STAT 516 Answers Homework 6 April 2, 2008 Solutions by Mark Daniel Ward PROBLEMS

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1 STAT 56 Aswers Homework 6 April 2, 28 Solutios by Mark Daiel Ward PROBLEMS Chapter 6 Problems 2a. The mass p(, correspods to either o the irst two balls beig white, so p(, 8 7 4/39. The mass p(, correspods to the irst ball beig red ad the secod ball 3 2 beig white, so p(, 8 5 /39. The mass p(, correspods to the irst ball beig 3 2 white ad the secod ball beig red, so p(, 5 8 /39. The mass p(, correspods 3 2 to both o the irst two balls beig white, so p(, 5 4 5/ b. Similarly, we have p(,, , ad p(,, p(,, p(,, , ad p(,, p(,, p(,,, ad ially p(,, a. The mass p(, correspods to the white balls umbered ad 2 ot appearig withi the three choices, so p(, ( 3 5/26. The mass p(, correspods to the white ( 3 3 balls umbered ot beig chose, ad the white ball umbered 2 gettig chose, withi the three choices, so p(, ( ( 2 5/26. The mass p(, correspods to the white balls ( 3 3 umbered 2 ot beig chose, ad the white ball umbered gettig chose, withi the three choices, so p(, ( ( 2 5/26. The mass p(, correspods to both o the white ( 3 3 balls umbered ad 2 appearig withi the three choices, so p(, (2 2( /26. ( 3 3 3b. Similarly, we have p(,, ( 3 ( 3 3 6, ad p(,, p(,, p(,, 43 ( ( 2 ( , ad p(,, p(,, p(,, ( ( ( 286 ( 3 3 5, ad ially p(,, 43 ( ( ( ( The oly modiicatio rom problem 3a above is that the balls are replaced ater each draw. Thus p(, ( 3 ( 3 33/297; also, p(, 3 2 ( ( ( 2 ( /297; similarly, p(, 397/297; ad ially, p(, 6 ( ( ( ( ( ( 3 2 ( 3 72/ We irst ote that N, N 2 4 ad N + N 2 5. There are ( 5 2 ways that the two deective items ca be chose, each equally likely, ad each way correspods to exactly oe pair, 2 satisyig the bouds metioed. Thus p(, 2 / i, 2 4 ad Otherwise p(, 2.

2 2 a. We irst compute Aother method is to compute P (X < Y P (X < Y y x e (x+y dx dy /2 e (x+y dy dx /2 Fially, a third method is to just otice that X ad Y are idepedet ad have the same distributio, so hal the time we have X < Y ad the other hal o the time we have Y < X. Thus P (X < Y /2. b. For a <, we have P (X < a. For a >, we compute P (X < a a (x, y dx dy a e (x+y dx dy e a Aother method is to simply ote that the joit desity o X ad Y shows us that, i this case, X ad Y must be idepedet expoetial radom variables, each with λ. So P (X < a e a or a >, ad P (X < a otherwise, sice this is the cumulative distributio uctio o a expoetial radom variable.. There are ( 5 2,,2 3 ways that the customers ca be split ito a pair who will buy ordiary sets, a sigle buyer who will buy plasma, ad a pair who will buy othig. The probability o each such choice is (.45 2 (.5 ( So the desired probability is (3( We write X or the locatio o the ambulece, ad Y or the locatio o the accidet, both i the iterval [, L]. The distace i betwee is D X Y. We kow that P (D < a or a < ad P (D < a or a L, sice the miimum ad maximum distaces or D are ad L, respectively. So D must be betwee ad L. Perhaps the easiest method or computig P (D < a with a L is to draw a picture o the sample space ad the divide the desired area over the etire area o the sample space; this method works sice the joit distributio o X, Y is uiorm. So the desired probability is 2 (L a2 L 2 2 (L a2 a(2l a/l 2. L 2 Aother possibility is to itegrate, ad we eed to break the desired itegral ito three regios: P (D < a a x+a dy dx + L2 L a x+a a x a L L dy dx + L2 L a x a dy dx a(2l a/l2 L2 5a. We ote that c dx dy, so /c dx dy area o regio R. R R 5b. We see that, or < a, b <, we have F X,Y (a, b P (X a, Y b a b a ( b ( /4 dy dx. Thus F 2 2 X,Y (x, y x ( y (. So we have successully actored the cumulative distributio uctio ito two parts: oe part is a uctio o 2 2 x ad the other part is a uctio o y, so X ad Y are idepedet, ad their cumulative distributio uctios have the required orms or uiorm radom variables o the iterval (,, i.e., F X (a F Y (a a ( or < a < ad F 2 X (a F Y (a otherwise. 5c. We itegrate X 2 +Y 2 (/4 dx dy area o the circle X2 + Y 2 area o the etire square π2 π/4. 4 6a. We see that A happes i ad oly i at least oe o the A i happe. So A i A i.

