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

Lecture 6.
•Maximum Likelihood
•Conditional Probability and two-stage experiments
•Markov Chains (introduction).
• Markov
Chains with Mathematica
•Bayes formula.
•Student’s presentation
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Two-stage experiments
Example 1. North flips 3 coins and South flips 2. North wins if the number of tails
he gets is more than the number South gets. What is the probability North will win?
Let N=" North wins ", S[n]=“South gets n tails", N[m] = “North gets m tails".
P(S[0])= 1/4. P(S[1])=1/2, P(S[2])=1/4
P(N[0]) = 1/8 = P(N[3]); P(N[1]))= 3/8 =P(N[2]).
N = [S0 (N[1]UN[2]UN[3])] U [S[1] (N[2]UN[3])] U [S[2] N[3]]
P(N)= P(S[0]) *[P(N[1])+P(N[2])+P(N[3]) ] + P(S[1]) * [ P(N[2])+P(N[3]) ] +
P(S[2])*P(N[3]) = 1/4*7/8+ 1/2*1/2+1/4*1/8= ½.
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Another form of the same equation:
P(N) = P(N*S[0]) + P(N*S[1]) + P(N*S[2]) =
Sum[ P(N*S[ n ] ), { n, 0, 2} ]
Probability that South gets n heads and North wins
= Sum[ P(N | S[ n ] )* P(S[ n ] ), { n, 0, 2}
Conditional Probability of N winning given that S has I heads
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We think of such an experiment as occurring in two stages. The first stage
determines which of B's occur, and when Bi occurs on the first stage, A occurs
with probability P(A|Bi) on the second.
Suppose that B1,B2,...Bk is a collection of k disjoined events whose union is .
Using the fact that the sets A∩Bn are disjoined , and the multiplication rule
for the conditional probability , P(AB) = P(A|B) P(B), we have
k
k
n 1
n 1
P( A)   P( A Bn )   P( A |Bn ) P( Bn ) (6.1)
This formula is sometimes called the law of the total probability
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Example 2. Roll a die and then flip that number of coins.
What is the probability of 3H= “We get exactly 3 Heads”?
Let Bi = “The die shows i”. P(Bi) = 1/6 for i=1,2,..6.
Now find the conditional probabilities.
P(3H|B1)=P(3H|B2)= 0. P(3H|B3) = 1/23;
P(3H|B4)= C4,3 /24; P(3H|B5)= C5,3 /25; P(3H|B6) = C6,3 /26.
So plugging into Eq. (1) we find:
P(3H) = 1/6 [ 1/8 + 4/16 + 10/32 + 20/64] = 1/6
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An example from Genetics
Each hereditary characteristic is carried by a pair of genes, so that each
offspring gets one chromosome from its mother and one from its father. We
will consider a case in which each gene can take only two forms called
alleles, which we denote by a and A. An example from a pioneering work
of Gregor Mendel is A="smooth skin" and a ="wrinkled skin" for the pea
plants. In this case A is dominant over a, which means that Aa individuals
will have a smooth skin.
Let us start from an idealized infinite population with following distribution
of genotypes
AA Aa
aa
0
0
0
where the proportions 0, 0 and 0 sum to 1.
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AA
Aa
If we assume that random mating occurs then each
0
0
new individual picks two parents at random from
population and picks allele at random from two carried by each
parent.
aa
0
Then, the first allele will be A with probability
p1=0+0/2 and a with probability 1-p1= 0+ 0/2
.
The second allele will be independent and have the same
distribution, so that the first generation of offspring will be
AA
Aa
aa
1=p12
1=2p1(1-p1)
1= (1-p1)2
Note that this distribution is determined by the relative weight p1 of A (or a)
in the population ( which can be expressed through 0 and 0 ) while the
original distribution (or the choice of 0 and 0) is arbitrary.
Let us now use this distribution to calculate the second generation of
offspring. A will now have a probability p2 = p12 + p1(1-p1) = p1 (!)
