Solutions to Problem Set 3

Transcription

1 UC Berkeley, CS 174: Combnatorcs and Dscrete Probablty (Fall 2010) Solutons to Problem Set 3 1. (MU 3.3) Suppose that we roll a standard far de 100 tmes. Let X be the sum of the numbers that appear over the 100 rolls. Use Chebyshev s nequalty to bound P[ X Let X be the number on the face of the de for roll. Let X be the sum of the dce rolls. Therefore X 100 X. By lnearty of expectaton, we wrte E[X 100 E[X. We can compute E[X jp[x j j1 j1 j(1/6) (1/6) 6(7) 2 7/2, where we use the fact that n j1 j n(n+1) 2. Then we have E[X 100(7/2) 350. To use Chebyshev s nequalty, the only remanng value we need to compute s the varance of X. By the ndependence of the dce rolls we have ( ) 100 Var(X) Var X Var(X ) To compute the varance of a sngle dce roll, we use Var(X ) E[X 2 + E[X2 E[X 2 j 2 P[X j j1 j 2 (1/6) j (7)(13) 6 91/6 where we use the fact that n 6. Now we can fnsh computng the varance of X as Var(X ) E[X 2 E[X 2 91/6 (7/2) 2 35/12. j1 j2 n(n+1)(2n+1) And the varance of X s Var(X) 100(231/4). Fnally, we can by Chebyshev s nequalty we have P[ X (35/12) / (MU 3.5) Gven any two random varables X and Y, by the lnearty of expectatons we have E[X Y E[X E[Y. Prove that, when X and Y are ndependent, Var[X Y Var[X + Var[Y. 1

2 From the defnton of varance, we wrte Var[X Y E[(X Y ) 2 E[X Y 2 E[X 2 2XY + Y 2 (E[X E[Y ) 2 E[X 2 2E[XY + E[Y 2 (E[X 2 2E[XE[Y + E[Y 2 ) E[X 2 E[X 2 + E[Y 2 E[Y 2, snce by ndependence E[XY E[XE[Y. Fnally, we see that Var[X Y Var[X + Var[Y. 3. (MU 3.6) For a con that comes up heads ndependently wth probablty p on each flp, what s the varance n the number of flps untl the kth head appears? The number of con flps untl a head s a geometrc random varable, X, wth parameter p. Let X be the number of con flps untl k heads. Then X k X. [ k Var[X Var X k Var[X k (1 p)/p 2 k(1 p)/p 2 4. (MU 3.19) Let Y be a non-negatve nteger-valued random varable wth postve expectaton. Prove E[Y 2 E[Y 2 P[Y 0 E[Y. Frst, we consder the upper bound. By Markov s nequalty, we have P[Y 0 P[Y 1 E[Y. Now, for the lower bound. Notce that one mght thnk of usng P[Y 0 1 P[Y 0, and upper boundng P[Y 0 by Chebyshev s nequalty. However, ths wll not work, because Jensen s nequalty, for convertng E[X 2 to E[X 2, cannot provde a bound n the proper drecton. (Note that the frst soluton provded to ths problem was ncorrect n attemptng to use proceed va Chebyshev s and Jensen s nequaltes.) A much more useful approach s to use condtonal expectaton to obtan an nequalty that contans P[Y 0 as one of the coeffcents. Recall that Jensen s nequalty tells us that E[X 2 E[X 2 2

3 Let X be a random varable derved from Y where X has sample space Ω Ω \ z Ω Y (z) 0 where Ω s the sample space for Y, and where the random varable X satsfes X(z) Y (z) z Ω. So, we have defned X such that X (Y Y 0). Then, the above Jensen s nequalty tell us that E[Y Y 0 2 E[Y 2 Y 0. Now we compute each sde of the above nequalty. For the left-hand, sde we have E[Y Y 0 ( P[Y Y ) 2 ( ) 2 P[Y, Y 0 P[Y 0 ( P[Y P[Y 0 E[Y 2 P[Y 0 2. ) 2 For the rght-hand sde, we have E[Y 2 Y 0 2 P[Y Y P[Y P[Y 0 E[Y 2 P[Y 0. Puttng everythng together, we have whch concludes the proof. 5. (MU 3.20) E[Y 2 P[Y 0 2 E[Y 2 P[Y 0 E[Y 2 E[Y 2 P[Y 0, (a) Chebyshev s nequalty uses the varance of a random varable to bound ts devaton from ts expectaton. We can also use hgher moments. Suppose that we have a random varable X and an even nteger k for whch E[(X E[X) k s fnte. Show that P [ X E[X > t k E[(X E[X) k 1 t k. 3

