北京大学：EL6303 Sample Final Exam with Solution

EL6303 Sample Final Exam with Solution (This was a real test in Fall 2013. For our ePoly class, I may change Part I to regular questions. That is no multiple choices. Part 2 will remain the same style. Part 1 X noisy channel channel Given P(X=0)=0.1, P(X=1=0.9 and for all Isksn P(X+=0X=0)=0.7,P(+1=1=0)=0.3 P(X1=0X=1)=0.2,P(CX1=1X=1)=08 ()P(X,=1)is most nearly (a)0.35(b)0.55(c)0.75(d)0.85(e)1 (2)P(X, =0) is most nearly (a)0.165(b)0.325(c)0.435(d)0.875(e) 3)P(X,=0JX=0) is most nearly (a)0(b)0.315(c)0.538(d)0.785(e) Solution: This is the binary transmission problem in Test 1 with multiple stages 01"0 0.7 0.3

1 EL6303 Sample Final Exam with Solution (This was a real test in Fall 2013. For our ePoly class, I may change Part 1 to regular questions. That is no multiple choices. Part 2 will remain the same style. ) Part 1 1. 1 X 2 X k X k 1 X noisy  channel noisy channel Given 1 P(X 0) 0.1, 1 P(X 1) 0.9 and for all 1k n, 1 ( 0| 0) 0.7 k k P X X     , 1 ( 1| 0) 0.3 k k P X X     1 ( 0| 1) 0.2 k k P X X     , 1 ( 1| 1) 0.8 k k P X X     . (1) 2 P(X 1) is most nearly ___ . (a) 0.35 (b) 0.55 (c) 0.75 (d) 0.85 (e) 1. (2) 3 P(X 0) is most nearly ___ . (a) 0.165 (b) 0.325 (c) 0.435 (d) 0.875 (e) 1. (3) 2 3 P(X 0| X 0) is most nearly ___ . (a) 0 (b) 0.315 (c) 0.538 (d) 0.785 (e) 1. Solution: This is the binary transmission problem in Test 1 with multiple stages. 0.8 0.7 0.2 0.3 0.9 "0" "0" "1" "1" 0.1

(1)P(X2=1)=P(X2=1,X1=0)or(X2=1X1=1) =P(X2=11X1=0)P(x1=0)+P(X2=1X1=1)P(X1=1) =(0.3)(0.1)+(0.8)(0.9)=0.75 P(X2=0)=1-P(X2=1)=1-0.75=0.25 2)P(X3=0)=P(X2=0,X2=0)or(X2=0,X2=1) P(X3=0,x2=0)+P(X3=0,x2=1) P(X3=0X2=0)P(X2=0)+P(X3=0X2=1)P(X2=1) (0.7)(0.25)+(0.2)(0.75)=0.325 P(X3=1)=1-P(X3=0)=1-0.325=0675 (3)P(x,=0X2=0)=P(x2=0.X=0)=Px=0x2=0x2=0 P(X3=0) P(X2=0 (0.7)(0.25)_0.175 ≈0.538 0.3250.325 -2,for-∞<x<-2 号x-1,for-2≤x<0 2. Y=g(r)with 8(x)= x+1,for0≤x<2 for 2< (4)For-2<y≤-1,f(y) (a)2f2(2y+2)(b)2fx(2y-2)(c)f2(y+)(d)fxy-)(e)None

2 (1) 2 2 1 2 1 P(X 1) P((X 1,X 0) or (X 1,X 1)) 2 1 1 2 1 1  P(X 1|X 0)P(X 0)P(X 1|X 1)P(X 1) (0.3)(0.1)(0.8)(0.9)0.75 2 2 P(X 0) 1P(X 1) 10.750.25 (2) 3 3 2 3 2 P(X 0) P((X 0,X 0) or (X 0,X 1)) 3 2 3 2  P(X 0,X 0) + P(X 0,X 1) 3 2 2 3 2 2  P(X 0|X 0)P(X 0)P(X 0|X 1)P(X 1) (0.7)(0.25)(0.2)(0.75)0.325 3 3 P(X 1) 1P(X 0) 10.3250.675 (3) 2 3 3 2 2 2 3 3 3 ( 0, 0) ( 0| 0) ( 0) 0| 0 ( 0) ( 0) ( ) P X X P X X P X X P X P X P X            (0.7) (0.25) 0.175 0.538 0.325 0.325     2. Y  g(X ) with 1 2 1 2 2, 1, 1, 2, ( ) x g x x          for 2 for 2 0 for 0 2 for 2 x x x x              . Then, (4) For 2  y  1, ( ) Y f y ____ ? (a) 2 (2 2) X f y (b) 2 (2 2) X f y (c) 1 2 ( 1) X f y (d) 1 2 ( 1) X f y (e) None

