北京大学：EL6303 Sample Final Exam with Solution

(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 1<z<oo F2(=)=P(Z≤-)=P{(C≤z)n(X>Y)}+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 ___