{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

ECE302HW5Soln_SP10

# ECE302HW5Soln_SP10 - ECE 302 Homework 5 Prof Saul Gelfand...

This preview shows pages 1–13. Sign up to view the full content.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: ECE 302: Homework 5 Prof. Saul Gelfand Spring 2010 Due March 26, 2010 Problem 1 a. We have that {Kt g k} is the “event that the number, Kt, of events prior to time t is g 1;.” Which means that this event is equivalently “the event that the time, Tic“, until the k + 1-th event is > t”, or {Tk+1 > t}. b. We have that the number, Kt, of events up to time t is a Poisson random variable, while the time, Tk of the k—th event is an Erlang random variable. We are given that the Erlang GDP is 1“ Mt)“ FTk(t) = Pr[T,c g t] :1- Z n' e—At n=0 By part (a), we have that k Tl P7“[Kt < k] = Pr[Tk+1 > t] = 1 — P7"[Tk+1 g t] = 2 (At) 6"“ By the hint, we have that Pr[{Kt g [5}] = Pr[{Kt S k — 1} U {Kt = [9}] = P7“[Kt g 16 ~— 1] + Pr[Kt = k] : pK,(lc) = Pr[Kt = k] = Pr[Kt g k] — Pr[Kt S k — 1] 19 19—1 _ 0‘75)” —/\t Qt)” —)\t — :0 n! e ‘2 n! e _ (At),C —/\t _ k! e J m HTm) = 1"" T'e‘ sow Awrmvdg, MT“) ~ x, nu w + = I~ F (r) .o _ 19 '1’ 2 ‘ [I (’IALY) : !- 0’21”” + y = —e 3!! = e" E 6—, w _: \LJ chT.>w)= j“) ﬁg) 6 i gt g 4:0 ‘1‘ 371.91% ‘ Vic—4%) -2 ‘3" -j + : ”I — .a e ”(‘17 ~+ 0° :le’~l‘ e T / (0 we" ‘6) Frame I T >r) : fr CTN”) : e 4 meyleu 790V}. 0 w J '” + Aim/aw»? Fr LT>I”IT>J’) = PAIN) = f‘" 4’ a yd? ' \ PFCT>J) ' e‘l \ -! .- e I “t j— ‘v 1- ram) w ﬁve 4+ koljfrCTi>W1L>W = P a j Pr (TJ>/0) \$Q—L _ (eh? 5‘ 7L ‘ MIL : "56’ ‘ I 6* %2 ‘dt 00 ,i‘ :-e r +6 We?) — —e rota/L ~§) I 6 ...i7. 09 no ”it =~e1§e ‘r/r + {1?— WV = [-15- :(O (f; EITL’TxNo] + 1: 4 - ‘37 f1 UHTLMO) ~ 7‘:ch ‘ 41% ‘two ' Pram) 52" E'CT,{IL>10]— f“? fft or) 79w) a/t - e‘ M * ~¥ ?? [0" 2. 0H- f __ e. u --'t ' v:[fooe ~rbTe; ”2+!" 6 raft ID -.L = 7:»: [JOOE‘l +fb0€ ~ “biog; Ch 1]] (hi— C’) “L ‘rf Abmnﬂ'?’ ?rCT=7-‘IT>Y)= MT“) : 7M)" ‘PrLT’A? £3;- PL “pi-1 2 RI 2— :a\ > 3 ”R k} b TIT 1] éL-HT 4] CZUtf: —L "F: 7‘ LJ‘MWO'JIQ‘ lob-ye? u— pﬁp '“ I? NOV-top’vo Ip‘ :- Pr £17"er ‘%*frCT=tIT>/J') = ”1:”, = C] s _ tn 1. (13/) Cr?) 7 M = W7 _ fay/'0 Par-(715,10) /- E. U::”)(,,r)T»-LP» 61:1. = 61—!) '0.0r. 0‘8"..." E T; = 1 I ’7‘”) ﬁlo-f Pr(T»=tI71>A03 = Muff. Problem 3: text 4.69 We need to ﬁnd the probability that the 7—th passenger comes after 10 minutes. Let T7 be a 7th order Erlang random variable. Pr[T7 > 10] = 1 — Pr[T7 g 10] = 1 — FT7(10) : 1; <1_ £103”) 6 nine-:0:1 _ T; n! 2 23:39 s 0.13 Problem 4: text 4.73 The time, TF between widget failures is exponential so that fTF(t) = /\e_’\t for t Z 0. We are given that E[TF] : i = 3 days so that A = % We have, then, that the widgets’ failures