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, So
the proportions of the second generation of offspring will be
2=p22=p12= 1
2=2p2(1-p2)= 1
2= (1-p2)2= 1
-exactly the same as in the first generation
We can see that the distribution of genotypes reaches its equilibrium in
one generation of offspring starting from an arbitrary distribution. It
means that if the proportion of dominant alleles A is p then the proportion
of genotypes (uniquely determined by p) will be
AA
p2
Aa
2p(1-p)
aa
(1-p)2
This result is called the Hardy-Weinberg Theorem.
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To illustrate its use suppose that in a population of pea plant 10% have
wrinkled skin (aa) and 90% have smooth skin (AA or Aa). Using the table
above, we can find the proportions of three genotypes.
The fraction p of alleles we find from the condition + =0.9 leading to
2p-p2=0.9, p=0.684. As a result, the proportions are:
 =p2=0.46 (AA),  =2p(1-p)= 0.44 (Aa), =(1-p)2=0.1(aa).
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Problems (working in groups):
1.
You are going to meet a friend in the airport. You know that the plane is late 70%
of the time when it rains, and 20 % when it does not rain. The weather forecast
calls for a 40% chance of rain. What is the probability that the plain will be late?
A=“plain will be late”; R=“it will rain”; S= “it won’t rain”
Solution:
Using the law of total probability, we find:
P(A)= P(A|R)*P(R) + P(A|S)P(S) = 0.7*0.4+ 0.2* 0.6= 0.28+0.12= 0.4.
2. How can 5 black and 5 white balls be put into two urns to maximize the
probability that a white ball is drawn when we draw from a randomly-chosen
urn (try solving it with Mathematica).
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See the solution in Mathematica/Class_2Boxes_solution.nb
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Two-state Markov Chains
In 1907, Russian mathematician A.A. Markov began study chance
processes in which the outcome of a given experiment can affect the
outcome of the next experiment. We consider here a simple example of
such process leaving a more detailed discussion for future
Suppose that there are two brands of bread, A and B, and that on each
purchase a customer changes brands with probability p(change)=2/7 ≡ p
and buys the same brand with probability p(stay)=5/7 ≡ 1-p. Let fk be the
fraction that buys brand A on their k-th purchase and suppose f1=1/3.
Compute f1,f2, …
Let Ak=" a customer bought brand A on the k-th step".
We do not know explicitly the probability P(Ak), but we know that it
depends on the previous step:
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stay
change
P(Ak)= P(Ak-1)(1-p)+P(Bk-1 )p = 5/7 P(Ak-1) + 2/7[1- P(Ak-1)].
Using the frequency definition of probability, we can present it as:
fk  (1  p)fk 1  p(1  fk 1 ) 
5
2
fk 1  (1  fk 1 )
7
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(Mark _ 1)
This is a typical example of a "recursive equation". It can be solved
step by step, but we prefer to use Mathematica
This is how the recursive equation is solved:
f[k_] := (5/7) f[k-1] + (1/7)(1 – f[k-1]);
f[1] = 1/3.;
a = Table[f[k], {k, 1, 15}]
{0.333333,0.428571,0.469388,0.48688,0.494377,0.49759,0.498967,0.499
557,0.49981,0.499919,0.499965,0.499985,0.499994,0.499997,0.499999}
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The population rapidly approaches 0.5.
0.5
0.475
0.45
0.425
0.4
0.375
0.35
0
2
4
6
8
10
12
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Group work:
Why is it 0.5? Does it depend on the initial value f1 ? On the
probability p? Solve the same equation with f1 =0.1, then change
p to p = ¼. Does it change the results? –Practice with Markov0
and Markov1
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As you noticed, the equilibrium population is always 0.5, regardless the
values of p and f[1].
Let us now generalize the model.
Consider a system that has two states: A and B, and changes its states
with probabilities:
Prob(AB )= p, Prob(A  A)=1-p, Prob(B  A )= q, Prob(B  B)=1-q.
