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 n  c i X i i = 1

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  c 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  c 2 2  … + c n 2  n 2 M X (c 1 t) M X (c 2 t) … M X (c n t) = 12n

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 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 n  X i i = 1 n N( ,  2 / n)

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:

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 Theorem (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

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 < –  

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 =

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 < 

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 =

(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.

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

X 1 – X 2 X 2 – X 1 ——— +——— = 2 22 X 1 – X 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 – ——— = (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 

3.-continued (c) Explain how the results from part (b) prove Theorem 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 requires matrix algebra.) S 2 —– =  2

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 and Theorem 3.6-2). (from Corollary 5.4-3).

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) =  (X i – 10) 2 i =  (X i – X) 2 i = 1 81 Q has adistribution.  2 (6) P( < Q < ) = Find P(W > 11.07). W has adistribution.  2 (5) P(W > 11.07) = 0.05

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

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

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  )

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.

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

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

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 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.)

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.

8. (a) (b) (c) (d) Suppose the random variable T has a t distribution with r degrees of freedom If 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) = 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 (8) = (e)(f) (g)(h) t (15) =–2.947 t 0.10 (8) =1.397 t 0.90 (20) =–1.325 t (57) =–2.576

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.