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A Bayesian nonparametric method for the LR assessment in case of rare type match

Giulia Cereda October 8, 2015

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Forensic Statistics

Ingredients:

Giulia Cereda () Short title October 8, 2015 2 / 33

Forensic Statistics

Ingredients: Crime case

Giulia Cereda () Short title October 8, 2015 2 / 33

Forensic Statistics

Ingredients: Crime case Evidence (E)

Giulia Cereda () Short title October 8, 2015 2 / 33

Forensic Statistics

Ingredients: Crime case Evidence (E) 2 Hypotheses of Interest: Hp vs Hd

Giulia Cereda () Short title October 8, 2015 2 / 33

Forensic Statistics

Ingredients: Crime case Evidence (E) 2 Hypotheses of Interest: Hp vs Hd Background (B)

Giulia Cereda () Short title October 8, 2015 2 / 33

Forensic Statistics

Ingredients: Crime case Evidence (E) 2 Hypotheses of Interest: Hp vs Hd Background (B) D = (E,B).

Giulia Cereda () Short title October 8, 2015 2 / 33

Forensic Statistics

Ingredients: Crime case Evidence (E) 2 Hypotheses of Interest: Hp vs Hd Background (B) D = (E,B). The court asks for the likelihood ratio

Giulia Cereda () Short title October 8, 2015 2 / 33

Forensic Statistics

Ingredients: Crime case Evidence (E) 2 Hypotheses of Interest: Hp vs Hd Background (B) D = (E,B). The court asks for the likelihood ratio Pr(Hp | D) Pr(Hd | D) = Pr(D | Hp) Pr(D | Hd)

• LR

Pr(Hp) Pr(Hd)

Giulia Cereda () Short title October 8, 2015 2 / 33

Example

Ingredients:

Giulia Cereda () Short title October 8, 2015 3 / 33

Example

Ingredients: Crime case: murder

Giulia Cereda () Short title October 8, 2015 3 / 33

Example

Ingredients: Crime case: murder Evidence (E): profile of the DNA trace found on the crime scene matches the suspect’s DNA profile.

Giulia Cereda () Short title October 8, 2015 3 / 33

Example

Ingredients: Crime case: murder Evidence (E): profile of the DNA trace found on the crime scene matches the suspect’s DNA profile. 2 Hypotheses of Interest:

Giulia Cereda () Short title October 8, 2015 3 / 33

Example

Ingredients: Crime case: murder Evidence (E): profile of the DNA trace found on the crime scene matches the suspect’s DNA profile. 2 Hypotheses of Interest:

Hp: The suspect left the stain

Giulia Cereda () Short title October 8, 2015 3 / 33

Example

Ingredients: Crime case: murder Evidence (E): profile of the DNA trace found on the crime scene matches the suspect’s DNA profile. 2 Hypotheses of Interest:

Hp: The suspect left the stain Hd: Someone else left the stain

Giulia Cereda () Short title October 8, 2015 3 / 33

Example

Ingredients: Crime case: murder Evidence (E): profile of the DNA trace found on the crime scene matches the suspect’s DNA profile. 2 Hypotheses of Interest:

Hp: The suspect left the stain Hd: Someone else left the stain

Background (B): database of DNA profiles from the population of possible perpetrators

Giulia Cereda () Short title October 8, 2015 3 / 33

DNA profiles

Giulia Cereda () Short title October 8, 2015 4 / 33

DNA profiles

A DNA profile is a list of integers h = (4 − 5 − 2 − 10) that code some characteristics in some portions of the DNA sequence of an individual: different persons can share the same profile.

Giulia Cereda () Short title October 8, 2015 4 / 33

DNA profiles

A DNA profile is a list of integers h = (4 − 5 − 2 − 10) that code some characteristics in some portions of the DNA sequence of an individual: different persons can share the same profile. For Hp the match is a sure event,

Giulia Cereda () Short title October 8, 2015 4 / 33

DNA profiles

A DNA profile is a list of integers h = (4 − 5 − 2 − 10) that code some characteristics in some portions of the DNA sequence of an individual: different persons can share the same profile. For Hp the match is a sure event, For Hd the match is a random event with probability ph = frequency of the profile h of the suspect in the population of possible perpetrators.

