Solutions to Homework 6

Transcription

1 Solutios to Homework Math 500-, Summer 00 July 7, 00 p. 0 07, #. Clearly, Y has a biomial distributio with parameters 3 ad p. That is, k 0 3 PfY kg ( )3 ( )3 3 ( )3 3 ( )3 `3 0 `3 `3 `3 3 E(Y ) 0 ˆ «+ ˆ 3 «+ ˆ 3 «+ 3 ˆ «4 3 Also, E(Y 4 ) 0 4 ˆ «+ 4 ˆ 3 «+ 4 ˆ 3 «+ 3 4 ˆ «3 33 Var(Y ) E(Y 4 ) ` E(Y ) 33 ` 3 5 p. 0 07, #3. Kow EX EY EZ ad VarX VarY VarZ. (a) E(X + 3Y ) E(X) + 3E(Y ) 5. (b) Var(X + 3Y ) Var(X) + Var(3Y ), by idepedece. Var(X + 3Y ) 4VarX + 9VarY

2 (c) We kow that if X ad X are idepedet, the E(X X ) E(X )E(X ). This ad iductio together prove that if X ; ; X are idepedet, the E(X X ) E(X ) E(X ). i particular, (d) We write E(XY Z) E(X) E(Y ) E(Z) Var(XY Z) E(X Y Z )`fe(xy Z)g E(X ) E(Y ) E(Z )` Now, Var(X) E(X ) ` (EX) E(X ) `. E(X ) + 3. Similarly, E(Y ) E(Z ) 3, ad cosequetly, Var(XY Z) 3 3 ` p. 0 07, #0. (a) Clearly, E(X k ) Ad k ˆ «+ + k ˆ «k + + k E ˆ(X + ) k k + + ( + ) k s(k ; + ) ` s(k ; ) (b) By the biomial theorem, (X + ) k X 0 k 0 + X k` + X k` + + X k 0 k X k + kx k` + X k` + + kx k` + X k` + + (X + ) k ` X k Take expectatios to fid that» k E kx k` + X k` + + s(k ; + ) ` ; ) `s(k The right-had side is times s(k ; + ) ` s(k ; ) ` ( + ) k `, whece follows the desired result. (c) E(X) ( + + ), ad it is easy to see that ( + ) + +

3 For istace, ote that ( + + ) ( + ) because we ca write ( + + ) as ( ` ) + + ; but sum i colums to see that every colum is +, ad there are colums. E(X) + (d) Apply (b) with k 3 to fid that Now, Recall that E ˆ3X + 3X + ( + )3 ` E(X) + ) 3E(X 3( + ) ) + + {z } E[3X +3X+] ( + )3 ` Solve to fid that E(X )» ( + )3 ` 3( + ) ` ` 3» ` ` 3» ` 3 3 ` 3 `» ˆ ( + )( + ) s( ;) + + Because the left-had side is, we have s( ; ) ( + )( + ). I other words, this gives a probabilistic proof of the followig classical idetity [due to the Archimedes] + + ( + )( + ) 3

4 (e) VarX E(X ) ` (EX). «( + )( + ) + VarX ` + + ` + «3 + «(4 + ) ` (3 + 3) (f) Directly check. + ˆ ` (g) Now we apply (b) with k 4 ` E ˆ4X 3 + X + 4X + ( + )4 ` () Because E(X) + ad E(X ) (+)(+), the lefthad side is 4E(X 3 ) + E(X ) + 4E(X) + s(3 ; ) 4 + ( + )( + ) + ( + ) + Plug this ito () to fid that s(3 ; ) 4 +( + )( + ) + ( + ) + {z } ( +3+)+(+) Equivaletly, s(3 ; ) ; which simplifies to s(3 ; ) 4 Thus, ( + + ) ( + )» ( + ) s(3 ; ) [s( ; )] This is aother famous formula [due to Al Karaji]» ( + ) ( + ) 4 ` {z }

5 p. 0 07, #3. Let 00 ad ff 0 respectively deote the mea ad the SD. (a) Select oe perso at radom; call his or her IQ score X. Now EX 00 ad SDX ff 0. Because (3 ˆ 0) + (3ff), P fx > 30g» P fjx ` j > 3ffg» Var X (3ff) 9 But P fx > 30g is the total umber of scores that exceed 30 divided by the populatio size. the umber of scores that exceed 30 is at most (9)th of the total populatio size. (b) By symmetry, P fx > 30g P fjx ` j > 3ffg» Var X (3ff) the umber of scores that exceed 30 is at most ()th of the total populatio size. (c) If X is approximately ormal, the we ca compute [istead of estimate, usig Chebyshev iequality], j ff X ` 00 P fx > 30g P > 3 ` (3) 0 ı ` the umber of scores that exceed 30 is approximately 03 percet of the total populatio size. p. 7, #5. Let X i deote the umber of tosses required for the ith perso to get his or her first heads. We kow that each X i is geometrically distributed where q i ` p i. P fx i kg q k` i p i for k ; ; ; (a) Mary is the secod perso. P fx > g k+ q k` p p k+ q k` q p ; ` q thaks to properties of geometric series. Because p ` q, it follows that P fx > g q. 5

