section 1.1 - a motivating example: predicting the weatherpredicting the weather with octave make...

Section 1.1 - A Motivating Example: Predictingthe Weather

Maggie MyersRobert A. van de Geijn

The University of Texas at Austin

Practical Linear Algebra – Fall 2009

http://z.cs.utexas.edu/wiki/pla.wiki/ 1

Let us assume that on any day, the following table tells us how theweather on that day predicts the weather the next day:

Todaysunny cloudy rainy

Tomorrowsunny 0.4 0.3 0.1cloudy 0.4 0.3 0.6rainy 0.2 0.4 0.3

This table is interpreted as follows: If today it is rainy, then theprobability that it will be cloudy tomorrow is 0.6, etc.




Example

If today is cloudy, what is the probability that tomorrow is

sunny?

cloudy?

rainy?




Example - Answer

If today is cloudy, what is the probability that tomorrow is

sunny? 0.3cloudy? 0.3rainy? 0.4




Example

If today is cloudy, what is the probability that the day aftertomorrow is sunny?




Example - Answer

If today is cloudy, what is the probability that the day aftertomorrow is sunny?We don’t know what the weather will be tomorrow. What we doknow is

The probability that it will be

sunny the day after tomorrow and sunny tomorrow is 0.4× 0.3.sunny the day after tomorrow and cloudy tomorrow is0.3× 0.3.sunny the day after tomorrow and rainy tomorrow is 0.1× 0.4.

The probability that it will be sunny the day after tomorrow if it iscloudy today is 0.4× 0.3 + 0.3× 0.3 + 0.1× 0.4 = 0.25.




Exercise

If today is cloudy, what is the probability that the day aftertomorrow is cloudy?




Exercise

If today is cloudy, what is the probability that the day aftertomorrow is rainy?




Exercise - Answer

If today is cloudy, what is the probability that a week from today itis sunny? cloudy? rainy? We will not answer this question now.

We insert it to make the point that things can get messy.


Notation

As is usually the case when we are trying to answer quantitativequestions, it helps to introduce some notation.

Let χ(k)s denote the probability that it will be sunny k days

from now.

Let χ(k)c denote the probability that it will be cloudy k days

from now.

Let χ(k)r denote the probability that it will be rainy k days

from now.

Here, χ is the Greek lower case letter chi, pronounced [kai] inEnglish.




The table can be translated into the following equations:

χ(k+1)s = 0.4× χ(k)

s + 0.3× χ(k)c + 0.1× χ(k)

r

χ(k+1)c = 0.4× χ(k)

s + 0.3× χ(k)c + 0.6× χ(k)

r

χ(k+1)r = 0.2× χ(k)

s + 0.4× χ(k)c + 0.3× χ(k)

r




The probabilities that denote what the weather may be on day kand the table that summarizes the probabilities is often representedas a (state) vector, x, and (transition) matrix, P , respectively:

x(k) =

χ(k)s

χ(k)c

χ(k)r

and P =

0.4 0.3 0.10.4 0.3 0.60.2 0.4 0.3

.


The transition from day k to day k + 1 is then written as thematrix-vector product (multiplication) χ

(k+1)s

χ(k+1)c

χ(k+1)r

=

0.4 0.3 0.10.4 0.3 0.60.2 0.4 0.3

χ

(k)s

χ(k)c

χ(k)r

,

or x(k+1) = Px(k)

This is simply a more compact representation (way of writing) theequations

χ(k+1)s = 0.4× χ(k)

s + 0.3× χ(k)c + 0.1× χ(k)

r

χ(k+1)c = 0.4× χ(k)

s + 0.3× χ(k)c + 0.6× χ(k)

r

χ(k+1)r = 0.2× χ(k)

s + 0.4× χ(k)c + 0.3× χ(k)

r


Example

Assume again that today is cloudy so that the probability that it issunny, cloudy, or rainy today is 0, 1, and 0, respectively:

x(0) =

χ(0)s

χ(0)c

χ(0)r

=

010

.

Then the vector of probabilities for tomorrow’s weather (dayk = 1), x(1), is given by χ

(1)s

χ(1)c

χ(1)r

=

0.4 0.3 0.10.4 0.3 0.60.2 0.4 0.3

χ

(0)s

χ(0)c

χ(0)r


But...

χ(1)s

χ(1)c

χ(1)r

=

0.4 0.3 0.10.4 0.3 0.60.2 0.4 0.3

χ

(0)s

χ(0)c

χ(0)r

=

0.4 0.3 0.10.4 0.3 0.60.2 0.4 0.3

010

=

0.4× 0 + 0.3× 1 + 0.1× 00.4× 0 + 0.3× 1 + 0.6× 00.2× 0 + 0.4× 1 + 0.3× 0

=

0.30.30.4

.


