Probbility d Stochstic Processes: A Friedly Itroductio for Electricl d Computer Egieers Roy D. Ytes d Dvid J. Goodm Problem Solutios : Ytes d Goodm,4..4 4..4 4..7 4.4. 4.4. 4..6 4.6.8 4.6.9 4.7.4 4.7. 4.7.6 4.8. d 4.8.4 Problem 4..4 () By defiitio, x is the smllest iteger tht is greter th or equl to x. This implies (b) By prt (), x x x x x x Tht is, x x x This implies x lim x lim x x Problem 4..4 f X x x bx x otherwise First, we ote tht d b must be chose such tht the bove PDF itegrtes to. Hece, b d our PDF becomes x bx dx b f X x x x For the PDF to be o-egtive for x, we must hve x for ll x. This requiremet c be writte s x x For x, the requiremet holds for ll. However, the problem is tricky becuse we must cosider the cses x d x seprtely becuse of the sig chge of the iequlity.
Whe x, we hve x d the requiremet is most striget t x where we require or. Whe x, we c write the costrit s x. I this cse, the costrit is most striget t x, where we must hve or 6. Thus our complete expressio for our requiremets re 6 b As we see i the followig plot, the shpe of the PDF f X x vries gretly with the vlue of.. = 6 = = = f X (x)....4.6.8 x Problem 4..7 fid the PDF of U by tkig the derivtive of F U u. The CDF d correspodig PDF re F U u () The expected vlue of U is u u 8 u 4 u 4 u 8 u u E U (b) The secod momet of U is The vrice of U is Vr U f U u u u f U u du 8 du u 6 u 6 E U u f U u du E U E U u 4 49 u 8 du 7. u 8 u 8 u u 8 u u u 8 du u 8 du
(c) Note tht U e l U. This implies tht u du e l u du l e l u u l The expected vlue of U is the E U u f U u du u u 8l 7 6l 8 du u 8 du u 8l Problem 4.4. For, we hve the fct E X λ tht is give i the problem sttemet. Now we ssume tht E X! λ. To complete the proof, we show tht this implies tht E X! λ. Specificlly, we write E X x λe λx dx Now we use the itegrtio by prts formul udv uv vdu with u x d dv λe λx dx. This implies du x dx d v e λx so tht E X x e λx x e λx dx x λe λx dx λ λ E X By our iductio hyothesis, E X Problem 4.4.! λ which implies E X! λ () Sice f X x! d x r over the etire itegrl, we c write r x f X x dx r r f X x dx rp X r (b) We c write the expected vlue of X i the form E X r x f X x dx r x f X x dx
Hece, rp X r" Allowig r to pproch ifiity yields r x f X x dx lim rp X r" E X$ lim r r r E X# r x f X x dx x f X x dx E X# E X Sice rp X r% for ll r, we must hve lim r rp X r (c) We c use the itegrtio by prts formul udv uv vdu by defiig u F X x d dv dx. This yields F X x & dx x F X x ')( x f X x dx By pplyig prt (), we ow observe tht x By prt (b), lim r rp X F X x ')(. lim r r F X r '* lim rp X r r r d this implies x F X x &)( F X x & dx x f X x dx E X. Thus, Problem 4..6 We re give tht there re me i the Uited Sttes d of them re t lest 7 feet tll, d the heights of U.S me re idepedet Gussi rdom vribles with me + + +. () Let H deote the height i iches of U.S mle. To fid σ X, we look t the fct tht the probbility tht P H 84 is the umber of me who re t lest 7 feet tll divided by the totl umber of me (the frequecy iterprettio of probbility). Sice we mesure H i iches, we hve Sice Φ x P H Φ x 84 Q x, Q 4 σ X Φ 4 7 84 σ X From Tble 4., this implies 4 σ X or σ X 4. (b) The probbility tht rdomly chose m is t lest 8 feet tll is P H 96 Q 96 7 4 Q 6 Ufortutely, Tble 4. does t iclude Q 6, lthough it should be ppret tht the probbility is very smll. I fct, Q 6 4,. 4
(c) First we eed to fid the probbility tht m is t lest 7 6. P H 9 Q 9 7 4 Q.- 7 Although Tble 4. stops t Q 4 99, if you re curious, the exct vlue is Q β 87 7. Now we c begi to fid the probbility tht o m is t lest 7 6. This c be modeled s,, repetitios of Beroulli tril with prmeter β. The probbility tht o m is t lest 7 6 is β / / 9 4, 4 (d) The expected vlue of N is just the umber of trils multiplied by the probbility tht m is t lest 7 6. E N Problem 4.6.8 good, tht is, o foul occurs. The CDF of D obeys Give the evet G, P D F D y y( G Of course, for y 6, P D y( G D. This implies β P D y( G P G$ P D y( G c P G c P X y 6 e y 6 y 6. From the problem sttemet, if the throw is foul, the P D y( G c where u deotes the uit step fuctio. Sice P G F D y Aother wy to write this CDF is F D y u y 7, we c write P G P D y( G* P G c P D y( G c u y y 6 7 e y 6 y 6 u y 7u y 6 e y 6 However, whe we tke the derivtive, either expressio for the CDF will yield the PDF. However, tkig the derivtive of the first expressio perhps my be simpler: f D y δ y y 6 7e y 6 y 6 Tkig the derivtive of the secod expressio for the CDF is little tricky becuse of the product of the expoetil d the step fuctio. However, pplyig the usul rule for the differettio of product does give the correct swer: f D y δ y 7δ y 6 e y 6 4 7u y 6 e y 6 δ y 7u y 6 e y 6 The middle term δ y 6 e y 6 dropped out becuse t y 6, e y 6.