4 Let Y (X E[X) k. By Markov s nequalty we have P[Y t k E[Y E[Y t k E[Y 1. Now, t k we have [ [ k P Y t k E[Y P Y t k E[Y P [ X E[X t k E[(X E[X) k where the frst step s true snce we take the kth root of both sdes of the nequalty, and the second step s true snce the kth root of a number, where k s even, s the absolute value. Puttng ths together wth the Markov s nequalty, we have P [ X E[X t k E[(X E[X) k 1 t k. (b) Why s t dffcult to derve a smlar nequalty when k s odd? Snce X s any random varable, the value (X E[X) k may be negatve for odd values k. Therefore Markov s nequalty would not apply. 6. (MU 3.21) A fxed pont of a permutaton π : [1, n [1, n s a value for whch π(x) x. Fnd the varance n the number of fxed ponts of a permutaton chosen unformly at random from all permutatons. Let X be an ndcator random varable for the event that π(), makng a fxed pont,.e. X 1 when s a fxed pont, and X 0 otherwse. We can easly compute the E[X. Let X n be the number of fxed ponts. Frst, we notce that Var[X E[X 2 E[X 2. Next, we compute the expectaton of the number of fxed ponts. Snce the E[X P[X 1/n, we have [ n n n E[X E X E[X (1/n) 1. Now, we compute the frst term n the varance, ( n ) 2 E[X 2 E X ( n ) E[X 2 + E[X X j j ( n ) E[X + E[X X j j 1 + P[X 1E[X X j X 1 j n (n 1) j 4

5 The thrd lne follows snce for ndcator varables X 2 X. The forth lne s obtaned by usng condtonal expectaton, condtonng on the event X 1. The ffth lne comes from knowng that P[X 1 1/n, and condtonng on X 1, there are n 1 choces for mappng element j, yeldng 1/(n 1) as the condtonal probablty of j beng a fxed pont. Puttng everythng together we have Var[X (Balls and Bns) Ths problem nvolves a balls and bns experment n whch m balls are tossed ndependently nto n bns wth each ball equally lkely to land n any bn. (a) Usng a Chernoff bound, estmate the probablty that f 3n ln n balls are tossed nto n bns, the maxmum number of balls n any bn s less than or equal to 2 ln n. Ths problem was not graded due to ssues wth the problem statement. (b) Wrte a smple program that smulates the balls and bns experment for gven values of m and n up to one mllon. You wll need a random number generator; the standard C lbrary functon drand48() s recommended. Consult the man page for detals. The most straght forward way to do the smulaton s to use an array of dmenson one mllon to record the number of balls that have landed n each bn. It s wasteful of space to use one mllon ntegers to store the bn loads. Snce you are very unlkely ever to see a load greater than 15, you can n fact use a sngle byte (.e., a character) to store each bn load. There s an alternate way to do the smulaton usng an nteger array whch records, n poston, the number of bns that have receved exactly balls. Ths elmnates the need for an array of sze one mllon. If you see the trck, explan how to do the smulaton ths way. Usng one of the above two methods, perform at least 20 (preferably 100) smulatons wth m n 10 6, and make a table of the dstrbuton of the maxmum loads. Now, consder the followng alternatve scheme: balls are agan thrown sequentally, but nstead of smply choosng a sngle bn at random, each ball now chooses two bns at random, nspects ther current loads, and goes to the less full of the two (breakng tes arbtrarly). Modfy your program to mplement ths scheme. Agan perform at least 20 smulatons wth m n 10 6, and make a table of the maxmum loads that you observe. Are you surprsed by the results? Fnally, would you expect an even more dramatc effect f the balls were allowed three choces rather than two? Modfy your program agan and see what happens. For one choce the dstrbuton s Max Load P[Max Load For two choces the dstrbuton s Max Load P[Max Load For three choces the dstrbuton s Max Load P[Max Load

6 ////////////////////////////////////// // // Code for Balls and Bns Problem // ////////////////////////////////////// #nclude <vector> #nclude <tme.h> #nclude <math.h> usng namespace std; nt man (nt argc, char **argv) f (argc! 2) prntf ("USAGE: %s <number of choces>\n", argv[0); ext(-1); nt choces ato(argv[1); long nt seed tme(null); srand48(seed); nt num_balls (nt) pow(10,6); nt num_bns num_balls; nt num_smulatons 100; prntf("choces %, Smulatons %, No. Balls %, No. Bns %\n\n", choces, num_smulatons, num_balls, num_bns); vector<nt> dstrbuton; for (nt 0; < num_smulatons; ++) vector<nt> bns(num_bns, 0); // throw the balls n the bns for (nt j 0; j < num_balls; j++) nt bn -1; for (nt c 0; c < choces; c++) double r drand48()* ; 6

7 nt b ((nt) r) % num_bns; f (bn -1) bn b; else f (bns[b < bns[bn) bn b; bns[bn++; // fnd the max load nt max_load 0; for (nt j 0; j < bns.sze(); j++) f (bns[j > max_load) max_load bns[j; whle (dstrbuton.sze() < max_load) dstrbuton.push_back(0); dstrbuton[max_load++; // prnt out the dstrbuton prntf("dstrbuton of maxmum load:\n"); for (nt x 0; x < dstrbuton.sze(); x++) double probablty_x dstrbuton[x/(double) num_smulatons; prntf("[% %f\n", x, probablty_x); 7