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DNA database

Database: a list of DNA profiles from a sample from the population of possible perpetrators

Giulia Cereda () Short title October 8, 2015 5 / 33

DNA database

Database: a list of DNA profiles from a sample from the population of possible perpetrators DATABASE of size 10 Person 1 (4 − 10 − 6 − 7) Person 2 (3 − 5 − 6 − 8) Person 3 (3 − 7 − 8 − 10) Person 4 (10 − 1 − 4 − 5) Person 5 (3 − 7 − 8 − 10) Person 6 (3 − 7 − 8 − 10) Person 7 (1 − 5 − 7 − 2) Person 8 (3 − 7 − 8 − 10) Person 9 (3 − 5 − 6 − 8) Person 10 (3 − 7 − 8 − 10)

Giulia Cereda () Short title October 8, 2015 5 / 33

DNA database

Database: a list of DNA profiles from a sample from the population of possible perpetrators DATABASE of size 10 Person 1 (4 − 10 − 6 − 7) Person 2 (3 − 5 − 6 − 8) Person 3 (3 − 7 − 8 − 10) Person 4 (10 − 1 − 4 − 5) Person 5 (3 − 7 − 8 − 10) Person 6 (3 − 7 − 8 − 10) Person 7 (1 − 5 − 7 − 2) Person 8 (3 − 7 − 8 − 10) Person 9 (3 − 5 − 6 − 8) Person 10 (3 − 7 − 8 − 10) The database is used to find out the rarity of the matching profile.

Giulia Cereda () Short title October 8, 2015 5 / 33

LR assessment in the rare type match case

My research focuses on the LR assessment in the rare type match case, that is:

Giulia Cereda () Short title October 8, 2015 6 / 33

LR assessment in the rare type match case

My research focuses on the LR assessment in the rare type match case, that is: A match between the suspect’s DNA profile and the crime stain’s DNA profile.

Giulia Cereda () Short title October 8, 2015 6 / 33

LR assessment in the rare type match case

My research focuses on the LR assessment in the rare type match case, that is: A match between the suspect’s DNA profile and the crime stain’s DNA profile. This profile is not contained in the database B.

Giulia Cereda () Short title October 8, 2015 6 / 33

LR assessment in the rare type match case

My research focuses on the LR assessment in the rare type match case, that is: A match between the suspect’s DNA profile and the crime stain’s DNA profile. This profile is not contained in the database B. Especially if the database is big, the profile seems to be rare.

Giulia Cereda () Short title October 8, 2015 6 / 33

LR assessment in the rare type match case

My research focuses on the LR assessment in the rare type match case, that is: A match between the suspect’s DNA profile and the crime stain’s DNA profile. This profile is not contained in the database B. Especially if the database is big, the profile seems to be rare. How rare?

Giulia Cereda () Short title October 8, 2015 6 / 33

Previous models

Giulia Cereda () Short title October 8, 2015 7 / 33

Previous models

Frequentist model: (Cereda 2015) Frequentist approach to LR assessment in case of rare haplotype match arXiv:1502.04083

Giulia Cereda () Short title October 8, 2015 7 / 33

Previous models

Frequentist model: (Cereda 2015) Frequentist approach to LR assessment in case of rare haplotype match arXiv:1502.04083 Bayesian model: (Cereda 2015) Full Bayesian approach to LR assessment in case of rare haplotype match arXiv:1502.02406

Giulia Cereda () Short title October 8, 2015 7 / 33

Previous models

Frequentist model: (Cereda 2015) Frequentist approach to LR assessment in case of rare haplotype match arXiv:1502.04083 Bayesian model: (Cereda 2015) Full Bayesian approach to LR assessment in case of rare haplotype match arXiv:1502.02406 (Cereda 2015) Nonparametric Bayesian approach to LR assessment in case of rare haplotype match arXiv:1506.08444

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Assumptions

Assumption 1

There are so many different DNA types that they may be considered infinite.

Giulia Cereda () Short title October 8, 2015 8 / 33

Assumptions

Assumption 1

There are so many different DNA types that they may be considered infinite. Parameter: p = (pt|t ∈ T), T an infinite countable set,pt > 0, pt = 1, to represent the (unknown) frequencies of all DNA types in Nature.