(5) For I2)=1-F(2) For-2<y<-1,f(y)=()=2f,(x)=2/,(2(y+) l g(x)l Forl<y<2,f()=(=2f(x)=2/(2(y-1) g'(x) For-2<y<-1,F(y)=P(Ysy)=P(X-1sy)=P(X≤2+1)=Fx(2(y+1) Fo1<y<2,F0y)=P(Ysy)=P(X+1sy)=P(Xs2(y-1)=F2(2(y-1) herefore

3 (5) For 1 y  2, ( ) Y F y ____ ? (a) (2 2) X F y (b) 1 2 ( 1) X F y (c) (2 2) X F y (d) 1 1 2 2 ( ) X F y (e) None. (6) At y  0, ( ) Y f y ____ ? (a) 0 (b) ( (1) ( 1)) ( ) X X F F   y (c) ( (2) ( 2)) ( ) X X F F   y (d) (0) ( ) X F  y (e) Something else. (7) At y  2, ( ) Y f y ____ ? (a) (1 (1)) ( 2) X F  y (b) (2) ( 2) X F  y (c) (1) ( 2) X F  y (d) (1 (2)) ( 2) X F  y (e) Something else. Solution: 2 2 1  2 1 x y 1 2 y  x  1 1 2 y  x  1 ( ) X f x x ( ) X F x x ( 2) ( 2) ( 2) X P Y  P X  F  , ( 2) ( 2) 1 (2) X P Y  P X   F For 2 y1, ( ) | ( )| ( ) 2 ( ) 2 (2( 1)) X Y X X f x g x f y f x f y      For 1 y2, ( ) | ( )| ( ) 2 ( ) 2 (2( 1)) X Y X X f x g x f y f x f y      For 2 y1, 1 2 ( ) ( ) ( 1 ) Y F y P Y  y P X   y ( 2( 1)) (2( 1)) X  P X  y  F y For 1 y2, 1 2 ( ) ( ) ( 1 ) ( 2( 1)) (2( 1)) Y X F y P Y  y P X   y P X  y F y Therefore

Fx(-2)5(y+2) 2fx(2(y+1) Fx(2(y+) f(y)=12/x(2(-1) FY()=E() (1-Fx(2)6(y-2) otherwise y≥2 F(-2) 1-F(2) f07少0)1 F(2) 3. X and y have joint density function f,(Y以」4(x+y0≤ysrs 0. otherwise (a)2 (b)3 (c)4 (d)8 (e) None (9)For(0,1),f(x)=? (b)5 (c)4x3 (d)3 (e) no (10)For0≤y≤x≤1,f(x|y)=? x (b) 2 2(x+y) 2(x+y) 1+2 1+2y-3y 1+2,(e) (11)For0≤x≤1,E{F|x}=? (a)3x (b)x(x+1)(c)2x(x+1)(d)(x+1)(e)No (12)E{XY}=? (a)1/2 (b)1/3 (c)1/4 (d)3/4 (e)None

4 ( 2) ( 2), 2 2 (2( 1)), 2 1 2 (2( 1)), 1 2 (1 (2)) ( 2), 2 0, otherwise ( ) X X Y X X F y y f y y f y y F y y f y                       0, 2 (2( 1)), 2 1 (0), 1 1 (2( 1)), 1 2 1, 2 ( ) X Y X X y F y y F y F y y y F y                   ( ) Y f y y 2 1 1 2 ( ) Y F y y 2 1 1 2 ( 2) X F  1 (2) X F 1 ( 2) X F  (0) X F (2) X F 3. X and Y have joint density function ( ), 0 1, ( , ) 0, otherwise. XY A x y y x f x y         (8) A=? (a) 2 (b) 3 (c) 4 (d) 8 (e) None. (9) For (0, 1), f (x)? (a) 2x (b) 4 5x (c) 3 4x (d) 2 3x (e) None. (10) For 0 y  x 1, f (x| y)? (a) 1 2 x y y   (b) 2 1 2 x  y (c) 2 2( ) 1 2 3 x y y y    (d) 2( ) 1 2 x y y   (e) None. (11) For 0 x 1, E{Y | x}? (a) 5 9 x (b) x(x1) (c) 2x(x1) (d) 1 2 (x1) (e) None. (12) E{XY}? (a) 1/2 (b) 1/3 (c) 1/4 (d) 3/4 (e) None

