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 I<y<2, Fr (a)Fx(2y+2)(b)Fx(y-)(c)Fx(2y-2)(d)Fx(2y-)(e (6)Aty=0,f,(y) (a)0 (b)(F2(1)-F2(-1)(y)(c)(Fx(2)-Fx(-2)5(y) (d) F(O)S()(e) Something else (7)Aty=2,f,(y)= (a)(-F2(1)(-2)(b)Fx(2)D(y-2)(c)Fx(D(y-2) (d)(1-Fr(2))S(-2)(e) Something else Solution (x) y X-1 P(Y=-2)=P(X<-2)=Fx(-2),P(Y=2)=P(X>2)=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 ?