Chapter 6: Vector Random Variables

... INSTRUCTOR’S SOLUTIONS MANUAL Probability, Statistics, and Random Processes for Electrical Engineering 6.3 6-32 Expected Values of Vector Random Variables 6.30 6.31 6.32 © 2008 Pearson Education, Inc., ... INSTRUCTOR’S SOLUTIONS MANUAL Probability, Statistics, and Random Processes for Electrical Engineering 6.2 6-22 Functions of Several Random Variables 6.18 6.19 © 2008 Pearson...

Ngày tải lên : 20/09/2017, 21:33

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Chapter 6: FUNCTIONS OFF RANDOM VARIABLES

Ngày tải lên : 17/12/2013, 15:18

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Independent And Stationary Sequences Of Random Variables - Chapter 6 pptx

... formula to (6. 1 5), we obtain the following local limit theorem if xm =o(n) as n-+oo, then b (m, n , p) - (2rcnpq) ' exp { - ixm - ( xm) } , (6 6) where ~(x) = nPq v=3 ~ xv Pv- 1-( -q)v-1 v(v-1) (npq) ... DEVIATIONS Chap Theorem 6. 1 If x>,0 and x = o (n ) as n + oo, then P(Z„>x) - exp x3 - 7x n2 \n ) 1-G(x) 1+O C (6. 1.9) x+l~ n2 and P (Zn < - x) G(-x) x3 = e...

Ngày tải lên : 02/07/2014, 20:20

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Chapter 4: RANDOM VARIABLES AND PROBABILITY DISTRIBUTION

Ngày tải lên : 17/12/2013, 15:18

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Independent And Stationary Sequences Of Random Variables - Chapter 1 pptx

... >O and f (2r - 1) ( Tk ) = v(2r- 1) (Zk) Then ~ 00 x 2(r -1 ) sin (2T'kx)dF(x )- f(2r-2) (0)-f(2r-2)(,tk)=2 ( -1 )r- - 00 = V (2r-2)( ) - v(2r-2)('Lk) - Arguing as before, 00 x rdF(x) - Co < cc and ... we use the fact that 1- k (t) = for I t1 < 2T, so that 11 h ( 1- k)ll =11 (f-g)u ( 1- k )11 , for any function u c V with the property that u(t) =11 -it for jtj...

Ngày tải lên : 02/07/2014, 20:20

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Independent And Stationary Sequences Of Random Variables - Chapter 2 ppt

... a(0)=(a- 1-1 ) +2( 1-a) 02+ b(0), where 00 ( b(~) _ - n )n n-3 {r(1-a)} In (2 35) substitute -z for to give -1 -1 )} Re )_(1-a) -2 r exp{-r(a Y'0 i rz{r(1 - a)} x xexp { -2 4 ' 2- rb[{r(1-a)}-Z~]} e -ir - (2 4.36) ... (2. 4 .29 ) p(x ;a, -1 )=0, and p(x ; a, 1)' {2n (1 -a)}-gal /2( 1-a)x- 1-1 /2( 1 -a) exp {-( 1-a)(a/x)"/( 1- )} x z (1+ a -n /2( 1-a) anx an...

Ngày tải lên : 02/07/2014, 20:20

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Independent And Stationary Sequences Of Random Variables - Chapter 4 ppt

... that Ih1(t)I SBN , so that (4. 4.15) implies that 00 lI < I BN exp{-c(2ma-2)} - ~Ih (t)I dt=o(1) l (4. 4.16) Combining (4. 4.13), (4 4. 14) , (4 4.16) we obtain (4. 4.11) Since S ... ()a m b n- m = m-na

Ngày tải lên : 02/07/2014, 20:20

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Independent And Stationary Sequences Of Random Variables - Chapter 5 pot

... a, and therefore, as x-+ oo, F„(-x) = o(x -s) , G(-x) = o(x -a) , 1- F„ (x) = o (x - a) , 1- G (x) = o (x - a) Taking so that p > S -1 > a -1 , we have IF„(x)-G(x)IP= o(Ixl -Pa) so that (Fn - ... 1

Ngày tải lên : 02/07/2014, 20:20

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Independent And Stationary Sequences Of Random Variables - Chapter 7 potx

... (- cn2 z) = exp (- cn2 Re z) < , (7. 2.15) I and therefore = O(exp(-nr7 (E ))) + (7. 2.16) JL on L2 and L4 Moreover, lexp (- un- z) I = exp (- cn - Re z) < , 1 (7 2.15) and therefore = O(exp(-ni ... (27r/nK"(z o))2 (1 +Bn -1 log8 n) Thus the first term on the right-hand side of (7. 2.25) is equal to a( 2itK"(zo)) -I- exp In (K(zo) - aTzo)}(1+Bn -1 log8 n) ,...

