 small trees of a certain

  moment will be at most 200 kip-ft?

  type were planted, and another 50 trees were planted in an

  72. I have three errands to take care of in the Administration

  area having clay soil. Let X 5 the number of trees planted

  Building. Let X i 5 the time that it takes for the ith errand

  in sandy soil that survive 1 year and Y 5 the number of

  (i 5 1, 2, 3), and let X 4 5 the total time in minutes that

  trees planted in clay soil that survive 1 year. If the prob-

  I spend walking to and from the building and between

  ability that a tree planted in sandy soil will survive 1 year

  each errand. Suppose the X i ’ s are independent, and nor-

  is .7 and the probability of 1-year survival in clay soil is

  mally distributed, with the following means and standard

  .6, compute an approximation to P(25 X 2 Y 5) (do

  deviations: m 1 5 15, s 1 5 4, m 2 5 5, s 2 5 1, m 3 5 8,

  not bother with the continuity correction).