Fourth time assignment

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

Table 4-9 probability models for exercise 5 and 6
Letter Probability
a10.2
a20.3
a30.5


Solution: According to test instructions, the realistic value tag is to find the real value label of sequence 113230
Due to Fx (k) =0,k≤0,fx (1) =0.2,fx (2) =0.5,fx (3) =1,fx (k) =1,k>3
Calculation formula: U (k) =l (k-1) + (U (k-1)-L (k-1)) FX (XK)
L (k) =l (k-1) + (U (k-1)-L (k-1)) FX (xk-1)

The first element of the sequence is 1, and the following updates are obtained:

U (1) =l (0) + (U (0)-L (0)) *fx (1) =0+ (1-0) *0.2=0.2
I (1) =l (0) + (U (0)-L (0)) *fx (0) =0+ (1-0) *0=0
So the range of the sequence label is [0,0.2].

The second element of the sequence is 1, and the following updates are obtained:

U (2) =l (1) + (U (1)-L (1)) *FX (1) =0+ (0.2-0) *0.2=0.04
L (2) =l (1) + (U (1)-L (1)) *fx (0) =0+ (0.2-0) *0=0
So the interval of the sequence label is [0,0.04].
The third element of the sequence is 3, and the following updates are obtained:
U (3) =l (2) + (U (2)-L (2)) *FX (3) =0+ (0.04-0) * * =0.04
L (3) =l (2) + (U (2)-L (2)) *fx (2) =0+ (0.04-0) *0.5=0.02
So the range of the sequence label is [0.02,0.04].

The fourth element of the sequence is 2, and the following updates are obtained:
U (4) =l (3) + (U (3)-L (3)) *fx (2) =0.02+ (0.04-0.02) *0.5 =0.03
L (4) =l (3) + (U (3)-L (3)) *fx (1) =0.02+ (0.04-0.02) *0.2=0.024
So the interval of the sequence label is [0.024,0.03].

The fifth element of the sequence is 3, and the following updates are obtained:
U (5) =l (4) + (U (4)-L (4)) *fx (3) =0.024+ (0.03-0.024) *1=0.03
L (5) =l (4) + (U (4)-L (4)) *fx (2) =0.024+ (0.03-0.024) *0.5=0.027
So the range of the sequence label is [0.027,0.03].

The sixth element of the sequence is 1, and the following updates are obtained:
U (6) =l (5) + (U (5)-L (5)) *fx (1) =0.027+ (0.03-0.027) *0.2=0.0276
L (6) =l (5) + (U (5)-L (5)) *fx (0) =0.027+ (0.03-0.027) *0=0.027
So the range of the sequence label is [0.027,0.0276].

The following tags are available for generating sequence 113231:
That is, TX (A1A1A3A2A3A1) = (0.027+0.0276)/2=0.0273

6, for the probability model shown in table 4-9, for a label 0.63215699 of the length of a sequence of 10 decoding.
Solution: The probability model shows that:
FX (k) =0, K≤0, FX (1) =0.2, FX (2) =0.5, FX (3) =1, k>3.
Set U (0) =1,l (0) =0
L (1) =0+ (1-0) Fx (x1-1) =fx (xk-1)
U (1) =0+ (1-0) Fx (x1) =fx (XK)
When Xk=3, 0.63215699 is in the interval [0.5,1];
L (2) =0.5+0.5fx (xk-1)
U (2) =0.5+0.5fx (XK)
When xk=2, 0.63215699 is in [0.6,0.75];
L (3) =0.6+0.15fx (xk-1)
U (3) =0.6+0.15fx (XK)
When xk=2, 0.63215699 is in [0.63,0.675];
L (4) =0.63+0.045fx (xk-1)
U (4) =0.63+0.045fx (XK)
When Xk=1, 0.63215699 is in [0.63,0.639];
L (5) =0.63+0.009fx (xk-1)
U (5) =0.63+0.009fx (XK)
When xk=2, 0.63215699 is in [0.6318,0.6345];
L (6) =0.6318+0.0027fx (xk-1)
U (6) =0.6318+0.0027fx (XK)
When Xk=1, 0.63215699 is in [0.6318,0.63234];
L (7) =0.6318+0.00054fx (xk-1)
U (7) =0.6318+0.00054fx (XK)
When Xk=3, 0.63215699 is in [0.63207,0.63234];
L (8) =0.63207+0.00027fx (xk-1)
U (8) =0.63207+0.00027fx (XK)
When xk=2, 0.63215699 is in [0.632124,0.632205];
L (9) =0.632124+0.000081fx (xk-1)
U (9) =0.632124+0.000081fx (XK)
When xk=2, 0.63215699 is in [0.6321402,0.6321645];
L (Ten) =0.6321402+0.0000243fx (xk-1)
U (Ten) =0.6321402+0.0000243fx (XK)
When Xk=3, 0.63215699 is in [0.63215235,0.6321645];
All in all: a sequence with a length of 10 labeled 0.63215699 is decoded to a sequence of a3a2a2a1a2a1a3a2a2a3.

Fourth time assignment