3 6b. Yes, the A i are mutually exclusive. We caot have more tha oe A i occur at the same time. 6c. Sice the A i are mutually exclusive, the P (A P ( i A i i P (A i. We see that P (A i /2, sice we require each o the other poits (besides the ith poit itsel, o course to be i the semicircle clockwise o the ith poit. So P (A i /2. 2 9a. The margial desity o X is X (x or x ad also or x. For < x <, the margial desity o X is x X (x x dy 9b. The margial desity o Y is Y (y or y ad also or y. For < y <, the margial desity o Y is 9c. The expected value o X is E[X] 9c. The expected value o Y is E[Y ] Y (y y x X (x dx y Y (y dy dx l(/y x (x( dx /2 y l(/y dy /4 To see this, use itegratio by parts, with u l(/y ad dv y dy. 2a. Yes, X ad Y are idepedet, because we ca actor the joit desity as ollows: (x, y X (x Y (y, where { { xe x x > e y y > X (x Y (y else else 2b. No; i this case, X ad Y are ot idepedet. To see this, we ote that the desity is ozero whe < x < y <. So the domai does ot allow us to actor the joit desity ito two separate regios. For istace, P ( < X < > sice X ca be i the rage 4 betwee /4 ad. O the other had, P ( < X < 4 Y 8, sice X caot be i the rage betwee /4 ad whe Y /8; istead, X must always be smaller tha Y. 23a. Yes, X ad Y are idepedet, because we ca actor the joit desity as ollows: (x, y X (x Y (y, where { { 6x( x < x < 2y < y < X (x Y (y else else 23b. We compute E[X] x X(x dx x6x( x dx /2. 23c. We compute E[Y ] y Y (y dy y2y dy 2/3. 23d. We compute E[X 2 ] x2 X (x dx x2 6x( x dx 3/. Thus Var(X 3 ( 2 2 /2. 3

4 4 23e. We compute E[Y 2 ] y2 Y (y dy y2 2y dy /2. Thus Var(Y 2 ( / Sice N is Biomial with 6 ad p / 6, the is large ad p is small, so N is well approximated by a Poisso radom variable with λ p. So P (N i e λ λ i e. i! i! 26a. Sice A, B, C are idepedet, we multiple their margial desities to get the joit desity. Each o these variables has desity o the iterval (, ad desity otherwise. So the joit desity is (a, b, c or < a, b, c < ad (a, b, c. So the joit distributio is F (a, b, c F A (af B (bf C (c, where F A (a, F B (b, ad F C (c are each the cumulative distributio uctios o a uiorm (, radom variable, i.e., each o these uctios has the orm F (x i x, or F (x x i < x <, or F (x i x. 26b. The roots o the equatio Ax 2 +Bx+C are give by x b± b 2 4ac; these roots are 2a real i ad oly i b 2 4ac, which happes with probability (a, b, c da db dc, b 2 4ac which is exactly the volume o the regio {(a, b, c b 2 4ac } divided by the volume o the etire regio {(a, b, c < a, b, c < }. The secod regio has volume. The irst regio has volume mi{,b 2 /(4c} dadbdc So the desired probability is approximately To see how to do the itegral above, we compute mi{,b 2 /(4c} } da db dc mi {, b2 db dc 4c which simpliies to ( /4 4c b 2 4c db + b 2 db dc + 4c /4 4c db dc l( The cumulative distributio uctio o Z X /X 2 is P (Z a or a, sice Z is ever egative i this problem. For < a, we compute ( ax2 X P (Z a P a P (X ax 2 λ λ 2 e (λ x +λ 2 x 2 dx dx 2 λ a X 2 λ a + λ 2 A alterative method o computig is to write ( ( X P (Z a P a P X 2 a X X 2 x /a λ λ 2 e (λ x +λ 2 x 2 dx 2 dx λ a λ a + λ 2 32a. We assume that the weekly sales i separate weeks is idepedet. Thus, the umber o the mea sales i two weeks is (by idepedece simply (2( The variace o sales i oe week is 23 2, so that variace o sales i two weeks is (by idepedece simply (2(23 2 5,8. So the sales i two weeks, deoted by X, has ormal distributio with mea 44 ad variace 5,8. So P (X > 5 P ( X 44 5,8 > P (Z >.84 P (Z ,8

