Transcript Lesson3

Lesson 3: Choosing from distributions
•
Choosing from distributions
• Discrete (Review)
• Continuous: Direct (Review)
• Continuous: Rejection
• Probability mixing
• Metropolis method
• Stratified sampling
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Overview of pdf and cdf (Review)
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Basic definition of probability distribution
function (p.d.f.):
 ( x)dx  Pr{xi lies in ( x, x  dx)}
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And its integral, the cumulative distribution
function (c.d.f.):
x
 x     ( x' )dx'  Pr{xi  x}

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Overview of pdf and cdf (2)
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Corollaries of these definitions:
x
 x     ( x' )dx'  Pr{ xi  x}

b
 b    a     ( x' )dx'  Pr{a  xi  b}
a


  ( x' )dx'  1.000

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Mapping x->x using p(x)
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Our basic technique is to use a unique
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y->x
y=x from (0,1) and x from (a,b)
We are going to use the mapping backwards
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Mapping (2)
Note that:
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x(a)=0
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x(b)=1
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Function is non-decreasing over domain (a,b)
Our problem reduces to:
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Finding x(x)
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Inverting to get x(x), a formula for turning pseudorandom numbers into numbers distributed according
to desired (x)
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Mapping (3)
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We must have:
Pr{x lies in (x , x  dx )}  Pr{ x lies in ( x, x  dx)}
dx   ( x)dx
dx ( x)
  ( x)
dx
x
x ( x)  x (a )    ( x' )dx'
a
x
x ( x)    ( x' )dx'   ( x), the C.D.F of (x)
a
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Resulting general procedure
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Form CDF:
x
( x)    ( x' )dx'
a
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Set equal to pseudo-random number:
x   (x)
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Invert to get formula that translates from x to x:
x   1 (x )
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Uniform distribution
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For our first distribution, pick x uniformly in range
(a,b):
 ( x) 
~ ( x)  1, x  a, b 
~ ( x)
1

b
~ ( x) dx


a
•
b
 1dx
1

ba
a
Step 1: Form CDF.
1
xa
( x)    ( x' ) dx'  
dx' 
ba
ba
a
a
x
x
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Uniform distribution (2)
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Step 2: Set pseudo-random number to CDF:
xa
x
ba
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Step 3: Invert to get x(x):
x  a  x(b  a)
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Example: Choose m uniformly in (-1,1):
m  1  x(1  (1))  2x  1
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Discrete distribution
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For a discrete distribution, we have N
choices of state i, each with probability pi ,
so:
 ( x)  p1 ( x1 )  p2 ( x2 ) x    pN  ( xN )
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Step 1: Form CDF:
0, x  x1
 , x  x  x
1 1
2

x
   2 , x2  x  x3
 ( x)    ( x' ) dx'   1



N
  i  1, xN  x
 i 1
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Discrete distribution (2)
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Step 2: Set pseudo-random number to
CDF: x   (x)
Step 3: Invert to get x(x):
 x1 , if 0  x   1
 x , if   x    
1
1
2
 2


