1. Section 5.5 Important Theorems in the Text: Let X 1, X 2, …, X n be independent random variables with respective N(  1,  1 2 ), N(  2,  2 2 ), …, N(  n,  n 2 ) distributions, and let c 1, c 2, …, c n be constants. Then the random variable Y = c 1 X 1 + c 2 X 2 + … + c n X n = has adistribution. (Proof of this theorem is addressed in Class Exercise #1.) Theorem 5.5-1 n  c i X i i = 1

2. 1. (a) Let X 1, X 2, …, X n be independent random variables with respective N(  1,  1 2 ), N(  2,  2 2 ), …, N(  n,  n 2 ) distributions, and let c 1, c 2, …, c n be constants. Define the random variable Y = c 1 X 1 + c 2 X 2 + … + c n X n =. n  c i X i i = 1 Find the m.g.f. for Y. Is it possible to tell from the m.g.f. what the distribution of Y is? From Theorem 5.4-1, we have that M Y (t) = e  1 c 1 t +  1 2 c 1 2 t 2 /2 e  2 c 2 t +  2 2 c 2 2 t 2 /2 … e =  n c n t +  n 2 c n 2 t 2 /2 e (c 1  1 + c 2  2 + … + c n  n )t + (c 1 2  1 2 + c 2 2  2 2 + … + c n 2  n 2 )t 2 /2 From this m.g.f., we recognize that Y must have a distribution. N(,) c 1  1 + c 2  2 + … + c n  n c 1 2  1 2 + c 2 2  2 2 + … + c n 2  n 2 M X (c 1 t) M X (c 2 t) … M X (c n t) = 12n

3. Let X 1, X 2, …, X n be independent random variables with respective N(  1,  1 2 ), N(  2,  2 2 ), …, N(  n,  n 2 ) distributions, and let c 1, c 2, …, c n be constants. Then the random variable Y = c 1 X 1 + c 2 X 2 + … + c n X n = has adistribution. (Proof of this theorem is addressed in Class Exercise #1.) Theorem 5.5-1 n  c i X i i = 1 N,N, n  c i  i i = 1 n  c i 2  i 2 i = 1 If X 1, X 2, …, X n are a random sample from a N( ,  2 ) distribution, then the sample mean X =has a distribution. Corollary 5.5-1 n  X i i = 1 n N( ,  2 / n)

4. 2W + 5X – 8Y has adistribution. W / 3 has adistribution. – W / 3 has adistribution. N(– 4, 2720) N(– 7/3, 4/9) N(7/3, 4/9) 1.-continued (b) Suppose the random variables W, X, and Y are independent with respective N(–7, 4), N(10, 16), and N(5, 36) distributions. Complete each of the following statements:

5. Let X 1, X 2, …, X n be a random sample from a N( ,  2 ) distribution, with sample mean X = and sample variance S 2 =. Then,(1) (2) (Proof of this theorem is addressed in Class Exercises #2 & #3.) Theorem 5.5-2 Theorem 5.5-3 (Proof of this theorem is addressed in Class Exercise #7.) n  X i i = 1 n n  (X i – X) 2 i = 1 n – 1 the sample mean X and sample variance S 2 are independent random variables, the random variable has a  2 (n – 1) distribution. (n – 1)S 2 ——— =  2 n  (X i – X) 2 i = 1 22

6. 2. (a) (b) The random variables X and Y are independent, and each has a N( ,  2 ) distribution. We define the random variables V = X + Y and W = X – Y. Find the joint p.d.f. of X and Y. Since X and Y are independent, their joint p.d.f. is f(x,y) =if –  < x < , –  < y <  2  2 exp (x –  ) 2 + (y –  ) 2 – ——————— 2  2 Use the change-of-variables method to find the joint p.d.f. of V and W. First, we find the space of V and W as follows: –  < x <  < v <  –  < y <  < w < –  

7. 2.-continued Then, we find the inverse transformation as follows: v = x + yx =  w = x – yy = v + w —— 2 v – w —— 2 Next, we find the Jacobian determinant as follows: J = det xx——vwyy——vwxx——vwyy——vw—— 1/2 = det 1/2 – 1/2 =

8. The joint p.d.f. of V and W is g(v, w) = 2  2 exp ([v + w]/2 –  ) 2 + ([v – w]/2 –  ) 2 – —————————————— 2  2 1 — = 2 4  2 exp ([v + w] – 2  ) 2 + ([v – w] – 2  ) 2 – ————————————— 8  2 if –  < v < , –  < w < 

9. 2.-continued (c) Show that the random variables V = X + Y and W = X – Y are independent with each having a normal distribution, and find the mean and variance for each normal distribution. First, we algebraically rewrite the joint p.d.f. of V and W as follows: 4  2 exp ([v – 2  ] + w) 2 + ([v – 2  ] – w) 2 – ————————————— 8  2 = 4  2 exp 2[v – 2  ] 2 + 2w 2 – ——————— 8  2 = 4  2 exp (v – 2  ) 2 + w 2 – —————— 4  2 =

10. (2  ) 1/2 2 1/2  exp (v – 2  ) 2 – ———— 4  2 (2  ) 1/2 2 1/2  exp w 2 – —— 4  2 This is the p.d.f. for a N(, ) distribution. 22 2222 0 2222 Consequently, V and W must be independent random variables with the respective normal distributions stated above, since the joint p.d.f. is the product of the normal p.d.f. for W and the normal p.d.f. for V.