Giulia Cereda () Short title October 8, 2015 8 / 33

Assumptions

Assumption 1

There are so many different DNA types that they may be considered infinite. Parameter: p = (pt|t ∈ T), T an infinite countable set,pt > 0, pt = 1, to represent the (unknown) frequencies of all DNA types in Nature.

Assumption 2

The particular list of integers that forms a DNA type is just a category: no structure assumed.

Giulia Cereda () Short title October 8, 2015 8 / 33

Assumptions

Assumption 1

There are so many different DNA types that they may be considered infinite. Parameter: p = (pt|t ∈ T), T an infinite countable set,pt > 0, pt = 1, to represent the (unknown) frequencies of all DNA types in Nature.

Assumption 2

The particular list of integers that forms a DNA type is just a category: no structure assumed. “DNA types” or “colors” is now the same.

Giulia Cereda () Short title October 8, 2015 8 / 33

Random partitions of [n]

Let [n] denote the set [n] = {1, 2, ..., n}.

Giulia Cereda () Short title October 8, 2015 9 / 33

Random partitions of [n]

Let [n] denote the set [n] = {1, 2, ..., n}. A partition of the set [n] will be denoted as π[n].

Giulia Cereda () Short title October 8, 2015 9 / 33

Random partitions of [n]

Let [n] denote the set [n] = {1, 2, ..., n}. A partition of the set [n] will be denoted as π[n]. Random partitions on the set [n] will be denoted as Π[n].

Giulia Cereda () Short title October 8, 2015 9 / 33

DNA database can be reduced

DATABASE of size 10 Person 1 (4 − 10 − 6 − 7) Person 2 (3 − 5 − 6 − 8) Person 3 (3 − 7 − 8 − 10) Person 4 (10 − 1 − 4 − 5) Person 5 (3 − 7 − 8 − 10) Person 6 (3 − 7 − 8 − 10) Person 7 (1 − 5 − 7 − 2) Person 8 (3 − 7 − 8 − 10) Person 9 (3 − 5 − 6 − 8) Person 10 (3 − 7 − 8 − 10)

Giulia Cereda () Short title October 8, 2015 10 / 33

DNA database can be reduced

DATABASE of size 10 Person 1 (4 − 10 − 6 − 7) Person 2 (3 − 5 − 6 − 8) Person 3 (3 − 7 − 8 − 10) Person 4 (10 − 1 − 4 − 5) Person 5 (3 − 7 − 8 − 10) Person 6 (3 − 7 − 8 − 10) Person 7 (1 − 5 − 7 − 2) Person 8 (3 − 7 − 8 − 10) Person 9 (3 − 5 − 6 − 8) Person 10 (3 − 7 − 8 − 10)

Giulia Cereda () Short title October 8, 2015 11 / 33

DNA database can be reduced

DATABASE of size 10 Person 1 (4 − 10 − 6 − 7) Person 2 (3 − 5 − 6 − 8) Person 3 (3 − 7 − 8 − 10) Person 4 (10 − 1 − 4 − 5) Person 5 (3 − 7 − 8 − 10) Person 6 (3 − 7 − 8 − 10) Person 7 (1 − 5 − 7 − 2) Person 8 (3 − 7 − 8 − 10) Person 9 (3 − 5 − 6 − 8) Person 10 (3 − 7 − 8 − 10) Assumption 2 → data can be replaces by the equivalence classes on the indices of the relation “to have the same DNA type”. This is a partition of the set [n] : {{1}, {2, 9}, {3, 5, 6, 8, 10}, {4}, {7}}

Giulia Cereda () Short title October 8, 2015 11 / 33

Reduced data

Giulia Cereda () Short title October 8, 2015 12 / 33

Reduced data

Data D is made of the database + 2 new observations

Giulia Cereda () Short title October 8, 2015 12 / 33

Reduced data

Data D is made of the database + 2 new observations D = π[n+2] partition of the set {1, 2, ..., n + 2}

