Probability and Stochastic Processes: A Friendly Introduction for Electrical and Computer Engineers Roy D. Yates and David J.

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1 Probability and Stochastic Procsss: A Frindly Introduction for Elctrical and Computr Enginrs Roy D. Yats and David J. Goodman Problm Solutions : Yats and Goodman, and Problm 4.3. f X 4 x 3 othrwis W rcogniz that X is a uniform random variabl from [-,3]. (a) E X and Var X (b) Th nw random variabl Y is dfind as Y h X X. Thrfor and Finally Problm h E X h E h X E X Var X E X E Y E h X E X 7 3 Var Y E X 4 E X 3 x 4 4 dx (a) W can find th xpctd valu of X by dirct intgration of th givn PDF. Th xpctation is f Y y E Y y y othrwis y dy 4 3 (b) E Y y 3 dy Var Y E Y E Y 4 3 9

2 Problm 4.3. Th CDF of Y is F Y y y y y y (a) W can find th xpctd valu of Y by first find th PDF by diffrntiating th abov CDF. f Y y y othrwis And E Y y dy (b) E Y y dy 3 Var Y E Y E Y 3 3 Problm 4.4. obsrv that an xponntial PDF Y with paramtr λ has PDF f Y y λ λy y othrwis In addition, th man and varianc of Y ar E Y (a) Sinc Var Y, w must hav λ. (b) Th xpctd valu of Y is E Y λ. (c) λ Var Y λ P Y f Y y dy y Problm (a) Th PDF of a continuous uniform random variabl distributd from! is f X x othrwis

3 (b) For x, F X. For x, F X. For x, th CDF is F X " Th complt xprssion for th CDF of X is (c) th xpctd valu of X is F X x f X τ dτ x x # x x x x dx x Anothr way to obtain this answr is to us Thorm 4.7 which says th xpctd valu of X is (d) Th fifth momnt of X is Th xpctd valu of X is E X x dx x $ x dx x 6 6 4% 84 Problm Givn that (a) f X P X x x othrwis x dx % 387 (b) Th CDF of X may b b xprssd as F X x x & x x dτ x x x x (c) X is an xponntial random variabl with paramtr a. By Thorm 4.9, th xpctd valu of X is E X a. (d) By Thorm 4.9, th varianc of X is Var X a 4. 3

4 Problm 4.. Givn that th pak tmpratur, T, is a Gaussian random variabl with man 8 and standard dviation w can us th fact that F T t Φ t µ T σ T and Tabl 4. on pag 4 to valuat th following P T ' P T F T Φ P T 6 ' Φ Φ % % 933 % Φ % Φ % $% 993 % 7 P 7 T ' F T ( F T 7 Φ % ) Φ % Φ % ) % 866 Problm 4..7 N µ! σ distribution, th intgral w wish to valuat is I (a) Using th substitution x f W w dw πσ w µ σ dw w µ σ, w hav dx dw σ and 8 I π x dx (b) Whn w writ I as th product of intgrals, w us y to dnot th othr variabl of intgration so that I π x dx π x y + dxdy π y dy (c) By changing to polar coordinats, x y r and dxdy rdrdθ so that I π π r rdrdθ π π r dθ π π dθ 4

5 Problm 4.6. (a) Using th givn CDF P X F X P X F X, 3 3 dnots th limiting valu of th CDF found by approaching from th lft. Whr F X Likwis, F X is intrprtd to b th valu of th CDF found by approaching from th right. W notic that ths two probabilitis ar th sam and thrfor th probability that X is xactly is zro. (b) P X ' F X 3 P X ' F X 3 Hr w s that thr is a discrt jump at X. Approachd from th lft th CDF yilds a valu of /3 but approachd from th right th valu is /3. This mans that thr is a non-zro probability that X, in fact that probability is th diffrnc of th two valus. P X P X - P X (c) P X. F X / F X 3 3 P X. F X / F X 3 3 Th diffrnc in th last two probabilitis abov is that th first was concrnd with th probability that X was strictly gratr thn, and th scond with th probability that X was gratr than or qual to zro. Sinc th th scond probability is a largr st (it includs th probability that X ) it should always b gratr than or qual to th first probability. Th two diffr by th probability that X, and this diffrnc is non-zro only whn th random variabl xhibits a discrt jump in th CDF. Problm good, that is, no foul occurs. Th CDF of D obys Givn th vnt G, F D y P D y G P G P D y G c P G c P D y G P X y 6 y 6+ y 6 Of cours, for y 6, P D y G. From th problm statmnt, if th throw is a foul, thn D. This implis P D y G c u y

6 whr u 3 dnots th unit stp function. Sinc P G % 7, w can writ Anothr way to writ this CDF is F D y P G P D y G - P G c P D y G c % 3u y y 6 % 3 % 7 y 6+ y 6 F D y % 3u y ( % 7u y 6 y 6+ Howvr, whn w tak th drivativ, ithr xprssion for th CDF will yild th PDF. Howvr, taking th drivativ of th first xprssion prhaps may b simplr: f D y % 3δ y y 6 % 7 y 6+ y 6 Taking th drivativ of th scond xprssion for th CDF is a littl tricky bcaus of th product of th xponntial and th stp function. Howvr, applying th usual rul for th diffrntation of a product dos giv th corrct answr: f D y % 3δ y ( % 7δ y 6 y 6+ ) % 7u y 6 y 6+ % 3δ y ( % 7u y 6 y 64 Th middl trm δ y 6 y 6+ droppd out bcaus at y 6, y 64. 6

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