is a Poisson process with parameter A = %, so that the number, Kt, of failures before time t has Poisson PMF pKt(k) = (téjyc e—t/S Since there are four widgets in stock, the event that there is still widgets in stock after 15 days is the same as the event that fewer than four failurs occur before time 15, or the event {K 15 < 4}. We have 3 5k _ _ Pr[K15 < 4] = ZpK15(k)= 2 He 5 = e 5(1 + 5 + 25/2 + 125/6) k=O k=O Problem 5: text 5.10 If we let H0 and HM be the number of heads tossed by Carlos and Michael, respectively, and we assume that P7"[M gets heads] 2 pM and Pr[C gets heads] : pg. In both cases, we have the same form of the PMF: 2 (1—P)2, hZO PHM(h) :pHC(h) = (h) ph(1—p)2‘h= 229(1-29), h: 1 h 2 Ho 2 0 Ho 2 1 Ho = HM = (H X = 0 X = 1 1 X = Y = 0 Y = 1 Y = 2 P7“ = (1 —10M)2(1 ‘PCV PT = 2(1-PM)2Pc(1 —Pc) PT = (1~PAII)2P% HM = X = 1 X = 0 X = Y =1 Y = 2 Y = 3 P7” = 210M(1 - pM)(1 - PC)2 P7" = 4PM(1 — PM)Pc(1 — Pa) P7” = 2PM1 — PM)P% HM = 2 X = 2 T X = 1 X = 0 Y = 2 Y = 3 Y = P1" = 1011 1 — 190)2 P1" = 219111741 - pa) _1 P7“ = 191% Table 1: Problem 4: pHM’HC(hM,hC) l— X: 0 l— X = 1 X = 2 Y=0 (1—1114)2(1—pc)2 0 ‘T 0 [ =1. 0 2(1 —19M)2pc(1 - pc)+ 0 2PM“ —PM)(1 —PC)2 0 = 21 4PM( (1 —PM) )Pc(1 — Pa) 0 (1 ‘PM)2P2C +P114(1 ‘PCV Y = 3. 0 2PM(1 ‘PMXUg‘JF 0 210141941 —1%) 0 J Y = 4 ‘ pMpC 0 O Table 2: Problem 4: pXJ/(zmy) where p refers to the appropriate pM or pc. We then have nine cases for the pair (HC,HM) as tabulated 1n Table 1. The probability Pr[{HC— — ha} (1 {HM— — hM}]— — pHC(hc )pHM(hM) is listed for each pair in the table. For each pair (HC,HM), we also specify both X: |HC — HMI and Y 2 H0 + HM. We then map the probability space of (HC,HM) Shown in Table 1 to that of (X ,Y) shown in Table 2. For each pair (X, Y) in Table 2, we indicate the probability of the pair occurring. a. We use the general joint PMF pX1y(a:,y) from Table 2 with pM 2 p0 = % to obtain the case of the joint PMF shown in Table 3 ? l X20 X21 X22 Y=0 i l 0 0 Y=1 0 i 0 Im a o i lY=3 0 I 11: 0 [1:4 ﬁ 0 L0 Table 3: Problem 4(a): pXY(\$>y) y=0 a 0 0 Y=1 0 1 3g 0 Y=2 1% 0 E Y=3 0 i l 0 y=4 g 0 I O Table 4: Problem 4(b): jug/(13,31) b. From Table 3, we may easily compute the marginals as follows: 4 g, x 0 px(\$)=pry(x,y) = 5, 3:: y=0 %, LI? 2 2 f5, 31: 2 g, y : pY(y)=ZPXY(\$73/) = §7 =2 (12:0 24;.) y = E) y = C. We now use the general joint PMF pX,y(:v,y) from Table 2 with pM = %, pa = 2 to obtain the case of the joint PMF shown in Table 4 From Table 4, we may easily compute the marginals as follows: 1 25 | l PXCT) = ZPXY(\$7 y) y=0 3—2, {5:2 PYQJ) = ZpXY<x>ZD — g—é, y: 2 i=0 g y:3 &, y=4 Problem 7: text 5.57 We have fX(;c) = Ale—A17“, for :1: > 0 and fy(y) = Age—A231, for y > 0. Since X and Y are independent, we have