Fourth time assignment

5. Given the probability model shown in table 4-9, the real value label of the sequence A1A1A3A2A3A1 is obtained.

Answer: by formula

Solution: According to the title p (A1) =0.2, P (A2) =0.3, P (A3) =0.5

Map out A1=1,a2=2,a3=3

From the probabilistic model, we know that Fx (k) =0,k≤0,fx (0) =0,fx (1) =0.2,fx (2) =0.5,fx (3) =1,fx (k) =1,k>3

Set L (0) =0,u (0) =1

by formula, L (n) =l (n-1) + (U (n-1)-L (n-1)) FX (xn-1) with U (n) =l (n-1) + (U (n-1)-L (n-1)) FX (xn) has

When N=1 (A1):

　　L (1) =l (0) + (U (0)-L (0)) *fx (0) =0+ (1-0) *0=0

U (1) =l (0) + (U (0)-L (0)) *fx (1) =0+ (1-0) *0.2=0.2

When n=2 (A1A1):

　　L (2) =l (1) + (U (1)-L (1)) *fx (0) =0+ (0.2-0) *0=0

U (2) =l (1) + (U (1)-L (1)) *FX (1) =0+ (0.2-0) *0.2=0.04

When N=3 (a1a1a3) is =:

L (3) =l (2) + (U (2)-L (2)) *fx (2) =0+ (0.04-0) *0.5=0.02 u (3) =l (2) + (U (2)-L (2)) *FX (3) =0+ (0.04-0) * =0.04

When N=4 (A1A1A3A2):

L (4) =l (3) + (U (3)-L (3)) *fx (1) =0.02+ (0.04-0.02) *0.2=0.024

U (4) =l (3) + (U (3)-L (3)) *fx (2) =0.02+ (0.04-0.02) *0.5 =0.03

When N=5 (A1A1A3A2A3):

L (5) =l (4) + (U (4)-L (4)) *fx (2) =0.024+ (0.03-0.024) *0.5=0.027 u (5) =l (4) + (U (4)-L (4)) *fx (3) =0.024+ (0.03-0.024) *1=0.03

When N=6 (A1A1A3A2A3A1):

L (6) =l (5) + (U (5)-L (5)) *fx (0) =0.027+ (0.03-0.027) *0=0.027 u (6) =l (5) + (U (5)-L (5)) *fx (1) =0.027+ (0.03-0.027) *0.2= 0.0276

Therefore, the real value tag of the sequence A1A1A3A2A3A1 is: Tx (A1A1A3A2A3A1) = (0.027+0.0276)/2=0.0273

6. For the probability model given in table 4-9, a sequence with a length of 10 labeled 0.63215699 is decoded.

Solution:

By title: Tx=0.63215699,fx (k≤0) =0, FX (1) =0.2, FX (2) =0.5, FX (3) =1, k>3.

Given l (0) =0,u (0) =1

Formula:

T*= (Tag-l (k-1))/(U (k-1)-L (k-1))

F (x1) =0.2,f (x2) =0.5,f (x3) =1.

First set U (0) =1,l (0) = 0, there is

① the upper and lower bounds after the first element of the sequence is X1:

L (1) =0+ (1-0) Fx (x1-1) =fx (x1-1)

U (1) =0+ (1-0) Fx (x1) =fx (x1)

If X1=1 is taken, the interval is [0,0.2]

If x1=2 is taken, the interval is [0.2,0.5]

If X1=3 is taken, the interval is [0.5,1]

Because 0.63215699 falls within the interval [0.5,1], the first element of the sequence is A3

② the upper and lower bounds of the second element x2 of the sequence are:

L (2) =0.5+ (1-0.5) Fx (x2-1) =0.5+0.5fx (x2-1)

U (2) =0.5+ (1-0.5) Fx (x2) =0.5+0.5fx (x2)

If X2=1 is taken, the interval is [0.5,0.6]

If x2=2 is taken, the interval is [0.6,0.75]

If X2=3 is taken, the interval is [0.75,1]

Because 0.63215699 falls within the interval [0.6,0.75], the second element of the sequence is A2.

③ the upper and lower bounds of the third element x3 of the sequence are:

L (3) =0.6+ (0.75-0.6) Fx (x3-1) =0.6+0.15fx (x3-1)

U (3) =0.6+ (0.75-0.6) Fx (x3) =0.6+0.15fx (x3)

If X3=1 is taken, the interval is [0.6,0.63]

If x3=2 is taken, the interval is [0.63,0.675]

If X3=3 is taken, the interval is [0.675,0.75]

Because 0.63215699 falls within the interval [0.63,0.675], the third element of the sequence is A2.

