Probabilities. Basic probability theory

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# Day 1. Probabilities.

## Solution to exercises

.instructors[
  **ECL707/807** - Dominique Gravel
]

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# Solution 1.1.

Marginal probability :

`$P(A=1) = P(A_{10})+P(A_{11}) = 0.15 + 0.05 = 0.2$`

Conditional probability :

`$P(A=1|B=0) = \frac{P(A=1,B=0)}{P(B=0)} = \frac{0.15}{1 - (0.20 + 0.05)} = 0.2$`

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# Exercise 1.3.

What is the probability there is not a single seed ?

`$P(s=0) = (1-0.01)\times(1-0.2)\times(1-0.17)\times(1-0.24)\times(1-0.06)$`

`$P(s=0) = 0.47$`

What is the probability to pick at least one seed in the trap ?

`$P(n>0) = 1-P(s>0) = 0.53$`

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# Exercise 1.4.

Under this hypothesis, sites and species are considered independent. The probability that a given species `$i$` is located at both sites is therefore obtained by the square of regional prevalence. We have to sum them across species to get `$N_{11}$`, such that we get :

`$N_{11} = 0.22*0.22 + 0.15*0.15+0.37*0.37 = 0.21$`

We could do the same with the complements to obtain `$N_{10}$` and `$N_{01}$`. Note these are symetrical :

`$N_{10} = 0.22*(1-0.22)+0.15*(1-0.15)+0.37*(1-0.37)=0.53$`

And the resulting expected Jaccard index is :

`$J(A,B) = \frac{0.21}{0.21+.53+.53}=0.17$`

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# Solution 1.4.

In this situation composition of the two sites will depend on `$D_{xy}$`. We will therefore define a function to compute the different quantities necessary for the Jaccard.

Occurrence of species `$i$` at two sites is a joint probability. We will consider that the occurrence probability at site `$A$` is given by regional prevalence and then use conditional probability (the function) to derive the joint probability :

`$P(A_i = 1, B_i = 1) = P(B_i = 1 |A_i = 1)P(A_i = 1)$`

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# Solution 1.4.

Then we need to get the complement, for only site A but not site B :

`$P(A_i = 1, B_i = 0) = P(B_i = 0 |A_i)P(A_i = 1)$`

And the inverse :

`$P(A_i = 0, B_i = 1) = P(B_i = 1 |A_i = 0)P(A_i = 0)$`

We finally follow the previous recipe to compute the Jacccard index.

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# Solution 1.5.

Marginal occurrence probability at optimal climate is `$P(X=1|E^+) = 0.2$`

Conditional occurrence probability at optimal climate and habitat is `$P(X1|E^+, H^+)=0.8$`

Conditional occurrence probability at optimal climate but poor habitat is `$P(X1|E^+, H^-)=0.05$`

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# Solution 1.5.

Using rules of probability, we get `$P(H^+$` as follows :

`$P(X=1|E^+) = P(X=1|E^+,H^+)P(H^+)+ P(X=1|E^+,H^-)P(H^-)$`

`$P(X=1|E^+) = P(X=1|E^+,H^+)P(H^+)+ P(X=1|E^+,H^-)(1-P(H^+))$`

`$P(H^+) = \frac{P(X|E^+)-P(X|E^+,H^-)}{P(X|E^+,H^+)-P(X|E^+,H^-)}$`

`$P(H^+) = \frac{0.20-0.05}{0.80-0.05}$`

`$P(H^+) = 0.2$`

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# Solution 1.6.

We first have to define the quantities properly. First thing, define what you are looking for :

`$P(link | obs)$` : the probability there is an interaction given the observation

And then the additional data given in the problem :

`$P(obs | link)$` : the probability of the observation given the hypothesis there is a link
`$P(link)$` : prior knowledge on the probability there is a link

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# Solution 1.6.

Using Bayes theorem, we have the following solution :

`$P(link | obs) = \frac{P(obs|link)P(link)}{P(obs)}$`

We define the occurrence of an interaction a success of a bernouilli trial. For the numerator, we use the binomial distribution with `$5$` observations to find that the probability of observing no interaction with associated probability `$P(link)$` is :

$P(obs|link) = 0.33 $

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# Solution 1.6.

The trick is often to compute the denominator. It's easy to conceive the first part :

$P(obs|link)*P(link) = 0.33 x 0.20 $

The second one needs more thought but it's obvious at the end :

`$P(obs|link = 0)P(link = 0) = 1 x 0.80$`

Which gives the denominator :

`$P(obs) = 0.33 x 0.20 + 1 x 0.80$`

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# Solution 1.6.

Collecting the pieces together, we obtain the following answer :

`$P(link | obs) = \frac{0.33 x 0.20}{0.33 x 0.20 + 1 x 0.8} = 0.076$`

This is the probability the two species interact given the observations derived using prior information.

**Congratulations, you discovered the essence of Bayes statistics without knowing it !**