5 5 Φ( b. The weekly sales Y has ormal distributio with mea 22 ad variace ,9. So, i a give week, the probability p that the weekly sales Y exceeds 2 is p P (Y > 2 ( Y 22 P > 52,9 P (Z >.87 P (Z <.87 Φ( ,9 The ( probability that weekly sales exceeds 2 i at least 2 out o 3 weeks is (approximately 3 2 p 2 ( p + ( 3 3 p a. Write X or Jill s bowlig scores, so X is ormal with mea 7 ad variace Write Y or Jack s bowlig scores, so Y is ormal with mea 6 ad variace So X is ormal with mea 7 ad variace Thus, Y X is omal with mea 6 7 ad variace So the desired probability is approximately ( Y X ( P (Y X > P > ( ( P Z > 2 5 ( P Z 2 5 Φ( Sice the bowlig scores are actually discrete iteger values, we get a eve better approximatio by usig cotiuity correctio P (Y X > P (Y X.5 ( Y X ( P > 625 P (Z >.42 P (Z.42 Φ( ( 625

6 6 33b. The total o their scores, X + Y, is omal with mea ad variace So the desired probability is approximately ( X + Y P (X + Y > 35 P > ( P Z > 4 5 P (Z.8 Φ( Sice the bowlig scores are actually discrete iteger values, we get a eve better approximatio by usig cotiuity correctio ( X + Y P (X + Y 35.5 P > P (Z >.82 P (Z.82 Φ( a. We recall rom problem 3 that Y ad Y 2 have joit mass p(, 5/26, p(, 5/26, p(, 5/26, ad p(, /26. 5/26 So the coditio mass o Y, give that Y 2, is p Y Y 2 ( 5 ad 5/26+/26 6 p Y Y 2 ( /26 5/26+/ b. The coditio mass o Y, give that Y 2, is p Y Y 2 ( p Y Y 2 ( 5/26 5/26+5/ a. For y x 5, the joit mass is p(x, y p(xp(y x 5 39b. The coditio mass o X, give Y i, is p X Y (x i p(x, i p Y (i p(x, i 5 x p(x, i 5x 5 x 5x. x 5x 5x 37/3 6 37x 5/26 5/26+5/ ad 4. First, ote that there is oly oe way to obtai X ad Y as the same value, but or X > Y, there are two ways to obtai X ad Y as the same value. So, or y < i, the coditioal mass o Y, give X i, is p Y X (y i p(i, y p X (i p(i, y i y p(i, y (2(/6(/6 (i (2(/6(/6 + (/6(/6 2 2i For y i, the coditioal mass o Y, give X i, is p Y X (y i p(i, y p X (i p(i, y i y p(i, y (/6(/6 (i (2(/6(/6 + (/6(/6 2i