Problem 4.6.9 The professor is o time d lectures the full 8 miutes with probbility.7. Tht is, P T 8 7. Likewise whe the professor is more th miutes lte, the studets leve d miute lecture is observed. Sice he is lte % of the time d give tht he is lte, his rrivl is uiformly distributed betwee d miutes, the probbility tht there is o lecture is P T The oly other possible lecture durtios re uiformly distributed betwee 7 d 8 miutes, becuse the studets will ot wit loger the miutes, d tht probbility must dd to totl of 7. So the PDF of T c be writte s f T t δ t t 7 7 8 7δ t 8 t 8 otherwise Problem 4.7.4 We c prove the ssertio by cosiderig the cses where d, respectively. For the cse where we hve F Y y PY Therefore by tkig the derivtive we fid tht y P X F X f Y y f X Similrly for the cse whe we hve F Y y PY y P X F X Ad by tkig the derivtive, we fid tht for egtive, f Y y f X A vlid expressio for both positive d egtive is f Y y ( ( f X Therefore the ssertio is proved. 6
7 7 7 Problem 4.7. Uderstdig this clim my be hrder th completig the proof. Sice F x 6, we kow tht U. This implies F U u for u d F U u for u. Moreover, sice F x is icresig fuctio, we c write for u, Sice F X x F U u P F X % u F x, we hve for u, F U u Hece the complete CDF of U is P X F u F F u u F X F u F U u u u u u Tht is, U is uiform rdom vrible. Problem 4.7.6 First, we must verify tht F u is odecresig fuctio. To show this, suppose tht for u u+, x F u d x+ F u+'. I this cse, u F x d u+ F x+&. Sice F x is odecresig, F x! F x+ implies tht x x+. Hece, we c write F X x P F U. x PU F x & F x Problem 4.8. W is f W w π e w () SiceW hs expected vlue µ, f W w is symmetric bout w. Hece PC. From Defiitio 4., the coditiol PDF of W give C is f W8 C w f W w * PC w C otherwise (b) The coditiol expected vlue of W give C is e w π w otherwise PW E W (C Mkig the substitutio v w f W8 C w dw 4 7 π w, we obti we w dw E W (C π e v dv 7 π 7
(c) The coditiol secod momet of W is E W (C w f W8C w dw We observe tht w f W w is eve fuctio. Hece E W (C Lstly, the coditiol vrice of W give C is w f W w dw w f W w dw w f W w dw E W σ 6 Vr W (C E W (C E W (C9 6 π 8 Problem 4.8.4 () To fid the coditiol momets, we first fid the coditiol PDF of T. The PDF of T is f T t e t t otherwise The coditioig evet hs probbility P T : f T t dt From Defiitio 4., the coditiol PDF of T is f T8 T; : t f T t P< T; : = t otherwise e t : e e t : t otherwise The coditiol me of T is E T ( T The substitutio τ t yields : t e t : dt E T ( T τ e τ dτ τ f T τ dτ E T (b) The coditiol secod momet of T is E T ( T : t e t : dt 8
The substitutio τ t yields E T ( T τ e τ dτ Now we c clculte the coditiol vrice. τ f T τ dτ E T Vr T ( T E T ( T E T ( T 9 E T E T 9 Vr T Vr T 9