Transcript Initial probability distribution for Sam´s sister child birth: singletons

Lecture 7.
Markov Model with Matrixes.
Introduction to HMM with the example of DNA analysis.
Distributions. Probability density and cumulative
distribution functions.
Poisson and Normal distributions.
Practice:
Distributions with Mathematica
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7.1 Reminder
We remind here some facts about the distributions of discrete
and continuous random variables.
The discrete random distribution can be characterized by a
probability function p(xi) assigning the probabilities to all possible
values xi of a random variable X. The probability function should satisfy
the following equations :
p( xi )  0
 p( xi )  1
i
P (E ) 
( 7.1)
 p( xi )
i E
Compare with the pages 31 … of Lect. 1
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Example:
Suppose that a coin is tossed twice, so that the sample space is
={HH,HT,TH,TT}.
Let X represent a number of heads that can come up. Find the
probability function p(x). As we know,
the probability function is thus given by the table:
x
0
1
2
p(x)
1/4
1/2
1/4
The graphical example presented below represents a typical way
of depicting the probability distribution for a discrete random
variable
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~p(xi)
0.3
1
2
3
4
5
6
7
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Possible values of the random variable, xi
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10 11
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Continuous distribution. Probability density function
(PDF).
Remember:
For a continuous variable we must assign to each outcome a
probability p(x )=0. Otherwise, we would not be able to fulfill the
requirement 7.1 (the second of three).
A random variable X is said to have a continuous distribution
with density function f(x) if for all a b we have
b
P (a  X  b )   f ( x )dx
( 7.4)
a
The analogs of Eqs. 7.2 and 7.3 for the continuous distributions would be
(7.5)
 f (x)  1

P (E )   f ( x )dx
E
( 7.6)
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P(E) is a probability that X belongs to E.
f(x)
P(a<X<b)
a
b
Geometrically, P(a<X<b) is the area under the curve f(x) between a and b.
The question: Can f(x) exceed 1? Please argue.
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7.2 A few new distributions and their properties
a.
Poission distribution
Poisson distribution is one of the most important discrete distributions. Its
probability function is
e   x
P( x) 
x!
Poisson distribution is a limiting case of the Binomial distribution P(pn,n), with
parameters pn and n such that
pn 0, n , pn n  
In other words, if we have a large number of independent events with small
probability, then number of occurrences has approximately Poisson distribution.
Let us introduce now a more intuitive definition of the Poisson distribution.
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Examples with Poisson distribution
1. Suppose that the probability of a defect in a foot of magnetic tape is
0.002. Use the Poisson distribution to compute the probability that 1500
feet roll will have no defects .
Exp[  ] x
P [ x _ .  _] :
;
x!
p  0.002; n  1500;   p n  0.002  1500  3;
p[0]  P [0, 3] // N  0.0498
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  pn  3.
This example helps to describe the Poisson Distribution in a new way by
noticing that  is the expected (average) value of the defects in 1500 feet of
the tape.
In other words, the PD gives the probability of n events happening in
some experiment if the expected (average) number of the events, 
, for this particular experiment, is known.
Attention: it is important to understand that  is the average
(expected) value for the interval (of time, or space) where the n
questioned events should occur. For instance, in the former example 
is the average value of defects per 1500 feet of tape (but not per a foot,
or per 1000 feet, etc).
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Example 2
An airline company sells 200 tickets for a plane with 198 seats,
knowing that a probability that a passenger will not show up for
the flight is 0.008. Use the Poisson approximation to compute
the probability that they will have enough seats for all the
passengers that will show up.
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Solution.
p=0.012, L=0.012*200=2.4 – the average number (out of 200
passengers ) that won’t show up for the flight.
p[x]= Exp[-2.4]2.4x/x!;
P[more than 1 person won’t show up] = 1-p[0]-p[1]=
~ 0.7. In other words, there is (only) a 30% of chance that they
more than 198 passengers will show up. This is quite a familiar
scenario: the company would typically offer you a free
additional ticket for a future flight if you agree on switching to
a later flight. 
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Example 3:
(working in groups)
10% of the tools produced in a certain manufacturing process
turns out to be defective. Find a probability that in a sample of
ten tools selected at random, exactly 2 will be defective, by
using (a) binomial and (b) Poisson distribution.
Open a Mathematica file, and find the probabilities
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b. The exponential distribution (this is the continuous
distribution, for the continuous random variable x).
  e  x ,
f ( x )  
0