7 For y > i, the coditioal mass o Y, give X i, is p Y X (y i. Also ote that X ad Y are depedet. For istace, P (Y > 3, because Y ca take the value 3. O the other had P (Y > 3 X 2. So X ad Y are depedet. Oce X is give, or istace, the Y ca be o larger tha X. 4a. The coditioal mass uctio o X give Y is p X Y (, p(, p Y ( /2 ad p X Y (2, The coditioal mass uctio o X give Y 2 is p X Y (, 2 p(, 2 p Y ( /3 ad p X Y (2, 2 p(2, p Y ( p(2, 2 p Y ( /2 2/3 4b. Sice the coditioal mass o X chages depedig o the value o Y, the the value o Y aects the various probabilities or X, so X ad Y are ot idepedet. 4c. We compute 7 ad ad P (XY 3 p(, + p(2, + p(, P (X + Y > 2 p(2, + p(, 2 + p(2, P (X/Y > p(2, /8 54a. We see that u g (x, y xy ad v g 2 (x, y x/y. Thus x h (u, v uv ad y h 2 (u, v u. The Jacobia is v J(x, y y x x y x y 2x y so J(x, y y. Thereore the joit desity o U, V is 2x U,V (u, v X,Y (x, y J(x, y y x 2 y 2 2x 2x 3 y 2( uv 3 u v y x y 2 2u 2 v 57. We see that y g (x, x 2 x + x 2 ad y 2 g 2 (x, x 2 e x. Thus x h (y, y 2 l(y 2 ad x 2 y l(y 2. The Jacobia is J(x, y e x ex so J(x, y e x. Thereore the joit desity o Y, Y 2 is Y,Y 2 (y, y 2 X,X 2 (x, x 2 J(x, x 2 λ λ 2 e λ x λ 2 x 2 e x λ λ 2 e ((λ +x +λ 2 x 2 λ λ 2 e ((λ + l(y 2 +λ 2 (y l(y 2 λ λ 2 y λ +λ 2 2 e y λ 2

8 8 THEORETICAL EXERCISES 2. Geeralizig Propositio 2. (there is othig special about two variables, we ote that radom discrete variables X,..., X are idepedet i ad oly i their joit mass (x,..., x ca be actored as (x (x ; i this case, oce each i is ormalized so that x i i (x i or each i, the the i s are the margial mass uctios o the X i s. Write X or the total umber o evets i the give time period, ad write X i as the umber o evets o type i. The we ca actor the joit mass (x,..., x o X,..., X by writig (x,..., x P (X x + + x P (X x,..., X x X + + X e λ λ x + +x ( x + + x p x p x (x + + x! x,..., x e λ λ x + +x (x + + x! p x p x (x + + x!x! x! e λ λ x λ x p x p x x! x! e λp (λp x x! e λp (λp x x! So e λp i(λp i x i x i! is the mass o X i, ad also X,..., X are idepedet. 5a. For a >, the cumulative distributio uctio o Z is F Z (a P (Z a P (X/Y a P (X ay ay X (x Y (y dx dy Y (y ay X (x dx dy Y (yf X (ay dy, or equivaletly, or z >, we have F Z (z Dieretiatig throughout with respect to z yields Z (z Y (yf X (zy dy Y (y X (zyy dy O course, or z, we have Z (z. Whe X, Y are idepedet expoetial radom variables with parameters λ, λ 2, this yields Z (z λ 2 e λ 2y λ e λ zy y dy λ λ 2 (λ 2 + zλ 2 or z > ; as beore, Z (z or z. 5b. For a >, the cumulative distributio uctio o Z is F Z (a P (Z a P (XY a P (X a/y a/y X (x Y (y dx dy Y (y a/y X (x dx dy Y (yf X (a/y dy, or equivaletly, or z >, we have F Z (z Y (yf X (z/y dy