Giulia Cereda () Short title October 8, 2015 12 / 33

Reduced data

Data D is made of the database + 2 new observations D = π[n+2] partition of the set {1, 2, ..., n + 2} Example: Database → π[10] = {{1}, {2, 9}, {3, 5, 6, 8, 10}, {4}, {7}}

Giulia Cereda () Short title October 8, 2015 12 / 33

Reduced data

Data D is made of the database + 2 new observations D = π[n+2] partition of the set {1, 2, ..., n + 2} Example: Database → π[10] = {{1}, {2, 9}, {3, 5, 6, 8, 10}, {4}, {7}} D → π[12] = {{1}, {2, 9}, {3, 5, 6, 8, 10}, {4}, {7}, {11, 12}}

Giulia Cereda () Short title October 8, 2015 12 / 33

Reduced data

Data D is made of the database + 2 new observations D = π[n+2] partition of the set {1, 2, ..., n + 2} Example: Database → π[10] = {{1}, {2, 9}, {3, 5, 6, 8, 10}, {4}, {7}} D → π[12] = {{1}, {2, 9}, {3, 5, 6, 8, 10}, {4}, {7}, {11, 12}} We can see the data as a random variable. In that case, D = Π[n+2].

Giulia Cereda () Short title October 8, 2015 12 / 33

The distribution of D = Π[n+2] depends on p. However, it does not depend on the order of the pi.

Giulia Cereda () Short title October 8, 2015 13 / 33

The distribution of D = Π[n+2] depends on p. However, it does not depend on the order of the pi. ↓ We can consider directly the ordered vector p ∈ ∇∞ = {(p1, p2, ....), p1 ≥ p2 ≥ ... > 0, pi = 1}.

Giulia Cereda () Short title October 8, 2015 13 / 33

The distribution of D = Π[n+2] depends on p. However, it does not depend on the order of the pi. ↓ We can consider directly the ordered vector p ∈ ∇∞ = {(p1, p2, ....), p1 ≥ p2 ≥ ... > 0, pi = 1}. For instance, p3= the frequency of the third most frequent DNA type in Nature.

Giulia Cereda () Short title October 8, 2015 13 / 33

Prior distribution on p ∈ ∇∞

Bayesian nonparametrics: we need a prior for the parameter p.

Giulia Cereda () Short title October 8, 2015 14 / 33

Prior distribution on p ∈ ∇∞

Bayesian nonparametrics: we need a prior for the parameter p. Two parameter Poisson Dirichlet distribution.

Giulia Cereda () Short title October 8, 2015 14 / 33

Prior distribution on p ∈ ∇∞

Bayesian nonparametrics: we need a prior for the parameter p. Two parameter Poisson Dirichlet distribution. Parameters: 0 < α < 1, θ > −α

Giulia Cereda () Short title October 8, 2015 14 / 33

The model (first part)

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The model (first part)

A, Θ ∇∞ ∋ P

The model (first part)

A, Θ ∇∞ ∋ P P|α, θ ∼ PD(α, θ)

The model (first part)

A, Θ ∇∞ ∋ P P|α, θ ∼ PD(α, θ) ... X1 N ∋ X2 Xn Xi = j → the i-th observation has the jth most common type in Nature.

Giulia Cereda () Short title October 8, 2015 15 / 33

The model (first part)

A, Θ ∇∞ ∋ P P|α, θ ∼ PD(α, θ) ... X1 N ∋ X2 Xn Xi = j → the i-th observation has the jth most common type in Nature.

Giulia Cereda () Short title October 8, 2015 15 / 33

The model (first part)

A, Θ P ∇∞ ∋ (A, Θ) ∼ f P|α, θ ∼ PD(α, θ) ... X1 N ∋ X2 Xn

The model (first part)

A, Θ P ∇∞ ∋ (A, Θ) ∼ f P|α, θ ∼ PD(α, θ) ... X1 N ∋ X2 Xn Suspect Xn+1

The model (first part)

A, Θ P ∇∞ ∋ (A, Θ) ∼ f P|α, θ ∼ PD(α, θ) ... X1 N ∋ X2 Xn Suspect Xn+1 X1, ..., Xn+1|p ∼i.i.d p

The model (first part)

