This class is based on old material and is not completely updated.

It will be updated later.

My apologies for any confusing material.

• Has 2 sides: ‘Head’ and ‘Tail’
• These options are exhaustive and exclusive \[\Pr(\text{'Head'}|Z) + \Pr(\text{'Tail'}|Z)= 1\]
• If the coin is symmetric then \[\Pr(\text{'Head'}|Z) = \Pr(\text{'Tail'}|Z) = 0.5\]

• Four possibilities: \(\text{'A'}, \text{'C'}, \text{'G'}\) and \(\text{'T'}.\)
• These outcomes are exhaustive and exclusive \[\Pr(\text{'A'}|Z)+ \Pr(\text{'C'}|Z)+ \Pr(\text{'G'}|Z)+\Pr(\text{'T'}|Z)= 1\]
• The values depend on the organism

• The formula depends on the mechanics of the process
  • We have to understand the physics of it
  • e.g. DNA has 4 nucleotides
• It also depends on some parameters
  • We need experimental data to find them
  • Chagraff rules
  • GC content

We throw two dice at the same time. What will be the sum of both numbers?

A <- 🎲
B <- 🎲
C <- A + B

\[\Pr(C\,|\, (B>3)\wedge Z)\]

\[\Pr(C\,|\, (B=3)\wedge Z)\]

A probability associates each event with a number

In the case of simple events we have \[\Pr(X=x)=\Pr(x)\]

More complex events can be evaluated by decomposing them into simpler ones \[\Pr(X\text{ is purine})=\Pr(X=\text{'A'})+\Pr(X=\text{'G'})\]

We throw two dice at the same time. What will be the sum of both numbers?

The probability of an event depends on our knowledge

Information about an event of \(C\) may change the probabilities of \(B\)

\(\Pr(B\vert C\wedge Z)\) may not be the same as \(\Pr(B\vert Z)\)

That is the general case. We should not expect that two events are a priori independent

The variables \(B\) and \(C\) are independent if knowing any event of \(C\) does not change the probabilities of \(B\) \[\Pr(B|C\wedge Z)=\Pr(B\vert Z)\] By symmetry, knowing events about \(B\) does not change the probabilities for \(C\) \[\Pr(C|B\wedge Z)=\Pr(C\vert Z)\] We can write \(B\perp C\)

If two experiments \(B\) and \(C\) are performed, we can study the probability of events on \(B\) and events on \(C\)

The probability for both events is then \[\Pr(B=a, C=b) = \Pr(B=a|C=b)\cdot\Pr(C=b)\] or in short \[\Pr(B, C) = \Pr(B|C)\cdot\Pr(C)\]

The probability of an event on \(B\) and on \(C\) can be seen in two parts

• The probability that the event \(C\) happens, and
• The probability that \(B\) happens, given that \(C\) happened \[\Pr(B, C) = \Pr(B|C)\cdot\Pr(C)\]

The joint probability is always \[\Pr(B, C\vert Z) = \Pr(B|C\wedge Z)\cdot\Pr(C\vert Z)\] If \(B\) and \(C\) are independent, then \[\Pr(B|C\wedge Z)=\Pr(B\vert Z)\] Replacing the second equation into the first, we have \[\Pr(B, C\vert Z) = \Pr(B\vert Z)\cdot\Pr(C\vert Z)\quad\text{ if }B\perp C\]

Imagine we have a test to determine if someone has HIV.

Let’s assume that:

• There are \(10^{5}\) persons tested
• The test has a precision of 99%
• The incidence of HIV in the population is 0.1%

\[\Pr(\text{HIV}_+)=0.001\]

\[\Pr(\text{test}_+, \text{HIV}_+)=\Pr(\text{test}_+ \vert \text{HIV}_+)\cdot\Pr(\text{HIV}_+)\] \[\Pr(\text{test}_+ \vert \text{HIV}_+)=0.99\]

\[\Pr(\text{test}_-, \text{HIV}_-)=\Pr(\text{test}_- \vert \text{HIV}_-)\cdot\Pr(\text{HIV}_-)\] \[\Pr(\text{test}_- \vert \text{HIV}_-)=0.99\]

\[\Pr(\text{test}_-, \text{HIV}_+)=\Pr(\text{test}_- \vert \text{HIV}_+)\cdot\Pr(\text{HIV}_+)\] \[\Pr(\text{test}_- \vert \text{HIV}_+)=0.01\]

\[\Pr(\text{test}_+)= \Pr(\text{test}_+, \text{HIV}_+)+ \Pr(\text{test}_+, \text{HIV}_-)\]

What is the probability of being sick given that the test is positive?

\[\Pr(\text{test}_+, \text{HIV}_+)=\Pr(\text{HIV}_+ \vert \text{test}_+)\cdot\Pr(\text{test}_+)\] \[\Pr(\text{HIV}_+ \vert \text{test}_+)=\frac{\Pr(\text{test}_+, \text{HIV}_+)}{\Pr(\text{test}_+)}=\frac{99}{1098} = 9\%\]