Ngày tải lên : 02/07/2014, 20:20

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Independent And Stationary Sequences Of Random Variables - Chapter 8 pps

... R-1 f V(x-z)e-hwdWn(z) _ §~ n +1 x z J - 00 dWn(z)e-hz J - 00 Making the substitution ~ =1 1- z, R "+1 x f- Go Jf dW" (z) e -hz J 00 x =Rn+1 (8 2 _8) e-h(''-z)dn e -hsdV(~) (( 8) becomes V (h-Z) ... exp(-hdn-Iy)dFP(y)=(2n )- exp(-hin#y-iy )dyJ0 -Q n (0)+h6n* Jo e xp(-hdnly)Qn (y)dy (8 11) The last integral in (8. 3 11) is o Bn - hQnz exp (-hin y) dy = Bn - , (8. 3 12)...

Ngày tải lên : 02/07/2014, 20:20

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Independent And Stationary Sequences Of Random Variables - Chapter 9 ppsx

... suitable so and ltl < so , (9 9) implies that 10(t)1 < 1- 4t2 (9 3.11) If we write so + < a (9. 3 12) and °1 - 2-a1 (9 3.13) - s0+3' then (9 11) shows that, for n - "`' < ~(t)1" < (1-n-2"`')"= B ... m and writing ,/, tr = r=so+3 Yr r -, , E r Kso+3 we have, from (9 4.5), p© < n - 41, 185 (9 14) (x) = n J -2 7r (9 2) and (9. 4.18), exp(-Znt +nK s0+3 (t)-...

Ngày tải lên : 02/07/2014, 20:20

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Independent And Stationary Sequences Of Random Variables - Chapter 10 potx

... have (cf (9 5.2)) pn (x) n-u n2 _ - 7r exp (-4 nt +nKm (t)-n+itx)dt+B exp(-E2 n2 a) f-n-P (10 10) „ Application of the method of steepest descents The integrand of (10 10) is an entire function, ... ( 10. 4.11) = Bn a -2 Moreover, since z= ±in Re [n(?z 2- •rz)] =2n(~2-n-2 -2 •t ~) < in' u= - n2« , (10. 4.12) since ~ < n - '`/ p (n) and i < n - °/ p (n) (10...

Ngày tải lên : 02/07/2014, 20:20

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Independent And Stationary Sequences Of Random Variables - Chapter 11 doc

... '"2du=Bx-1 eXp ( - x 2) 11 - ES (11 11 11) x In view of this, (11 11 9) gives P(n - ZS n >x) >, (I +o(1))(2rc )-2 e -2 " du (11 11 12) fx If in (11 11 7) we replace n a -2 by -n 2" -2 and reverse ... under (11 11 2), P(n S©+n -2 00 Y©>x) e - -j "z du 2m -l ( )z fx (11 11 8) Then, from (11 11 6) and (11 11 7), P(n -2 S©>x) (1+0(1))~ ~ e -2 "...

Ngày tải lên : 02/07/2014, 20:20

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Independent And Stationary Sequences Of Random Variables - Chapter 12 ppsx

... jIzI>n~' -e Fn (x- z)g l (z) = B exp(-Znl-2E2) If x l

Ngày tải lên : 02/07/2014, 20:20

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Independent And Stationary Sequences Of Random Variables - Chapter 13 pdf

... give - J0 e -hvn 1/zv dFn (v) = (27r )-' x- (1 + 9C 11 h) (13 4.2) Since r=n -2 x, the substitution of (13 4.2) into (13 4.1) gives x3 1-Fn (x) _ (2~) -2 x -1 e -2 x exp ~,[2KI x x n2 n2 x (1+Bn -2 )(1+Bh)+6p ... (z, t) d- , l Izn-ina - '/2/P (n) zo+ina - '/2/P1(n) En (Z, t) dZ 12 = - 2761 in" - '/z/p j (n) zo - in a- '/2/P1(n) E n (z, t) dz , I3...

Ngày tải lên : 02/07/2014, 20:20

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