A, Θ P ∇∞ ∋ (A, Θ) ∼ f P|α, θ ∼ PD(α, θ) ... X1 N ∋ X2 Xn Suspect Xn+1 X1, ..., Xn+1|p ∼i.i.d p Crime stain Xn+2 H {Hp, Hd} ∋

The model (first part)

A, Θ P ∇∞ ∋ (A, Θ) ∼ f P|α, θ ∼ PD(α, θ) ... X1 N ∋ X2 Xn Suspect Xn+1 X1, ..., Xn+1|p ∼i.i.d p Crime stain Xn+2 H {Hp, Hd} ∋ Xn+2|p, H, xn+1 ∼

• δxn+1

if H = Hp p if H = Hd

Giulia Cereda () Short title October 8, 2015 16 / 33

The model (first part)

A, Θ P ∇∞ ∋ (A, Θ) ∼ f P|α, θ ∼ PD(α, θ) ... X1 N ∋ X2 Xn Suspect Xn+1 X1, ..., Xn+1|p ∼i.i.d p Crime stain Xn+2 H {Hp, Hd} ∋ Xn+2|p, H, xn+1 ∼

• δxn+1

if H = Hp p if H = Hd

Giulia Cereda () Short title October 8, 2015 16 / 33

The model (first part)

A, Θ P ... X1 X2 Xn Xn+1 Xn+2 H

Giulia Cereda () Short title October 8, 2015 17 / 33

Random partitions

Some notation: Given X1, ..., Xn ∈ N, random variables, Π[n](X1, X2, ..., Xn) is the random partition defined by the equivalence classes of i ∼ j iff Xi = Xj.

Giulia Cereda () Short title October 8, 2015 18 / 33

Random partitions

Some notation: Given X1, ..., Xn ∈ N, random variables, Π[n](X1, X2, ..., Xn) is the random partition defined by the equivalence classes of i ∼ j iff Xi = Xj. X1, ..., Xn − → Π[n] = πDb

[n]

X1, ..., Xn, Xn+1 − → Π[n+1] = πDb+

[n+1]

X1, ..., Xn, Xn+1, Xn+2 − → Π[n+2] = πDb++

[n+2]

Giulia Cereda () Short title October 8, 2015 18 / 33

Random partitions

Some notation: Given X1, ..., Xn ∈ N, random variables, Π[n](X1, X2, ..., Xn) is the random partition defined by the equivalence classes of i ∼ j iff Xi = Xj. X1, ..., Xn − → Π[n] = πDb

[n]

X1, ..., Xn, Xn+1 − → Π[n+1] = πDb+

[n+1]

X1, ..., Xn, Xn+1, Xn+2 − → Π[n+2] = πDb++

[n+2]

X1, ..., Xn are not observed, but generates the same partition as the

• riginal database.

Data can be defined as D = Π[n+2].

Giulia Cereda () Short title October 8, 2015 18 / 33

The complete model

A, Θ H P X1 X2 Xn Xn+1 Xn+2 D

Giulia Cereda () Short title October 8, 2015 19 / 33

Pitman sampling formula

Giulia Cereda () Short title October 8, 2015 20 / 33

Pitman sampling formula

P ∼ PD(α, θ) X1, X2, ..., Xn|P = p ∼i.i.d p

Giulia Cereda () Short title October 8, 2015 20 / 33

Pitman sampling formula

P ∼ PD(α, θ) X1, X2, ..., Xn|P = p ∼i.i.d p then Π[n] = Π[n](X1, ..., Xn) has the following distribution:

Giulia Cereda () Short title October 8, 2015 20 / 33

Pitman sampling formula

P ∼ PD(α, θ) X1, X2, ..., Xn|P = p ∼i.i.d p then Π[n] = Π[n](X1, ..., Xn) has the following distribution: Pr(Π[n] = π[n]|α, θ) = Pn

α,θ(π[n]) = [θ + α]k−1;α

[θ + 1]n−1;1

k

• i=1

[1 − α]ni−1;1,

Giulia Cereda () Short title October 8, 2015 20 / 33

Pitman sampling formula

P ∼ PD(α, θ) X1, X2, ..., Xn|P = p ∼i.i.d p then Π[n] = Π[n](X1, ..., Xn) has the following distribution: Pr(Π[n] = π[n]|α, θ) = Pn