11. 3. (a) If X 1, X 2, …, X n are a random sample from a N( ,  2 ) distribution, then X = is the sample mean, and S 2 = is the sample variance. Suppose n = 2, that is, the random sample is X 1, X 2. n  X i i = 1 n n  (X i – X) 2 i = 1 n – 1 Let V = X 1 + X 2 and W = X 1 – X 2, and show that X = and S 2 =. V — 2 W 2 — 2 X = X 1 + X 2 ——— = 2 V — 2 S 2 = (X 1 – X) 2 + (X 2 – X) 2 ———————— = 2 – 1 X 1 + X 2 X 1 + X 2 X 1 – ——— + X 2 – ——— = 2 2 2

12. X 1 – X 2 X 2 – X 1 ——— +——— = 2 22 X 1 – X 2 ——— = 2 2 2 W 2 — 2 S 2 = (X 1 – X) 2 + (X 2 – X) 2 ———————— = 2 – 1 X 1 + X 2 X 1 + X 2 X 1 – ——— + X 2 – ——— = 2 2 2 (b)Complete the following statements: From Class Exercise #2, we find that V = X 1 + X 2 and W = X 1 – X 2 are independent random variables with respective and distributions. From Theorem 3.6-1, we find that has a distribution. From Theorem 3.6-2, we find that has a distribution. N(2 , 2  2 ) N(0, 2  2 ) W 2 —– 2  2 N(0,1)  2 (1) W ——  2 

13. 3.-continued (c) Explain how the results from part (b) prove Theorem 5.5-2 when n = 2. Since V and W are independent, then it seems clear that X = V / 2 and S 2 = W 2 / 2 are independent. (It is actually somewhat complicated to write a rigorous proof that if V and W are independent random variables, then u 1 (V) and u 2 (W) are independent random variables, even though this result seems obvious and can be extended to any number of random variables.) (1) (2) (n – 1)S 2 ———— =  2 W 2 —– 2  2 has a  2 (1) =  2 (n – 1) distribution. (A complete proof of Theorem 5.5-2 requires matrix algebra.) S 2 —– =  2

14. 4. (a) (b) (c) (d) (e) (f) The random variables X and Y are independent with respective N(10, 16) and N(5, 36) distributions. Use an appropriate previous result to find the distribution for each of the following random variables: X + Y has adistributionN(15, 52)(from Theorem 5.5-1). X – Y has adistributionN(5, 52) X – 3Y has a distributionN(–5, 340) (X – 10) / 4 has a distributionN(0, 1) (X – 10) 2 / 16 has a distribution  2 (1) (X – 10) 2 / 16 + (Y – 5) 2 / 36 has a distribution  2 (2) (from Theorem 5.5-1). (from Theorem 3.6-1). (from Theorem 3.6-1 and Theorem 3.6-2). (from Corollary 5.4-3).

15. 5. (a) (b) The random sample X 1, X 2, …, X 6 is taken from a N(10, 81) distribution, and the following random variables are defined: Q =W = Find constants a and b such that P(a < Q < b) = 0.98. 6  (X i – 10) 2 i = 1 81 6  (X i – X) 2 i = 1 81 Q has adistribution.  2 (6) P( < Q < ) = 0.98 0.87216.81 Find P(W > 11.07). W has adistribution.  2 (5) P(W > 11.07) = 0.05

16. The random variables X and Y are independent, and each has a N(0, 1) distribution. The random variables R and  are defined to be the polar coordinates of the point (X,Y). 6. (a) (b) Find the joint p.d.f. of X and Y. Since X and Y are independent, their joint p.d.f. is f(x,y) = if –  < x < , –  < y <  22 e – (x 2 + y 2 ) / 2 Use the change-of-variables method to find the joint p.d.f. of R and . First, we find the space of R and  as follows: –  < x <  r <  –  < y <   < 0  0 22

17. 6.-continued Then, we find the inverse transformation as follows: r =x =   =y = (x 2 + y 2 ) 1/2 angle between vector (x,y) and positive x axis r cos  r sin  Next, we find the Jacobian determinant as follows: J = det  x —  r   y —  r  = cos  det – r sin  sin  r cos  = r