n-1
n

1
x   x    xn , if   i  ξ    i

i 1
i 1


N-1

 x N , if   i  ξ  1
i 1

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Discrete distribution (3)
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Example: Choose among 3 states with
relative probabilities of 4, 5, and 6.
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Continuous distribution: Direct
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This fits the “pure” form developed before.
Form CDF:
x
( x)    ( x' )dx'
a
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Set equal to pseudo-random number:
x   (x)
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Invert to get formula that translates from x to x:
x   1 (x )
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Continuous: Direct (2)
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Example: Pick x from:
~( x)  e x , 0  x  
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Testing your selection
There are two simple ways to check a routine that is used to
choose from a give distribution: binning or moments
Binning involves dividing the domain (or part of it) into (usually
equal-sized) regions and then counting what fraction of chosen
values fall in the region.
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The expected answer for a bin that goes from a to b is
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b
Fraction in (a,b)    i ( x ') dx '
a
This will be approximately equal to (and close enough for our
purposes) the midpoint value times the width:
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Fraction in (a,b)   i (
ab
)(b  a)
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The text notes (and Public area) have a Java routine that will
perform a bin testing
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Hint: Do NOT code this with a IF test for each bin. Instead use the integer value
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(chosen value)/(total width)*(number of bins)+1 to identify the bin that x goes into.
Continuous: Rejection
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Basis of rejection approach:
Pr{x chosen AND kept}
 Pr{x chosen}  Pr{x kept x chosen}
Usual procedure (using a flat x distribution):
1. Find a value  max   ( x) for all x
1
2. Choose x  a,b  using  ( x) 
i
ba
 ( xi )
x
3.
Keep i iff x 
~max
Otherwise, return to 1.
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Continuous: Rejection (3)
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Example: Use rejection to pick x from:
x
~
 ( x)  e , 0  x  2
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Basic idea of probability mixing
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Situations arise in which you have multiple
distributions involved in a single decision:
 ( x)  p1 1 x   p2 2 x     p N  N x 
where
N
p
i 1
i
1
and each of the  i x  is a valid p.d.f.
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Probability mixing procedure
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Real problems do not present themselves
so cleanly and you have to figure it out:
~ ( x)  ~1 ( x)  ~2 ( x)    ~N ( x), x  a, b 
~ ( x)
 ( x)  b
~ ( x' ) dx'


a
b~


   i ( x' ) dx' 

~
N
a
  i ( x) 
  b
 b

i 1
 ~ ( x' ) dx'  ~i ( x' ) dx' 

 

a
 a

N
  pi i ( x)
i 1
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Probability mixing procedure (2)
Procedure:
1. Form and normalize the pi
b
~
pi   ~i ( x ' ) dx '
a
pi 
~
pi
N
~
 pj
j 1
2.
Choose the distribution i using these pi
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Probability mixing procedure (3)
Procedure:
3. Form the p.d.f. for distribution i:
 i ( x) 
~i ( x)
b
~ ( x' ) dx'

 i

~i ( x)
~
pi
a
4.
Choose xi using  i (x)
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Probability mixing procedure (3)
Example:Use probability mixing to select x from:
~( x)  x 2  e x , x  1,2
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Metropolis
This is a very non-intuitive procedure that falls
under the category of Markov Chain MC
It will ULTIMATELY deliver a consistent series of
x’s distributed according to a desired functional
form (which does NOT have to be normalized nor
do you need to know a maximum value)
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It has many advantages for certain physical problems in
which the relative probability of a chosen point can be
determined even if a closed form of the PDF is not
available
The main disadvantage is that it is very hard to tell when
the procedure has “settled in” to the point that the stream
of x’s can be trusted to deliver a consistent distribution
This method was (supposedly) worked out as part
of an after-dinner conversation in Los Alamos after
WWII
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Metropolis (2)
In its simplest form, the procedure is:
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Choose x according to a distribution that has certain
properties. We will not go into the details except to say
that a uniform distribution has all the properties.
Evaluate the PDF at the chosen x
Decide whether to use the new point according to these
rules:
1.
IF the PDF evaluates higher than the PREVIOUSLY
chosen point’s PDF, then use the new x
2.
IF the PDF evaluates less than the previous point’s
PDF, then pull another random number between 0 and
1
ä
If the new random number is LESS than the ratio of
(new point’s PDF)/(old point’s PDF), then use the new
x
ä
If the previous test fails, then REUSE the old x
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“Other”: Two alternate
n
~
Choose x from  ( x)  x , x  0,1
using:
xi  max x1 , x2 ,..., xn1 
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1  x2 2
 ( x) 
e
,   x  
Choose x from
2
(Gaussian/normal) using:
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xi   x j  6
j 1
(Why 12?)
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Homework from text
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Homework from text
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Homework from text
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