α,θ(π[n]) = [θ + α]k−1;α

[θ + 1]n−1;1

k

• i=1

[1 − α]ni−1;1, In our model Pr(D|α, θ, h) = Pr(Π[n+2] = πDb++

[n+2] |α, θ, h) =

• Pn+2

α,θ (πDb++ [n+2] )

if h = Hd Pn+1

α,θ (πDb+ [n+1])

if h = Hp

Giulia Cereda () Short title October 8, 2015 20 / 33

The model, simplified

A, Θ H D

Giulia Cereda () Short title October 8, 2015 21 / 33

The model, simplified

A, Θ H D D = Π[n+2].

Giulia Cereda () Short title October 8, 2015 21 / 33

Lemma

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Lemma

A H X Y

Lemma

Given four random variables A, H, X and Y , as above, the likelihood function for h, given X = x and Y = y, satisfies lik(h | x, y) ∝ E(p(y | x, A, h) | X = x).

Giulia Cereda () Short title October 8, 2015 22 / 33

Lemma

A, Θ H Π[n+1] Π[n+2]

Giulia Cereda () Short title October 8, 2015 23 / 33

Lemma

A, Θ H Π[n+1] Π[n+2] lik(h | π[n+1], π[n+2]) ∝ E(p(π[n+2] | π[n+1], A, Θ, h) | Π[n+1] = π[n+1]).

Giulia Cereda () Short title October 8, 2015 23 / 33

Likelihood ratio

LR = p(π[n+2]|Hp) p(π[n+2]|Hd) = p(π[n+1], π[n+2]|Hp) p(π[n+1], π[n+2]|Hd) = lik(Hp|π[n+1], π[n+2]) lik(Hd|π[n+1], π[n+2])

Giulia Cereda () Short title October 8, 2015 24 / 33

Likelihood ratio

LR = p(π[n+2]|Hp) p(π[n+2]|Hd) = p(π[n+1], π[n+2]|Hp) p(π[n+1], π[n+2]|Hd) = lik(Hp|π[n+1], π[n+2]) lik(Hd|π[n+1], π[n+2]) Lemma allows to write LR = E(

1

• p(π[n+2] | π[n+1], A, Θ, Hp) | Π[n+1] = π[n+1])

E(p(π[n+2] | π[n+1], A, Θ, Hd)

• 1−A

n+1+Θ

| Π[n+1] = π[n+1])

Giulia Cereda () Short title October 8, 2015 24 / 33

Likelihood ratio

LR = p(π[n+2]|Hp) p(π[n+2]|Hd) = p(π[n+1], π[n+2]|Hp) p(π[n+1], π[n+2]|Hd) = lik(Hp|π[n+1], π[n+2]) lik(Hd|π[n+1], π[n+2]) Lemma allows to write LR = E(

1

• p(π[n+2] | π[n+1], A, Θ, Hp) | Π[n+1] = π[n+1])

E(p(π[n+2] | π[n+1], A, Θ, Hd)

• 1−A

n+1+Θ

| Π[n+1] = π[n+1]) = 1 E

• 1−A

n+1+Θ | Π[n+1] = π[n+1]

.

Giulia Cereda () Short title October 8, 2015 24 / 33

Likelihood ratio

LR = p(π[n+2]|Hp) p(π[n+2]|Hd) = p(π[n+1], π[n+2]|Hp) p(π[n+1], π[n+2]|Hd) = lik(Hp|π[n+1], π[n+2]) lik(Hd|π[n+1], π[n+2]) Lemma allows to write LR = E(

1

• p(π[n+2] | π[n+1], A, Θ, Hp) | Π[n+1] = π[n+1])

E(p(π[n+2] | π[n+1], A, Θ, Hd)

• 1−A

n+1+Θ

| Π[n+1] = π[n+1]) = 1 E

• 1−A

n+1+Θ | Π[n+1] = π[n+1]

.