9 Dieretiatig throughout with respect to z yields Z (z Y (y X (z/y y dy O course, or z, we have Z (z. Whe X, Y are idepedet expoetial radom variables with parameters λ, λ 2, this yields Z (z λ 2 e λ 2y λ e λ z/y y dy or z >, but I do ot see a easy way to simpliy this expressio; as beore, Z (z or z. 6. Method. My solutio does NOT use iductio. I the X i are idepedet ad idetically distributed geometrics, each with probability p o success, the P (X + + X i (( p x p (( p x p x + +x i p ( p x + +x i p ( p x + +x i ( p x + +x ( p i p ( p ( pi ( p i ( pi ( p ( i p ( p i x + +x i usig Propositio 6. o Chapter ad thus X + + X has a egative biomial distributio with parameters p ad. Method 2. Here is a solutio with o derivatio eeded. Do cosecutive experimets. I each experimet, lip a coi with probability p o ladig heads. The X + + X is the total umber o lips required. O course, the total umber o lips is a egative biomial radom variable, because the experimet eds whe we see the th success, or i other words, whe we see the appearace o the th head. 9. For coveiece, write Y mi(x,..., X. For a, we kow that F Y (a, sice Y is ever egative. For a >, we have F Y (a P (Y a P (Y > a P (mi(x,..., X > a P (X > a, X 2 > a,..., X > a P (X > ap (X 2 > a P (X > a (e λa (e λa (e λa e λa. Thus Y is expoetially distributio with parameter λ.. The lashlight begis with 2 batteries istalled ad 2 replacemet batteries available. Whe oe battery dies, the battery is immediately replaced, ad because o the memoryless property o expoetial radom variables, both batteries istalled (the old ad ew are just as good as ew, ad the waitig begis agai. A dead battery is replaced a total o 2 times. Upo the ( st battery dyig, we do ot have eough batteries let to ru the lashlight aymore. So the legth o time that the lashlight ca operate is X + + X, 9

10 where the X i s are idepedet, idetically distributio expoetial radom variables, each with parameter λ. [So the legth o time that the lashlight ca operate is a gamma radom variable with parameters (, λ; see, or istace, Example 3b o page 282. I you did ot get this last setece, do t worry, because you are ot required to uderstad gamma radom variables i my course.] 4a. We thik o X as the umber o lips o a coi util the irst appearace o heads, ad Y as the umber o additioal lips o a coi util the secod appearace o heads; here, heads appears o each toss with probability p. Give that X + Y, we kow that lips were required to reach the secod head. So ay o the irst lips could be the irst head, ad all such possibilities are equally likely, so P (X i X + Y or i ad P (X i X + Y otherwise. 4b. To veriy this, we ote that i i or i, we must have P (X i X +Y, because X, Y are positive radom variables. For i, we compute 5. Method. We compute P (X i ad X + Y P (X i X + Y P (X + Y P (X i, Y i P (X + Y P (X ip (Y i j P (X jp (Y j ( p i p( p i p j ( pj p( p j p ( p 2 ( ( p 2 P (X i ad X + Y m P (X i X + Y m P (X + Y m P (X i, Y m i P (X + Y m P (X ip (Y m i m j P (X jp (Y m j ( i p i ( p i( m i p m i ( p m+i m ( j j pj ( p j( m j p m j ( p m+j ( i( m i ( j( m j ( i( m i ( 2 m m j

11 Method 2. As i Ross s hit, lip 2 cois, let X deote the umber o heads i the irst sequece o lips ad let Y deote the umber o heads i the secod sequece o lips. I m lips are heads altogether, there are ( ( 2 m equally likely possibilities, exactly ( i m i o which have i heads i the irst sequece o lips ad the other m i heads i the secod sequece o lips. So P (X i X + Y m ( i( m i, ad thus, give X + Y m, we ( 2 m see that the coditioal distributio o X is hypergeometric. 8a. Give the coditio U > a, the U is uiormly distributed o the iterval (a,. To see this, just cosider ay b with a < b <. The P (U < b U > a P (a<u<b b a. P (U>a a Dieretiatig with respect to b yields the coditioal desity o U, amely,, which is a costat (sice a is ixed i this problem. So the coditioal distributio o U is uiorm o the iterval (a,, as we stated at the start. 8b. Give the coditio U < a, the U is uiormly distributed o the iterval (, a. To see this, just cosider ay b with < b < a. The P (U < b U < a P (U<b b. P (U<a a Dieretiatig with respect to b yields the coditioal desity o U, amely,, which is a costat (sice a is ixed i this problem. So the coditioal distributio o U is uiorm o the iterval (, a, as we stated at the start.

Problems from 9th edition of Probability and Statistical Inference by Hogg, Tanis and Zimmerman:

Problems from 9th edition of Probability and Statistical Inference by Hogg, Tanis and Zimmerman: Math 224 Fall 2017 Homework 4 Drew Armstrog Problems from 9th editio of Probability ad Statistical Iferece by Hogg, Tais ad Zimmerma: Sectio 2.3, Exercises 16(a,d),18. Sectio 2.4, Exercises 13, 14. Sectio