their joint PDF is fxy(9:,y) = fx(:v)fy(y) :5 EHX — Yl] /\1)\2€_)‘1x_)‘2y / / tx—ylfxy<m,y>dxdy / / lx—ylAlkge—A1x_’\zyda:dy 0 O y 00 AlAg/ {My [/ (y — 2:)e'Awdx +/ (a: — {me—Alma] dy 0 0 y H 00 why 1 _ y _)\ 6 :AA *29—1—2 ”1 2 — d / e [A5 e >+ A1 (2131)] y = )‘2 /00 ye-Azy + ie—y(/\1+A2)dy 0 /\1 2 A; A1()\1+)\2) ‘ Z 6 Problem 8: text 5.65 In problem 5.26, we have that ny(a:, y) = k:(a:+y). We ﬁrst ﬁnd k: / / fXY(\$a3/)d\$dy = 1 _°° 7° 1 =>k/ /(:I:+y)da:dy = 1 o 0 11 :>/c/ —+ydy = 1 0 2 :>k = 1. We now ﬁnd the correlation E [X Y]: E[XY] = [O [0 mmmmdmdy ll 0“th R to Q + H {d {\D Q a 9.‘ Cd We have that the covariance Cov[X, Y] = E [X Y] —— E[X]E[Y], so we ﬁnd E [X] and E[ [Y] as follows: /01/01x(ac + y) )dydm X]=/: /: foy(:3,ydyd:3 A1562 + gdmzl—Z then by symmetry, E [Y] = T72 Therefore CO’U[X, Y] = EX[ Y]~ E[X]E [Y] = l + {1,3—4— — % We ﬁnd that X and Y are notindependent, since f(X:l:=fo:ofX1/.3,—y)dy — 13+ 5, and Slmllarlx fY(y ) — 31+ 5- 811109 fXCE )fy(y ) 7E ooin/(Q? y) then X and Y are not independent. X and Y are not orthogonal since E[X Y] # 0. X and Y are not uncorrelated since COU[X, Y] 75 0. Problem 9: text 5.28 a. We have for each of the uniform distributions depicted, j'X,y(a:, y) = k, Where k is 1 divided by the area of the region (so that f; ff; fX,y(IL', y)dxdy : 1) (a) The area of the circle is 27r, so that k = 517;. (b) The area of the diamond is 2, so that k = %. (c ) The area of the triangle is 1, so that k = 2 b. We have that the marginal PDF fX(x =f°:o f,Xy(:3 y)dy Since each of the uniform distributions shown are symmetric about the line 37 = y, then fil/(Z/ )2 fX(y ) (a) ﬁg y(.’l7, y): iwinside the circle so that for :3 6 [0,1] 00 W 1 =/ f’xy(93,y)dy = / —dy ll By symmetry, we have that fX(a:) = V 17:93 for a: 6 [—1,0), so that . _ 1:62, m 6 [—1,1] fX<x> # { 0, else, and My) = fx(y)- (b) fX,y(a:, y) = % inside the square so that for a: 6 [0,1], fxt'r): /_ fx,y<:c,y>dy = f gt 2 1—x. By symmetry, we have that fX(x) = 1 + x for m 6 [—1,0), so that Mac) ={ (i,— m’ iii—1’ 1] and My) = fx(y)- (c) fX,y(x, y) = 2 inside the triangle so that for x E [0, 1] 00 1—90 jx<m)=/_ xx,y(x,y>dy = /0 My 2—290, and My) = fx(y)- C. We have that P[X > O,Y > 0] = fooo fooo f'X,y(a:,y)d\$dy. Since fX,y(.'E, y) = k within the drawn shapes, P[X > 0, Y > 0] is thus ls times the area of the shape within the ﬁrst quadrant, or the fraction of the total area which falls in the ﬁrst quadrant. (a) P[X > 0,Y > 0] = (b) P[X > 0,Y > 0] = (C) P[X > 0,Y > 0] = 1—“ Ali—I J>IH ...
View Full Document

{[ snackBarMessage ]}

### Page1 / 13

ECE302HW5Soln_SP10 - ECE 302 Homework 5 Prof Saul Gelfand...

This preview shows document pages 1 - 13. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online