Let fk be a share of the total population belonging to the state A on the
k-th step of evolution. The recursive equation for this model becomes
f[p_,q_,k_]:=(1-p)f[p,q,k-1]+q(1-f[p,q,k-1]);
f[p_,q_,1]=1/3;
Mark_2
Mark_3
(see Markov2.nb)
Solve this equation for ten steps, k = (1,10) , and with various
combinations of probabilities: {p,q}= {0.2,0.4},{0.2,0.5},{0.3,0.6}, {0.3,0.3}.
Try to figure out how the asymptotic value of fk depends on p , q and f1
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You will find (Markob2.nb) that fkq/(p+q). This result can be derived
analytically.
Assuming in Eq. Mark_2
fk-1=fk= r, we find:
r = (1-p)r +(1-r)q which leads to
r= q/(p+q).
Final comments.
The 2-state Markov chain is a simplest example of the Markov Chains.
In general, if there are m states, the transition between the states is
described by a “transition matrix” pij.. For a uniform Markov process, pij does
not depend on the step number (on the “discrete time coordinate”), although
in the reality its not always the case.
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Bayes Formula
Conditioning usually implies that the probability of the
following step can be found given that the previous step
has occurred. Very often, however, the question is
reversed: given the result, find the probability that its
certain precondition occurred.
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Suppose that B can only occur in combination with one of m events
A1,A2,....,Am any two of which are mutually exclusive. Suppose that P(Ai) and
P(B| Ai) are known. We then have B=BA1+BA2+...+B Am and
Then,
m
P(B)=  P(A )P(B | A )
k
k
k=1
P( An )P(B|An )
P(An|B)=P( AnB)/ P(B) 
 P( Ai )P(B|Ai )
(2)
This is the Bayes Formula.
It allows us to estimate probabilities of occurring of any of the
events Ak each of whom can lead to the event B, given that B
occurred.
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Graphical interpretation of Bayes formula.
There are m different roots leading to B. The probability of reaching B through the
n-th root equals P(An)P(B|An), the probability of choosing this root by the
probability that it was traveled successfully. We assume that all P(An) and
P(B|An) are known. Assuming that B was reached, the Byes formula allows to
calculate a probability that it was done through a certain root:
P(n-th root was chosen | B was reached)=
A1
A2
P(getting to B by the n-th root)
P(getting to B by any one of m roots)
B
P(B|An)
An
P(An|B)
Am
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Example 1.
Box 1 contains 2 red and 3 blue balls; Box 2, 4 red and
one blue; and Box 3, 3 red and 4 blue.
A box is selected at random, and a ball drawn at random
is found to be red. Find the probability that box 1 was
selected.
1
2
Each box can be chosen with P(Box n)= 1/3.
We are looking for the probability P(Box1|A) that Box1 was
selected given A = “the ball was red”.
We will be using Eq. 2 translated into our events
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P(Box1|A)=P(Box1 A)/ P( A)  P(Box1)P( A|Box1)
 P(Boxj )P( A|Boxj )
First, find P(A) = 1/3*2/5 + 1/3*4/5+1/3*3/7=0.54
Second, find P(Box1*A)= 1/3*2/5 =0.13
Finally, P(Box1|A) =0.13/0.54~ 0.25
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Example 2: Exit Polls.
In the California gubernatorial elections in 1982, several TV stations
predicted, on the basis of exit polls analysis, that Tom Bradley, would win
the election. When the votes were counted, however, he lost by a
considerable margin.
What happened?
•Suppose we chose a person at random. Let B = “The person votes for Bradley”,
and suppose P(B) = 0.45 (this is the real probability which is hidden from the
analyst). Then, the probability that he voted for his opponent, P(Bc) = 0.55.
•Suppose now that some voters are reluctant to answer the questions. Let A =
“The voter stops and answers a question about how she voted”, and suppose
that P(A|B)=0.4 and P(A|Bc)=0.3. That is, 40% of Bradley’s voters will respond
compared to 30% of his opponent’s voters.