The vector of probabilities for the day after tomorrow (day k = 2),x(2), is given by χ

(2)s

χ(2)c

χ(2)r

=

0.4 0.3 0.10.4 0.3 0.60.2 0.4 0.3

χ

(1)s

χ(1)c

χ(1)r

=

0.4 0.3 0.10.4 0.3 0.60.2 0.4 0.3

0.30.30.4

=

0.4× 0.3 + 0.3× 0.3 + 0.1× 0.40.4× 0.3 + 0.3× 0.3 + 0.6× 0.40.2× 0.3 + 0.4× 0.3 + 0.3× 0.4

=

0.250.450.30

.


Exercise

Given that today is cloudy, predict the weather for the next week.

Note

Don’t get busy doing this by hand. Let’s let the computer do theheavy lifting.


Predicting the Weather with Octave

Make sure you install octave if not yet installed.

Start the octave application.

On linux, type octave on the command line of a commandwindow.On Mac OSX, start the X11 application and type octave onthe command line of a command window or start the Octaveapplication from the Applications folder.On Windows, start the Octave application.


Enter the transition matrix, P

octave:1> P = [> 0.4 0.3 0.1> 0.4 0.3 0.6> 0.2 0.4 0.3> ]P =

0.40000 0.30000 0.100000.40000 0.30000 0.600000.20000 0.40000 0.30000


Enter the vector that described the weather today (cloudy)

octave:2> x0 = [> 0> 1> 0> ];

The “;” means that M-script does not automatically print out theresult. To see the contents of variable (vector) x0 type

Display x(0) (in variable x0)

octave:2> x0x0 =

010


Compute x(1) = Px(0) (probability vector for tomorrow)

octave:4> x1 = P * x0x1 =

0.300000.300000.40000

Compute x(2) = Px(1) (day after tomorrow)

octave:5> x2 = P * x1x2 =

0.250000.450000.30000


Exercise

Given that today is cloudy, predict the weather for the next week.

Answer

Continue on to compute x(3), . . . , x(7):

k0 1 2 3 4 5 6 7

x(k) =

0@ 010

1A 0@ 0.30.30.4

1A 0@ 0.250.450.30

1A 0@ 0.2650.4150.320

1A 0@ 0.26250.42250.3150

1A 0@ 0.263250.420750.31600

1A 0@ 0.263120.421120.31575

1A 0@ 0.263160.421040.31580

1A


Exercise

Given that today is cloudy, predict the weather for the next month.

Note

Even though we used Octave to compute x(0), . . . , x(7), theprevious approach, computing x0, x1, etc., would get quite tiring.Let’s compute x(0), . . . , x(7) with a loop.


Initialize variable x to x(0).

octave:7> x = [> 0> 1> 0> ]x =

010


Use a for loop

octave:8> for k=1:7> x = P * x> endx =

0.300000.300000.40000

x =

0.250000.450000.30000

x =

etc.http://z.cs.utexas.edu/wiki/pla.wiki/ 25

Or we could compute the table...

Create a 3× 8 array in which to store the table and set the firstcolumn to x(0).

octave:9> x = zeros( 3, 8 )

x =

0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0

octave:10> x( :, 1 ) = [

> 0

> 1

> 0

> ]

x =

0 0 0 0 0 0 0 0

1 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0


Or we could compute the table... (continued)octave:11> for k=1:7

> x( :, k+1 ) = P * x( :, k );

> end

octave:12> x

x =

0.00000 0.30000 0.25000 0.26500 0.26250 0.26325 0.26312 0.26316

1.00000 0.30000 0.45000 0.41500 0.42250 0.42075 0.42112 0.42104

0.00000 0.40000 0.30000 0.32000 0.31500 0.31600 0.31575 0.31580

Here x( :, k ) and x( :, k+1 ) equal the kth and (k + 1)stcolumns of array x, respectively. The “;” ensures that thesubresults aren’t printed out. (Indeed: type the loop in again, butleave off the “;” and see what happens.)These numbers can then be used to fill the below table:

k0 1 2 3 4 5 6 7

x(k) =

0@ 010

1A 0@ 0.30.30.4

1A 0@ 0.250.450.30

1A 0@ 0.2650.4150.320

1A 0@ 0.26250.42250.3150

1A 0@ 0.263250.420750.31600

1A 0@ 0.263120.421120.31575

1A 0@ 0.263160.421040.31580

1A


Review of Octave commands

Try the following Octave commands:octave 13> helpoctave 14> help *octave 15> help :octave 16> help zerosoctave 17> help for


section 1.1 - a motivating example: predicting the weatherpredicting the weather with octave make...

Documents