We want to train a mouse to follow a particular path in the maze (see image 1). To measure the effectiveness of the training, we want to compare the behaviour of the trained mouse to that of a mouse that runs through the maze at random. (a) Construct the matrix T describing the behaviour of the untrained mouse, where tij gives the probability of going to room i when the mouse is in room j. (Note that since the mouse moves randomly if it is in front of two doors, it has a 1/2 probability of going through each of these doors.) (b) Suppose the mouse begins its journey through the maze with a vector of probabilities P(0) = (see image 3) where pi(0) represents the probability of being in room i at the start; so of course we have p1(0)+p2(0)+p3(0)= 1). Show that the probabilities of being in room i after a change of room P(1) are given by P(1)=TP(0).

Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter4: Polynomial And Rational Functions
Section4.1: Polynomial Functions Of Degree Greater Than
Problem 54E
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We want to train a mouse to follow a particular path in the maze (see image 1). To measure the effectiveness of the training, we want to compare the behaviour of the trained mouse to that of a mouse that runs through the maze at random.

(a) Construct the matrix T describing the behaviour of the untrained mouse, where tij gives the probability of going to room i when the mouse is in room j. (Note that since the mouse moves randomly if it is in front of two doors, it has a 1/2 probability of going through each of these doors.)

(b) Suppose the mouse begins its journey through the maze with a vector of probabilities P(0) = (see image 3) where pi(0) represents the probability of being in room i at the start; so of course we have p1(0)+p2(0)+p3(0)= 1).


Show that the probabilities of being in room i after a change of room P(1) are given by P(1)=TP(0).


Indications: proceed with a proof on the one hand and on the other hand: start by calculating the probabilities of being in parts 1,2 or 3 after a change using your matrix T to complete the probability tree begins below down. Compare the distribution obtained with that calculated by the formula TP(0). (see image 2)

Consequently, we will have that the probabilities after n changes of pieces will be given by: P(n) = TP(n-1) = TnP(0)
(You don't have to show it, it's a direct consequence of your result)

(c) It can be assumed that after several changes of pieces, the probabilities will stabilize, which means they will remain unchanged after a change of pieces. Find, if they exist, these stable probabilities P= (see image 4).

Give a clear and complete approach. Start by posing the system of linear equations to be solved and use one of the resolution methods learned (of your choice) in class to find the solution. Make sure you answer the question correctly. Do not forget that we want to find a distribution, therefore percentages whose sum gives 100%: p1+p2+p3= 1.


Image 1
Image 2
1
room at
the start
(0)
Pi
(0)
P2
20 (0)
P3
3
2
3
room after change
1
2
3
2
3
2
3
p(0)
||
Image 3
P
(0)
Image 4
P₁
(0)
P₂
(0)
P3
-(
P1
P2
P3
Transcribed Image Text:Image 1 Image 2 1 room at the start (0) Pi (0) P2 20 (0) P3 3 2 3 room after change 1 2 3 2 3 2 3 p(0) || Image 3 P (0) Image 4 P₁ (0) P₂ (0) P3 -( P1 P2 P3
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