④ the upper and lower bounds of the fourth element x4 of the sequence are:

L (4) =0.63+ (0.675-0.63) Fx (x4-1) =0.63+0.045fx (x4-1)

U (4) =0.63+ (0.675-0.63) Fx (x4) =0.63+0.045fx (x4)

If X4=1 is taken, the interval is [0.63,0.639]

If x4=2 is taken, the interval is [0.639,0.6525]

If X4=3 is taken, the interval is [0.6525,0.675]

Since 0.63215699 falls into [0.63,0.639], the fourth element of the sequence is A1.

⑤ the upper and lower bounds of the Fifth Element X5 of the sequence are:

L (5) =0.63+ (0.639-0.63) Fx (x5-1) =0.63+0.009fx (x5-1)

U (5) =0.63+ (0.639-0.63) Fx (X5) =0.63+0.009fx (X5)

If X5=1 is taken, the interval is [0.63,0.6318]

If x5=2 is taken, the interval is [0.6318,0.6345]

If X5=3 is taken, the interval is [0.6345,0.639]

Because 0.63215699 falls within the interval [0.6318,0.6345], the fifth element of the sequence is A2.

⑥ the upper and lower bounds of the sixth element x6 of the sequence are:

L (6) =0.6318+ (0.6345-0.6318) Fx (x6-1) =0.6318+0.0027fx (x6-1)

U (6) =0.6318+ (0.6345-0.6318) Fx (X6) =0.6318+0.0027fx (X6)

If X6=1 is taken, the interval is [0.6318,0.63234]

If x6=2 is taken, the interval is [0.63234,0.63315]

If X6=3 is taken, the interval is [0.63315,0.6345]

Because 0.63215699 falls within the interval [0.6318,0.63234], the sixth element of the sequence is A1.

⑦ the upper and lower bounds of the seventh element x7 of the sequence are:

L (7) =0.6318+ (0.63234-0.6318) Fx (x7-1) =0.6318+0.00054fx (x7-1)

U (7) =0.6318+ (0.63234-0.6318) Fx (x7) =0.6318+0.00054fx (X7)

If X7=1 is taken, the interval is [0.6318,0.631908]

If x7=2 is taken, the interval is [0.631908,0.63207]

If X7=3 is taken, the interval is [0.63207,0.63234]

Because 0.63215699 falls within the interval [0.63207,0.63234], the seventh element of the sequence is A3.

⑧ the upper and lower bounds of the eighth element x8 of the sequence are:

L (8) =0.63207+ (0.63234-0.63207) Fx (x8-1) =0.63207+0.00027fx (x8-1)

U (8) =0.63207+ (0.63234-0.63207) Fx (x8) =0.63207+0.00027fx (x8)

If X8=1 is taken, the interval is [0.63207,0.632124]

If x8=2 is taken, the interval is [0.632124,0.632205]

If X8=3 is taken, the interval is [0.632205,0.63234]

Because 0.63215699 falls within the interval [0.632124,0.632205], the eighth element of the sequence is A2.

⑨ the upper and lower bounds of the nineth element x9 of the sequence are:

L (9) =0.632124+ (0.632205-0.632124) Fx (x9-1) =0.632124+0.000081fx (x9-1)

U (9) =0.632124+ (0.632205-0.632124) Fx (x9) =0.632124+0.000081fx (x9)

If X9=1 is taken, the interval is [0.632124,0.6321402]

If x9=2 is taken, the interval is [0.0.6321402,0.6321645]

If X9=3 is taken, the interval is [0.6321645,0.63234]

Because 0.63215699 falls within the interval [0.0.6321402,0.6321645], the nineth element of the sequence is A2.

⑩ the upper and lower bounds of the tenth element x10 of the sequence are:

L (Ten) =0.0.6321402+ (0.6321645-0.0.6321402) Fx (x10-1) =0.0.6321402+0.0000243fx (x10-1)

U (Ten) =0.0.6321402+ (0.6321645-0.0.6321402) Fx (x10) =0.0.6321402+0.0000243fx (x10)

If X10=1 is taken, the interval is [0.6321402,0.63212886]

If x10=2 is taken, the interval is [0.63212886,0.63215325]

If X10=3 is taken, the interval is [0.63215325,0.6321645]

Because 0.63215699 falls within the interval [0.63215325,0.6321645], the tenth element of the sequence is A3.

So, get the sequence as a3a2a2a1a2a1a3a2a2a3.

Fourth time assignment