18. The joint p.d.f. of R and  is g(r,  ) = if 0 < r < , 0 <  < 2  22 r e – r 2 / 2 (c) Find the marginal p.d.f. for R and the marginal p.d.f. for . g 1 (r) = 0 22 d  =  = 0 22  = r e if 0 < r 22 r e – r 2 / 2 22 r e – r 2 / 2 R has p.d.f.  has p.d.f. g 2 (  ) = 0  dr = r = 0  = 1 — if 0 <  < 2  2  22 r e – r 2 / 2 22 – e – r 2 / 2 We recognize that  has a distribution.U(0, 2  )

19. g(r,  ) = if 0 < r < , 0 <  < 2  22 r e – r 2 / 2 g 1 (r) = r e if 0 < r – r 2 / 2 g 2 (  ) = 1 — if 0 <  < 2  2  We note that g(r,  ) = g1(r)g1(r) g 2 (  ) which implies that the random variables R and  are independent.

20. The random variables Z and U are independent and have respectively a N(0, 1) distribution and a  2 (r) distribution. 7. (a) (b) Find the joint p.d.f. of (Z, U). Since Z and U are independent, their joint p.d.f. is (2  ) 1/2 e – z 2 / 2  (r/2) 2 r/2 u r/2 – 1 e – u/2 if –  < z < , 0 < u <  f(z,u) = Define the random variable T =. Use the distribution function method to find the p.d.f. of T by completing the steps outlined, and making use of the two facts from calculus reviewed in Class Exercise #7 in Section 5.2. Z ———  U / r The space of T = is {t | –  < t <  } Z ———  U / r

21. The distribution function of T = is G(t) = P(T  t) = Z ———  U / r P Z ———  t =  U / r P( Z  t  U / r ) = 0  –  t  u / r 1 ————  (r/2)  2 (r+1)/2 e – z 2 / 2 dzu r/2 – 1 e – u/2 du

22. The p.d.f. of T is g(t) = G / (t) = 7. - continued 0  1 ————  (r/2)  u r/2 – 1 e – u/2 du = 2 (r+1)/2 e – (u / 2)(t 2 / r)  u / r 0  1 ————— (  r) 1/2  (r/2) u (r+1)/2 – 1 du 2 (r+1)/2 e – (u / 2)(1 + t 2 / r) (To simplify this p.d.f., we could either (1) make an appropriate change of variables in the integral, as is done in the proof of Theorem 5.5-3 in the textbook, or (2) do some algebra to make the formula under the integral a p.d.f. which we know must integrate to (one) 1, as we shall do here.)

23. 0  (  r) 1/2  (r/2) du [2 / (1 + t 2 / r)] (r+1)/2 eu (r+1)/2 – 1 u – —————— 2 / (1 + t 2 / r) (1 + t 2 / r) (r+1)/2  [(r + 1)/2] This is the p.d.f. for a distribution, so the integral must equal (one) 1. gamma[(r + 1)/2, 2 / (1 + t 2 / r)] This must be the p.d.f. we seek, for –  < t < . 0  1 ————— (  r) 1/2  (r/2) u (r+1)/2 – 1 du = 2 (r+1)/2 e – (u / 2)(1 + t 2 / r) This is the p.d.f. for a random variable having a Student’s t distribution with r degrees of freedom. This distribution is important in some future applications of the theory of statistics.

24. 8. (a) (b) (c) (d) Suppose the random variable T has a t distribution with r degrees of freedom. 0.975If r = 15, then P(T < 2.131) = If r = 15, then P(T < –2.131) =0.025 If r = 15, then P(–1.753 < T < 2.602) =0.99 – 0.05 = 0.94 If r = 8, then find a constant c such that P(|T| < c) = 0.99. P(|T| < c) = 0.99P(– c < T < c) = 0.99 P(T < c) – P(T < – c) = 0.99P(T < c) – (1 – P(T < c)) = 0.99 P(T < c) = 0.995c = t 0.005 (8) = 3.355 (e)(f) (g)(h) t 0.995 (15) =–2.947 t 0.10 (8) =1.397 t 0.90 (20) =–1.325 t 0.995 (57) =–2.576

25. If X 1, X 2, …, X n are a random sample from a N( ,  2 ) distribution, then the sample mean X =has a distribution. n  X i i = 1 n N( ,  2 / n) According to Corollary 5.5-1: This motivates the following question, which turns out to be extremely important in the development of statistical analysis: If X 1, X 2, …, X n are a random sample from a non-normal distribution, then what type of distribution does the sample mean X = have? n  X i i = 1 n In order to thoroughly answer this question, we shall cover Sections 10.5 and 10.6 before continuing in Chapter 5 with Section 5.6.