| << Chapter < Page | Chapter >> Page > |
The following simple random variable is in canonical form:
.
Express the events , , and in terms of , and D .
Random variable X , in canonical form, is given by .
Express the events , , and , in terms of , and E .
The class is a partition. Random variable X has values on C 1 through C 10 , respectively. Express X in canonical form.
T = [1 3 2 3 4 2 1 3 5 2];[X,I] = sort(T)X = 1 1 2 2 2 3 3 3 4 5
I = 1 7 3 6 10 2 4 8 5 9
The class in [link] has respective probabilities 0.08, 0.13, 0.06, 0.09, 0.14, 0.11, 0.12, 0.07, 0.11, 0.09. Determinethe distribution for X .
T = [1 3 2 3 4 2 1 3 5 2];pc = 0.01*[8 13 6 9 14 11 12 7 11 9];[X,PX] = csort(T,pc);disp([X;PX]')1.0000 0.2000
2.0000 0.26003.0000 0.2900
4.0000 0.14005.0000 0.1100 A wheel is spun yielding on an equally likely basis the integers 1 through 10. Let C i be the event the wheel stops at i , . Each . If the numbers 1, 4, or 7 turn up, the player loses ten dollars; if the numbers 2, 5,or 8 turn up, the player gains nothing; if the numbers 3, 6, or 9 turn up, the player gains ten dollars; if the number 10 turns up, the player loses one dollar. The randomvariable expressing the results may be expressed in primitive form as
p = 0.1*ones(1,10);
c = [-10 0 10 -10 0 10 -10 0 10 -1];
[X,PX]= csort(c,p);
disp([X;PX]')
-10.0000 0.3000-1.0000 0.1000
0 0.300010.0000 0.3000
Pneg = (X<0)*PX'
Pneg = 0.4000Ppos = (X>0)*PX'
Ppos = 0.300 A store has eight items for sale. The prices are $3.50, $5.00, $3.50, $7.50, $5.00, $5.00, $3.50, and $7.50, respectively.A customer comes in. She purchases one of the items with probabilities 0.10, 0.15, 0.15, 0.20, 0.10 0.05, 0.10 0.15. Therandom variable expressing the amount of her purchase may be written
Determine the distribution for X (a) by hand, (b) using MATLAB.
p = 0.01*[10 15 15 20 10 5 10 15];c = [3.5 5 3.5 7.5 5 5 3.5 7.5];[X,PX] = csort(c,p);disp([X;PX]')3.5000 0.3500
5.0000 0.30007.5000 0.3500 Suppose in canonical form are
The are 0.3, 0.6, 0.1, respectively, and the are 0.2 0.6 0.2. Each pair is independent. Consider the random variable . Then on , on , etc. Determine the value of Z on each and determine the corresponding . From this, determine the distribution for Z .
A = [2 3 5];B = [1 2 3];a = rowcopy(A,3);
b = colcopy(B,3);Z =a + b % Possible values of sum Z = X + Y
Z = 3 4 64 5 7
5 6 8PA = [0.3 0.6 0.1];PB = [0.2 0.6 0.2];pa= rowcopy(PA,3);
pb = colcopy(PB,3);P = pa.*pb % Probabilities for various values
P = 0.0600 0.1200 0.02000.1800 0.3600 0.0600
0.0600 0.1200 0.0200
[Z,PZ]= csort(Z,P);
disp([Z;PZ]') % Distribution for Z = X + Y
3.0000 0.06004.0000 0.3000
5.0000 0.42006.0000 0.1400
7.0000 0.06008.0000 0.0200 
Notification Switch
Would you like to follow the 'Applied probability' conversation and receive update notifications?
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|