“An Essay towards solving a Problem in the Doctrine of Chances” is a work on the mathematical theory of probability by the Reverend Thomas Bayes, published in 1763, two years after its author’s death

The use of the Bayes theorem has been extended in science and in other fields

Since \[\Pr(B, C\vert Z) = \Pr(B|C,Z)\cdot\Pr(C\vert Z)\] and, by symmetry \[\Pr(B, C\vert Z) = \Pr(C|B,Z)\cdot\Pr(B\vert Z)\] then \[\Pr(B|C) = \frac{\Pr(C|B,Z)\cdot\Pr(B\vert Z)}{\Pr(C\vert Z)}\]

It can be understood as \[\Pr(B|C\wedge Z) = \frac{\Pr(C|B\wedge Z)}{\Pr(C\vert Z)}\cdot\Pr(B\vert Z)\] which is a rule to update our opinions

• \(\Pr(B\vert Z)\) is the a priori probability
• \(\Pr(B|C\wedge Z)\) is a posteriori probability

Bayes says how to change \(\Pr(B\vert Z)\) when we learn \(C\)

Another point of view is \[\Pr(B|C\wedge Z) = \Pr(C|B\wedge Z)\cdot\frac{\Pr(B\vert Z)}{\Pr(C\vert Z)}\] which is a rule to invert the conditional probability

This is the view we will use now

We have two variables:

• The DNA sequence: \(\mathbf{X}=(X_1,\ldots,X_m)\)
• The presence or absence of a binding site: \(B_+\) or \(B_-\)

We do an “experiment” and get a short DNA sequence \(\mathbf{x}=(s_1,\ldots,s_m)\)

We want \(\Pr(B_+|\mathbf{X}=\mathbf{x})\)

We want \(\Pr(B_+|\mathbf{X}=\mathbf{x})\)

Applying Bayes’ theorem we have \[\Pr(B_+|\mathbf{X}=\mathbf{x})= \frac{\Pr(\mathbf{X}=\mathbf{x}|B_+)\cdot\Pr(B_+)}{\Pr(\mathbf{X}=\mathbf{x})}\] so we need to find them

We have a matrix \(\mathbf{M}\) with the empirical frequencies of nucleotides in \(n\) sequences

\(\mathbf{M}\) has 4 rows (A, C, T, G) and \(m\) columns

\(M_{ij}=\) number of times nucleotide \(i\) is at position \(j\)

The sum of each column of \(\mathbf{M}\) is \(n\)

We assume that these sequences are outcomes of a probabilistic process

That is, the sequences follow some probability

We don’t know the exact probability

But we can approximate \[\Pr(X_j=i|B_+)=M_{ij}/n\] for \(i\in\{A,C,T,G\}\)

We also assume that the probabilities of each \(X_j\) are independent

In such case we have \[\Pr(\mathbf{X}=\mathbf{x}|B_+)= \Pr(X_1=s_1|B_+) \cdots \Pr(X_m=s_m|B_+)\] or, in short \[\Pr(\mathbf{X}=\mathbf{x}|B_+)= \prod_{j=1}^m\Pr(X_j=s_j|B_+)\]

Using the same hypothesis of independence, we have \[\Pr(\mathbf{X}=\mathbf{x})= \Pr(X_1=s_1) \cdots \Pr(X_m=s_m)\] or, in short \[\Pr(\mathbf{X}=\mathbf{x})= \prod_{j=1}^m\Pr(X_j=s_j)\] Usually \(\Pr(X_j=i)\) is approximated by the frequency of each nucleotide in the complete genome \[\Pr(X_j=i)=\frac{N_i}{L}\]

We got “good” guesses of \(\Pr(\mathbf{X}=\mathbf{x}|B_+)\) and \(\Pr(\mathbf{X}=\mathbf{x})\)

We need \(\Pr(B_+)\)

How do we get it?

There is no easy answer for that

Let’s say \(\Pr(B_+)=K\) and later we solve it

Applying Bayes’ theorem we have \[\Pr(B_+|\mathbf{X}=\mathbf{x})=\prod_{j=1}^m \frac{\Pr(X_j=s_j|B_+)}{\Pr(X_j=s_j)}\cdot K\]

Can it be simpler?

…are made to change multiplications into sums

\[\log\Pr(B_+|\mathbf{X}=\mathbf{x})=\sum_{j=1}^m \log\frac{\Pr(X_j=s_j|B_+)}{\Pr(X_j=s_j)} + mK\]

For each sequence \(\mathbf{x}\) we calculate the score

\[\mathrm{Score}(\mathbf{x}) =\sum_{j=1}^m Q_{s_j,j} =\sum_{j=1}^m\log\frac{\Pr(X_j=s_j|B_+)}{\Pr(X_j=s_j)}+Const\]

We prepare a matrix \(\mathbf{Q}\) for each binding site \[Q_{i,j}=\log\frac{M_{ij}}{N_i}\]

Write a program (in any computer language) to calculate the score of each position of a genome