Solution (6A4(x+y)y)dx=46(6(x+y))dr=A6(0y+6ya x+y)=x2+x=3xb=2=1=A=2 f(x)=2x+y)y=20xd+6)=2(x2+2x2)=3x2,0≤x≤1 f(y)=2(x+1y)bx=2(x+y)x=2y(xd+y)=2(2x2)y+y1-y)=-3y2+2y+10≤ys f(xy) f(x,y)_2(x+y) 0≤y≤x≤1 f(y)-3y2+2y+1 2(x+y) E{x}=5y22=36(x+y 32(x5ydy+y2小)=3(2x2+3x3)=3x,0≤xs1 E{XY2=(x2x+y))dk=2((xyxy2)小)=卡 4. X and y have joint density function mIr LX y>0.z-max(X, n and W n(X, r) 0. otherwise min(x, y) max(, y) (13)The valid definition region of z is )(-∞,∞)(b)[1∞) )[-1,∞)(d)[0,1](e)[0,∞) (14)In its definition region, f(2)=? 8 2 3(1+z)3 (1+2)3 (e)no (15)For0<w<1,f(w)=?

5 Solution: 1 1 1 0 0 0 0 0 0 0 ( ( ) ) ( ( ) ) ( ) x x x x  A x y dy dx A  x y dy dx A  xdy ydy dx    1 1 2 1 2 3 1 2 0 0 0 0 2 2 0 2 ( ) ( ) 1 2 x x A A  A x dy ydy dx  A x  x dx   x dx    A 2 1 2 2 0 0 0 2 ( ) 2( ) 2( ) 2( ) 3 , 0 1 x x x f x   x y dy   xdy ydy  x  x  x  x 1 1 1 1 ( ) 2( ) 2 ( ) 2 ( ) y y y y f y  x y dx  x y dx  xdx ydx     1 1 2 2 2 2( | (1 )) 3 2 1, 0 1 y  x y  y   y  y  y , 2 ( , ) 2( ) ( | ) ( ) 3 2 1 0 1 f x y x y f x y f y y y y x          . 2 2 2 0 2 3 0 2( ) { | } ( ) 3 x x x x y E Y x y dy yx y dy x       2 2 2 2 2 1 3 1 3 5 3 0 0 3 2 3 9 ( ) ( ) , 0 1 x x x x  x ydy  y dy  x  x  x x  1 1 2 2 1 4 2 1 4 1 0 0 0 0 0 3 0 3 { } ( 2( ) ) 2 ( ( ) ) x x E XY    xy x y dy dx   x yxy dy dx  x dx  x dx 4. X and Y have joint density function ( ) , 0, 0 ( , ) 0, otherwise XY x y e x y f x y         , max( , ) min( , ) X Y X Y Z  and min( , ) max( , ) X Y X Y W  . (13) The valid definition region of Z is ___ . (a) (, ) (b) [1, ) (c) [1, ) (d) [0, 1] (e) [0, ). (14) In its definition region, ( ) Z f z ? (a) 3 8 3(1 ) z  z (b) 2 4 (1 z ) (c) 3 8 (1 z) (d) 2 2 (1 z) (e) None. (15) For 0w1, ( ) W f w ?

(d) (e) None (1 (16) EW is most nearly (a)0.2 (b)0.3 (c)04(d0.5(e)0.6 (17)E{ZW}=? (a (b)2 C (e)No Solution: Z-max(X, r)_X/y,X>r min(X, r)Y/X, Xsy We have 1Y)}+P{(Z≤z)∩(X≤Y)} =P(X/≤2X>Y)+P(YX≤z,X≤Y) P(X≤zY,X>Y)+P(Y≤zX,X≤Y) X≤zY,X>Y} ≤zX,X≤H} y y -X y ya X=v (ax>r (b)X≤Y The shaded region in the above figure(c)represents the desired total area. Thus F(E)=JoJ-f(x, y)dydxSoe(e )x-dxoeGete)dx b(e-(1+#1+1))dx1+2 e-(+2)x1-1e+l-)x1=++1=-1,z>