6 (b) Let Y deote the miimum of X, X, ad X 3. We are asked to fid P fy > g. But P fy > g P fx > ; X > ; X 3 > g P fx > g P fx > g P fx 3 > g; by idepedece. fid that Plug i the probabilities [from (a)] to P fy > g q q q 3 (q q q 3 ) (c) Because P fy g + P fy > g P fy > ` g, we have P fy g P fy > ` g ` P fy > g (q q q 3 ) ` ` (q q q 3 ) (q q q 3 ) ` [ ` q q q 3 ] I other words, the radom variable Y distributio with parameter ` q q q 3! (d) We wat P fx > X ; X 3 > X g. Oce agai, P fx > X ; X 3 > X g has a geometric P fx ; X > ; X 3 > g P fx gp fx > gp fx 3 > g q ` p q q 3 This expressio ca be simplified as follows P fx > X ; X 3 > X g p q q 3 (q q q 3 ) ` p q q 3 (q q q 3 ) k p q q 3 ` q q q 3 p. 7, #0. We will eed, for this problem, two idetities that were discussed i the lectures. Namely, that if 0 < p <, the kp k` d dp p k d dp «` p q ; ()

7 ad because k k(k ` ) + k, k p k` k(k ` )p k` + d dp p k + p kp k` kp k` q 3 + pq (3) (a) First, ote that P fx g P (S F ) + P (F S ) pq + qp [ pq] Also, P fx 3g P (S S F ) + P (F F S 3 ) p q + q p. Ad ow we keep goig to fid that P fx kg p k` q + q k` p for all k (b) We follow the defiitio of expectatio E(X) k `p k` q + q k` p k q kp k` + p kq k` k Eq. () above tells us that kp k` ) q k k kp k` q ` kp k` q ` Similarly, P k kqk` p` ` Because p + q, it follows that E(X) q ` q + p ` p q + p ` pq ` (c) We compute E(X ) k `p k` q + q k` p k q k p k` + p k q k` k k pq k p k` + pq k q k` k k 7

8 Thaks to eq. (3), k p k` q + 3 pq ` k Similarly, k E(X ) p q k q k` p 3 + qp ` q k p k` q 3 + pq ` p + q ` q + q p + p ` p Now p + q ad p` + q` (p + q)pq (pq)`. E(X ) p q + q p + pq ` (p3 + q 3 «) + p q pq ` This ca be simplified eve further By the biomial theorem, (p + q) 3 p 3 + 3p q + 3pq + q 3 p 3 + q 3 ` 3p q ` 3pq ` 3pq(p + q) ` 3pq Cosequetly, Ad E(X ) ( ` 3pq) + p q Var(X) ( ` 3pq) + p q «pq ` ««pq ` ` pq ` Let (pq) to see that E(X) `, ad Var(X) ` + ( ` ) ` ( ` ) ` 3 ` p. 7, #. (a) Let q i ` p i to fid that P fw W g P fw W kg k k q k` p q k` p p p p p ` q q P fw kgp fw kg k (q q ) k` p p (q q ) j k j0

9 (b) Oce agai, P fw < W g P fw k < W g k k q k` p q k p q P fw kgp fw > kg k (q q ) k` p q ` q q k (c) P fw > W g is the same as P fw > W g but with the roles of (p ; q ) ad (p ; q ) reversed. That is, P fw > W g p q ` q q (d) Let M mi(w ; W ). We saw i #5 that M is geometric with parameter ` q q. Explicitly said P fm > g P fw > gp fw > g q q (q q ) P fm g P fm > `g`p fm > g (q q ) ``(q q ) Factor to fid that P fm g (q q ) [ ` (q q )] This is of the form q ` p; therefore, M has a geometric distributio with parameter ` q q. (e) Let M max(w ; W ). The, o P M < k P fw < k ; W < kg P fw < kgp fw < kg; by idepedece. Now, P fw < kg ` P fw > k + g ` q k+ ; see #5(a). Similarly, P fw < kg ` q k+. for all k, P M < k o ` q k+ ` q k+ From this we fid the distributio of M as follows P fm kg P fm < k + g ` P fm < kg (why?), whece for all k, P o M k ` q k+ ` q k+ ` ` q k+ ` q k+ 9

10 p. 33 3, #. Suppose that pulses arrive idepedetly i time; the the Radom Scatter Theorem (p. 30) tells us that the total umber of pulses i a give half-miute period is distributed accordig to the Poisso distributio with 5. p. 33 3, #0. For parts (a) ad (b) it might help to recall that E(X) ad Var(X). (a) E(3X + 5) 3E(X) (b) Var(3X + 5) 9VarX 9. (c) This part requires a direct computatio E «+ X e` e` «e` k + k k! e` k (k + )! j [j k + ] j! j j j! ` A e` `e ` ` e` j0 0