x0
otherwise
(7.8)
Those who know how to integrate can verify that (7.8) satisfies (7.5)
(the total area under the curve f(x) equals 1.
Note: In Matematica, the integral of a function f[x] can be found as:
Integrate[f[x],{x,x1,x2}] , Shift+Enter.
Here x1 and x2 are the limits of integration.
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A typical example of the exponential distribution results from the
discussion of the waste products of the nuclear power plant. If at time t=0
there are N(0) identical unstable particles, and the number of particles dN(t)
decaying in time dt is proportional to dt and to the number of particles, then
we have
dN(t)= - N(t)dt
This is so called differential equation. Here is how it is solved with
Mathematica.
DSolve[{n’[t] + G n[t] == 0, n[0] == n0},n[t],t];
n[t]-> n0 Exp[-Gt];
As a result we came up with the exponential distribution.
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Let’s introduce the “half-time” T , such that N(T)=N0/2.
Then we find : T=ln2=0.693.
c. The standard normal distribution
f(x)=(2)-1/2 exp(-x2/2)
(7.12)
A.
Using Mathematica, check that this PDF satisfies the normality condition
(7.5). Make a plot of (7.12).
If a random variable y is related to x as y=ax+b, how the distribution
function f(y) looks like? (we assume that x is distributed according to (7.12).
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More generally, X is said to have a normal (,2)
distribution if it has density function
f(x)=(2 2)-1/2 exp[-(x- )2/2 2]
(7.12’)
2 is called “variance” and  is the “mean” or the
“expectation”.
Try to analyze, assigning
different numeric values to 
and 2 how they affect the
shape
of f(x). For instance, how the
parameters for the green and
red
curves are related? Green and
blue? See the Mathematica
file
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7.3 Probability distribution function
( also called “cumulative distribution function”= CDF)
1.
Continuous random variable
From the “outside”, random distributions are well described by the probability
distribution function (we will use CDF for short) F(x) defined as
x
F ( x )  P(  X  x )   f ( y )dy

(7.13)
This formula can also be rewritten in the following very useful form:
P(a  X  b)  F (b)  F (a)
Question: Can F(x) exceed 1? Argue it.
(7.14)
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To see what the distribution functions look like, we return to our examples.
1. The uniform distribution (7.7):
1
,
 b a
f (x)  
0

ax b
otherwise
Using the definition (7.13) and Mathematica, try to find F(x) for the
uniform distribution. Prove that
F(x)=0 for x a; (x-a)/(b-a) for a  x  b; 1 for x>b.
Draw the CDF for several a and b. Consider an important special case
a=0, b=1. How is it related to the spinner problem? To the balanced die?
2. The exponential distribution (7.8):
  e  x ,
f ( x )  
0

x0
otherwise
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Use Mathematica to prove that
F(x)= 0 for x  0; 1-exp(-x) for x >0. (7.15)
“Lack of memory” for the exponential distribution
Suppose that X has an exponential distribution (7.8). The probability that
the event (such as the radioactive decay) did not happen in t units of time
is P(X>t) = 1-F(x). According to (7.15) it results in P(X>t)= exp(-t) .
Let’s find now a probability that we will have to wait some additional
time s given that we have been waiting t units of time:
P(X>t+s|T>t) = P(X > t+s)/P(X > t) = exp[-(t+s)]/ exp[-t)]=
exp[-s].
As we see, the result depends only on s and does not depend on the
previous waiting time. The probability you must wait additional s units of
time till decay occurs is the same as if you had not been waiting at all.
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The standard normal distribution
Using Mathematica and Eq. (7.12), find F[x] for the snd.
Use NIntegrate[f[t],{t,-,x}] and Plot[…] functions.
2. CDF for discrete random variables
For discrete variables the integration is substituted for summation:
F ( x )  P( X  x )   p( x )
i
x x
i
(7.16)
It is clear from this formula that if X takes only a finite number of values,
the distribution function looks like a stairway.
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F(x)
p(x4)
1
p(x3)
p(x2)
p(x1)
x1
x2
x3
x4
x
Draw F(x) for the example in page 7.
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This problem is for your practice
Predicting the rate of mutation based on the Poisson
probability distribution function.
The evolutionary process of amino acid substitutions in proteins is
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