6 (a) 2 4 (1w ) (b) 2 2 (1w) (c) 3 8 3(1w) (d) 3 8 (1 ) w w (e) None. (16) E{W} is most nearly ___ . (a) 0.2 (b) 0.3 (c) 0.4 (d) 0.5 (e) 0.6. (17) E{ZW}? (a) 1 (b) 2 (c) 4 (d) 8 (e) None. Solution: max / , ( , ) min( , ) / , X Y X Y X Y X Y Y X X Y Z        . We have 1 z . FZ (z)P(Z  z)P{(Z  z)(X Y)}P{(Z  z)(X Y)}  P(X /Y  z,X Y)P(Y / X  z,X Y)  P(X  zY,X Y)P(Y  zX,X Y) x  y x yz x y {X  zY,X Y} x  y x y {Y  zX,X Y} y  xz x  y x y y  xz x yz (a) X Y (b) X Y (c) The shaded region in the above figure (c) represents the desired total area. Thus 0 / 0 / ( ) ( , ) ( ) Z x y xz xz x z x z F z f x y dydx e e dx         / 0 ( ) x xz x z e e e dx         (1 ) (1 1/ ) 0 ( ) z x z x e e dx          1 (1 ) 1 (1 1/ ) 1 1 1 , 1 0 1 1/ 0 1 1 1/ 1 | | 1 z x z x z z z z z z e e z                   

f(2)=dF(=)={+)2.2> 10. otherwise W=min(x, y, then W=1/z, 0<w<l and w===w2 max(X, r 2 dhn1y2+m2s0+)0< so that fw(w)=/(/w)_I 0. otherwise eZo odz does not converge E{W}=,-"、,h=1/3 E{∠W}=E{ r max(X, r) min(X, y) nin(X, r) max(rri= 5. X and y are independent with exponential densities fx(xr=e-u(x), f(=e-u(y). Z=X+Y and W=X/. (18)f2(x)=? (a)e-l(=) (b) (-1) u(z (c)zeu(z) (d)(z-1)e(2-(z-1) (19)fm()=?

7 2 2 , 1, (1 ) 0, otherwise. ( ) ( )/ Z Z z f z dF z dz z        min( , ) max( , ) X Y X Y W  , then W 1/Z, 0w1 and 2 21 | | | | dw w dz z   so that 1 2 2 2 2 (1 1/ ) 2 , 0 1, (1/ ) (1 ) | / | 0, otherwise. ( ) W w w f w w w dw dz f w            1 2 2 (1 ) { } zz E Z dz    does not converge. 10 2 2 (1 ) { } 1/3 ww E W  dw    . E{ZW}E{ max( , ) min( , ) X Y X Y min( , ) max( , ) X Y X Y  }1. 5. X and Y are independent with exponential densities ( ) ( ), ( ) ( ). x y X Y f x e u x f y e u y     Z=X+Y and W=X/Y. (18) ( ) Zf z ? (a) ( ) z e u z  (b) ( 1) ( 1) z e u z    (c) ( ) z ze u z  (d) ( 1) ( 1) ( 1) z z e u z     (e) None. (19) ( ) Wf w ?

(b) 2 (+1) (+1) (c)we u(w) d) u(] + (20)fm(=,w)=? (a)e-u(z) (b)e-(=)-1 l(4 (+1) 1+)n2(-1)(e)None (2 1)Are Z and w independent? (a)Y (b)N Solution 1m=x20y20=0m≥0 z=x+y v=x/y→+1 z+2-2z =2/(+1) We see that is the unique solution y=2/(+ y ax

8 (a) 2 1 ( ) ( 1) u w w (b) 2 2 ( 1) ( 1) u w w   (c) ( ) w we u w  (d) 2 8 ( 1) (1 ) u w w   (e) None. (20) ( , ) ZW f z w ? (a) 2 2 ( ) ( 1) ( 1) z e u z u w w     (b) 2 1 ( ) ( ) ( 1) z e u z u w w    (c) 2 1 ( ) ( ) ( 1) z ze u z u w w    (d) 2 8 ( ) ( 1) (1 ) z ze u z u w w     (e) None. (21) Are Z and W independent? (a) Yes. (b) No. Solution: , / , z x y w x y       x0, y0z0, w0 . 1 / 1 x y x y y y y y w z z w x y w y            w 1 w 1 w 1 z zw z z zw x z y z             We see that /( 1) is the unique solution. /( 1) x wz w y z w        2 1 1, 1, , / z z w w x x y x y y y z x y w x y                    

