# Assignment for the fourth time

Source: Internet
Author: User

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

 Table 4-9 The probabilistic model of Exercise 5, exercise 6 Letter Probability A1 0.2 A2 0.3 A3 0.5

Solution:

P (A1) =0.2, P (A2) =0.3, P (A3) =0.5

FX (0) =0,FX (1) =0.2, FX (2) =0.5, FX (3) =1.0, U (0) =1, L (0) =0

by X (AI) =i x (A1) =1,x (A2) =2,x (A3) =3

by formula, L (n) =l (n-1) + (U (n-1)-L (n-1)) Fx (xn-1)

U (n) =l (n-1) + (U (n-1)-L (n-1)) Fx (xn)

When the A1 first appears:

L (1) =l (0) + (U (0)-L (0)) Fx (0) =0

U (1) =l (0) + (U (0)-L (0)) Fx (1) =0.2

When A1 appears for the second time:

L (2) =l (1) + (U (1)-L (1)) Fx (0) =0

U (2) =l (1) + (U (1)-L (1)) Fx (1) =0.04

When A3 appears for the third time:

L (3) =l (2) + (U (2)-L (2)) Fx (2) =0.02

U (3) =l (2) + (U (2)-L (2)) Fx (3) =0.04

Fourth time when A2 appears:

L (4) =l (3) + (U (3)-L (3)) Fx (1) =0.024

U (4) =l (3) + (U (3)-L (3)) Fx (2) =0.03

Fifth time when A3 appears:

L (5) =l (4) + (U (4)-L (4)) Fx (2) =0.027

U (5) =l (4) + (U (4)-L (4)) Fx (3) =0.03

Sixth time when A1 appears:

L (6) =l (5) + (U (5)-L (5)) Fx (0) =0.027

U (6) =l (5) + (U (5)-L (5)) Fx (1) =0.0276

Therefore, the real value tag of the sequence A1A1A3A2A3A1 is: T (113231) = (L (6) + U (6))/2=0.0273.

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

Solution:

Obtained from table: p (A1) =0.2, P (A2) =0.3, P (A3) =0.5

FX (0) =0,FX (1) =0.2, FX (2) =0.5, FX (3) =1.0, U (0) =1, L (0) =0

by X (AI) =i x (A1) =1,x (A2) =2,x (A3) =3

by formula, L (n) =l (n-1) + (U (n-1)-L (n-1)) Fx (xn-1)

U (n) =l (n-1) + (U (n-1)-L (n-1)) Fx (xn)

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

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

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

if x1=1, then the interval is [0,0.2]

if x1=2, then the interval is [0.2,0.5]

if x1=3, then the interval is [0.5,1]

1. because 0.63215699 is in [0.5,1] , x1=3

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, then the interval is [0.5,0.6]

if x2=2, then the interval is [0.6,0.75]

if x2=3, then the interval is [0.75,1]

2. because 0.63215699 is in [0.6,0.75] , x2=2

L (3) =0.5+ (0.75-0.6) Fx (x3-1) =0.6+0.15fx (x3-1)

U (3) =0.5+ (0.75-0.6) Fx (x3) =0.6+0.15fx (x3)

if x3=1, then the interval is [0.6,0.63]

if x3=2, then the interval is [0.63,0.675]

if x3=3, then the interval is [0.675,0.75]

3. because 0.63215699 is in [0.63,0.675] , x3=2

L (4) =0.63+ (0.675-0.625) Fx (x4-1) =0.63+0.045fx (x4-1)

U (4) =0.63+ (0.675-0.625) Fx (x4) =0.63+0.045fx (x4)

if x4=1, then the interval is [0.63,0.639)

if x4=2, then the interval is [0.639,0.6525]

if x4=3, then the interval is [0.6525,0.675]

4. because 0.63215699 is in [0.63,0.639) , so x4=1

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, then the interval is [0.63,0.6318]

if x5=2, then the interval is [0.6318,0.6345]

if x5=3, then the interval is [0.6345,0.639]

5. because 0.63215699 is in [0.6318,0.6345] , x5=2

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, then the interval is [0.6318,0.63234]

if x6=2, then the interval is [0.63234,0.63315]

if x6=3, then the interval is [0.63315,0.63345]

6. because 0.63215699 is in [0.6318,0.63234) , so x6=1

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, then the interval is [0.6318,0.631908]

if x7=2, then the interval is [0.631908,0.63207]

if x7=3, then the interval is [0.63207,0.63234]

7. because 0.63215699 is in [0.63207,0.63234] , x7=3

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, then the interval is [0.63207,0.632124]

if x8=2, then the interval is [0.632124,0.632205]

if x8=3, then the interval is [0.632205,0.63234]

8. because 0.63215699 is in [0.632124,0.632205] , x8=2

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, then the interval is [0.632124,0.6321402]

if x9=2, then the interval is [0.6321402,0.6321645]

if x9=3, then the interval is [0.6321645,0.632205]

9. because 0.63215699 is in [0.6321402,0.6321645] , x9=2

L (Ten) =0.6321402+ (0.6321645-0.6321402) Fx (x10-1) =0.6321402+0.00006243fx (x10-1)

U (Ten) =0.6321402+ (0.6321645-0.6321402) Fx (x10) =0.6321402+0.00006243fx (x10)

if x10=1, then the interval is [0.6321402,0.63214506]

if x10=2, then the interval is [0.63214506,0.63215235]

if x10=3, then the interval is [0.63215235.0.6321645]

because 0.63215699 is in [0.63215235,0.6321645] , x10=3

so the sequence is 3221213223 that is a3a2a2a1a2a1a3a2a2a3

Assignment for the fourth time

Related Keywords:

The content source of this page is from Internet, which doesn't represent Alibaba Cloud's opinion; products and services mentioned on that page don't have any relationship with Alibaba Cloud. If the content of the page makes you feel confusing, please write us an email, we will handle the problem within 5 days after receiving your email.

If you find any instances of plagiarism from the community, please send an email to: info-contact@alibabacloud.com and provide relevant evidence. A staff member will contact you within 5 working days.

## A Free Trial That Lets You Build Big!

Start building with 50+ products and up to 12 months usage for Elastic Compute Service

• #### Sales Support

1 on 1 presale consultation

• #### After-Sales Support

24/7 Technical Support 6 Free Tickets per Quarter Faster Response

• Alibaba Cloud offers highly flexible support services tailored to meet your exact needs.