Yang Yaju-140705010013-fourth time job

5. given the probability model shown in table 4-9, find out the sequence a1a1a3 a 2 a 3 a 1 the real value label.

table 4-9 exercise 5. Exercise 6 the probabilistic model

Letters	Probability
A1	0.2
A2	0.3
A3	0.5

Solution: The above table indicates : FX (k) =0, K≤0, FX (1) =0.2, FX (2) =0.5, FX (3) =1, k>3.

①A1: using the update formula, you get:

　　　　L (1) =0+ (1-0) Fx (0) =0

　　　　U (1) =0+ (1-0) Fx (1) =0.2

The interval for the label of the sequence a1a1 is [0,0.2]

②A1 : using the update formula, you get:

　　　　L (2) =0+ (0.2-0) Fx (0) =0

　　　　U (2) =0+ (0.2-0) Fx (1) =0.04

The interval for the label of the sequence a1a1 is [0,0.04]

③A3 : using the update formula, you get:

　　　　L (3) =0+ (0.04-0) Fx (2) =0.02

　　　　U (3) =0+ (0.04-0) Fx (3) =0.04

The interval for the label of the sequence a1a1a3 is [0.02,0.04]

④A2 : using the update formula, you get:

　　　　L (4) =0.02+ (0.04-0.02) Fx (1) =0.024

　　　　U (4) =0.02+ (0.04-0.02) Fx (2) =0.03

The interval for the label of the sequence a1a1a3a2 is [0.024,0.03]

⑤A3 : using the update formula, you get:

　　　　L (5) =0.024+ (0.03-0.024) Fx (2) =0.027

　　　　U (5) =0.024+ (0.03-0.024) Fx (3) =0.03

The interval for the label of the sequence a1a1a3a2a3 is [0.027,0.03]

⑥A1 : using the update formula, you get:

　　　　L (6) =0.027+ (0.03-0.027) Fx (0) =0.027

　　　　U (6) =0.027+ (0.03-0.027) Fx (1) =0.0276

the label for the sequence A1A1A3A2A3A1 can be generated is: Tx (A1A1A3A2A3A1) = (0.027+0.0276)/2=0.0273

6. for the probability model given in table 4-9, decode a sequence with a length of ten labeled 0.63215699 .

Solution: known 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

　　①l (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];

　　②l (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];

　　③l (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];

　　④l (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];

　　⑤l (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];

　　⑥l (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];

　　⑦l (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];

　　⑧l (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];

　　⑨l (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];

　　⑩l (=0.6321402+0.0000243FX) (xk-1)

　　U (Ten) =0.6321402+0.0000243fx (XK)

when xk=3, 0.63215699 is in [0.63215235,0.6321645];

　　The resulting sequence is:a3a2a2a1a2a1a3a2a2a3.

Yang Yaju-140705010013-fourth time job