•We are interested in computing P(B|A)- a fraction of voters in our sample that
voted for Bradley (this is the measured <apparent> probability available to the
analyst).
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Here is this apparent probability of winning by Bradley.
In other words, the probability of winning by Bradley given that the voter
stops and answer the question
P (B | A) 
P ( A B )
P ( A)

P (B )P ( A | B )
0.45  0.4

 0.52
C
C
P (B )P ( A | B )  P (B )P ( A | B ) 0.45  0.4  0.55  0.3
In reality, as we know, P(B)=0.55.The reason why the prediction was wrong is
that we in fact measured a conditional probability that person voted for
Bradley given that he agreed to answer the question. In fact, Bradley
supporters are more likely to answer the question, and that is why our
prediction overestimates Bradley's chances.
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Example 3
Mr. X travels from France to London. Let B="Mr. X left from France
and successfully arrived to London".
Suppose, there are only three ways to get to London from France
A1="to take a train", A2="to take a flight", A3="to cross La'Manche in a
kayak". These events are mutually exclusive. Suppose that the
probabilities of different X’s choices are:
P(A1)=0.1, P(A2)=0.2, P(A3)=0.7.
Suppose also that due to the chances of various possible accidents,
P(B|A1)=0.98, P(B|A2)=0.99, P(B|A3)=0.4.
If X arrived to London, what are the posteriori probabilities of him
choosing A1, A2 or A3.
P(A1|B)=0.1*0.98/(0.1*0.98+0.2*0.99+0.7*0.4) = 0.17,
P(A2|B)=0.34, P(A3|B)=0.49.
And what if X did not arrive to London? In such a case
P(A1|BC)=0.047 = P(A2|BC), P(A3|BC)=0.991 (check if these values are
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correct).
Example 4 (genetics)
A woman has a brother with hemophilia, but her two parents do not have
a decease. It’s known that hemophilia is caused by a recessive gene h on
the X chromosome, which implies that mother is the carrier. In such case,
the mother has h on one of here X chromosomes and the healthy gene H
on the other X chromosome.
Since the woman received one X from her mother and one from her
father, there is 50% chance that she is a carrier and 50% chance that her
sons will have a decease.
If she has 2 sons without a decease, what’s the probability she is the
carrier?
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B=“She is a carrier”
A=“She has two healthy sons”
P(B|A) = P(AB)/P(A)
she is a carrier AND has two
healthy sons
P(AB)= P(B)*P(A|B) = ½*1/4=1/8.
P(A) = P(AB) +
P(ABC);
she is not a carrier AND has two
healthy sons
P(ABC)= P(BC)*P(A| BC)= 1/2*1 (if she is not a carrier, then
her sons 100% won’t have hemophilia)
P(A) = P(AB) + P(ABC) )=(1/2)(1/2)2 + (1/2) 1= 5/8
P(B|A)=1/8 /(5/8) = 1/5
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Home assignment
If you have any questions about the HW, please contact me.
1. Read the lecture and work on the problems at pp. 9, 20-25. Make
sure you understand the solutions.
2. Solve “Self-Test 6” problems, compare with the solutions.
3. Solve the problems posted on Webct (week 6)
4. Practice with Mathematica. Make sure that everyone goes
through the Steps 1-6 and the Markov0-2 files. Based on your
HWs submitted lately I can tell that most of you became quite
proficient with Mathematica, although some of you practically did
not take time to learn this amazing tool and still are walking in
darkness.
5. Read and practice with the Maximum Likelihood file (see the link on
the page).
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5. Extra credit problems:
5.1 Suppose we arrange the 26 letters of the alphabet in random order.
What is the probability p that the x and the y will be next to each
other?
5.2 In the same setting, what is the probability that x, y and z are next to
each other?
Solve these problems analytically and test your solutions by writing
Mathematica code and running the random experiments.
PS Please submit it only if you think that at least one problem is solved.
It should contain both analytical and Mathematical parts. Do not
submit to show that you tried. It does not count .
6. Work on your projects.
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