J(x,y)|= =|-(x/y2+1/y)l 1/y -x/y (1+1 卩+1)2/z fzy(, w)=frr(x, y)J ,e-=ze-l()1l() (v+1) (+1) We see that Z, w are independent. It is easy to check that ze-u(=)andou(w)are (+1) density functions. They are Gamma and Cauchy respectively 6. Given Y(0)+5r(t)+6r(t=X(), for all t, and S(o)=60 X(t) r(t) H(o) WSS WSS 22)H(j) Then, H(o=? X(jo (a) 60 (o)2+5jo+6 (o)2+5jo+6 (e) None (j0)2+5 (o)2+3jo+2 (23)Rx(r)=?(a)60(b)606(x)(c)6(r)(d)sinc(r)(e)None (24) Then, the power spectral density S(o)of r(t)is

9 2 2 1 1 | ( , )| | | | | | ( / 1/ )| 1/ / z z x y w w x y J x y x y y y x y               2 2 2 1 2 1 ( 1) 1 ( 1) / x y y w z z zw w w  w z         , indep ( ) 2 2 ( , ) ( , )/| | ( 1) ( 1) X Y x y z ZW XY z z f z w f x y J e e w w         2 1 ( ) ( ) ( 1) z ze u z u w w     . We see that Z, W are independent. It is easy to check that ( ) z ze u z  2 1 and ( ) ( 1) u w w are density functions. They are Gamma and Cauchy respectively. 6. Given Y(t)5Y(t)6Y(t) X(t), for all t, and ( ) 60 X S   . X(t) Y(t) H( j) WSS WSS (22) ( ) ( ) ( ) Y j X j H j     . Then, H( j)? (a) 2 1 ( j) 5 j6 (b) 2 60 ( j) 5 j6 (c) 2 60 ( j) 5 j1 (d) 2 1 ( j) 3 j2 (e) None. (23) ( ) RX  ? (a) 60 (b) 60 ( ) (c)  ( ) (d) sinc( ) (e) None. (24) Then, the power spectral density ( ) Y S  of Y(t) is ___

60 (o2+92+4) (O2+92+4) 60(02+4) )2+4) (25)R2(z)=? (a)-2e2n+e(b) (c (e)n Solution: Taking fourier transform both sides of r(0)+5r(0+6r(0=X(), for all t we get (@Y(o)+5joY(@) +6Y(jo)=X(o) H(o Yo (jo)2+5j0+6 S(O)=H(o)2S(o)=60H(0)=60H(10)H℃o) 60 (o)2+5jo+6}{(-o)2-5j+6}{-o2+5j0+6}{-o-2-50+6} (6-02)计+5ji}{(6+o3)-5ji0}(6-02)2+25 60 1302+36(o2+9)2+4) 60 B_AO2+4)+B(O2+9) (o2+9)o2+4)o2+9o2+4(o2+9)(2+4) A(o2+4)+B(2+9)=60→(A+B)o2+(4A+9B)=60 Comparing coefficients, we get

10 (a) 2 2 1 ( 9)( 4) (b) 2 2 60 ( 9)( 4) (c) 2 2 2 60( 1) ( 9)( 4)       (d) 2 2 2 60( 4) ( 9)( 1)       (e) None. (25) ( ) RY  ? (a) 2| | 3| | 2e e       (b) 2| | 3| | 2e 3e       (c) 3| | 2| | e 2e       (d) 3| | 2| | 2e 3e       (e) None. Solution: Taking Fourier transform both sides of Y(t)5Y(t)6Y(t) X(t), for all t, we get 2 ( j) Y( j)5 jY( j)6Y( j) X( j). ( ) ( ) ( ) Y j X j H j     2 1 ( j) 5 j6  2 2 * ( ) | ( )| ( ) 60| ( )| 60 ( ) ( ) Y X S   H j S   H j  H j H j 2 2 2 2 1 1 1 1 {( ) 5 6} {( ) 5 6} { 5 6} { 5 6} 60 60 j j j j  j  j                2 2 2 2 2 1 1 60 {(6 ) 5 } {(6 ) 5 } (6 ) 25 60  j  j            4 2 2 2 60 60  13 36 ( 9)( 4)   2 2 2 2 2 2 2 2 60 ( 4) ( 9) ( 9)( 4) 9 4 ( 9)( 4) A B A  B                    2 2 A( 4)B( 9)60  2 (AB) (4A9B)60 Comparing coefficients, we get