Giulia Cereda () Short title October 8, 2015 24 / 33

LR = 1 E

• 1−A

n+1+Θ | Π[n+1] = π[n+1]

• Giulia Cereda ()

Short title October 8, 2015 25 / 33

LR = 1 E

• 1−A

n+1+Θ | Π[n+1] = π[n+1]

• By defining the random variable Φ = n

1−A n+1+Θ we can write the LR as

Giulia Cereda () Short title October 8, 2015 25 / 33

LR = 1 E

• 1−A

n+1+Θ | Π[n+1] = π[n+1]

• By defining the random variable Φ = n

1−A n+1+Θ we can write the LR as

LR = n E(Φ | Π[n+1] = π[n+1]).

Giulia Cereda () Short title October 8, 2015 25 / 33

LR = 1 E

• 1−A

n+1+Θ | Π[n+1] = π[n+1]

• By defining the random variable Φ = n

1−A n+1+Θ we can write the LR as

LR = n E(Φ | Π[n+1] = π[n+1]). We are interested in the distribution of Φ, Θ|Π[n+1]

Giulia Cereda () Short title October 8, 2015 25 / 33

LR = 1 E

• 1−A

n+1+Θ | Π[n+1] = π[n+1]

• By defining the random variable Φ = n

1−A n+1+Θ we can write the LR as

LR = n E(Φ | Π[n+1] = π[n+1]). We are interested in the distribution of Φ, Θ|Π[n+1] p(φ, θ | π[n+1]) ∝ p(π[n+1] | φ, θ)f (φ, θ)

Giulia Cereda () Short title October 8, 2015 25 / 33

LR = 1 E

• 1−A

n+1+Θ | Π[n+1] = π[n+1]

• By defining the random variable Φ = n

1−A n+1+Θ we can write the LR as

LR = n E(Φ | Π[n+1] = π[n+1]). We are interested in the distribution of Φ, Θ|Π[n+1] p(φ, θ | π[n+1]) ∝ p(π[n+1] | φ, θ)f (φ, θ)

Giulia Cereda () Short title October 8, 2015 25 / 33

Log likelihood with φ and θ

log10 p(π[n+1] | φ, θ)

θ φ

− 1 −100 −80 −80 −60 − 6 −40 − 4 −20 −20

150 200 250 0.40 0.45 0.50 0.55

− 2 . 9 9 5 7 3 2 − 4 . 6 5 1 7

(φMLE, θMLE)

• Dutch Y-STR database, 7 loci, N=18,925

Giulia Cereda () Short title October 8, 2015 26 / 33

Log likelihood with φ and θ

log10 p(π[n+1] | φ, θ)

θ φ

− 1 −100 −80 −80 −60 − 6 −40 − 4 −20 −20

150 200 250 0.40 0.45 0.50 0.55

− 2 . 9 9 5 7 3 2 − 4 . 6 5 1 7

(φMLE, θMLE)

• −100

−100 −80 −80 −60 −60 −40 −40 −20 −20 −2.995732 −4.60517

logL(φ,θ|π[n+1]) logN((φMLE, θMLE), Iobs

−1 (φMLE, θMLE))

95% confidence interval 99% confidence interval

Dutch Y-STR database, 7 loci, N=18,925

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Log likelihood as a function of φ and θ

θ φ

150 200 250 0.40 0.45 0.50 0.55

(φMLE, θMLE)

• −

logL(φ,θ|π[n+1]) logN((φMLE, θMLE), Iobs

−1 (φMLE, θMLE))

95% confidence interval 99% confidence interval

p(π[n+1] | φ, θ) ≈ N((φMLE, θMLE), I −1

MLE)

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Log likelihood as a function of φ and θ

θ φ

150 200 250 0.40 0.45 0.50 0.55

(φMLE, θMLE)

• −

logL(φ,θ|π[n+1]) logN((φMLE, θMLE), Iobs

−1 (φMLE, θMLE))

95% confidence interval 99% confidence interval

p(π[n+1] | φ, θ) ≈ N((φMLE, θMLE), I −1

MLE)

p(φ, θ | π[n+1]) ∝ p(π[n+1] | φ, θ)f (φ, θ)

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Log likelihood as a function of φ and θ

θ φ

150 200 250 0.40 0.45 0.50 0.55

(φMLE, θMLE)

• −

logL(φ,θ|π[n+1]) logN((φMLE, θMLE), Iobs

−1 (φMLE, θMLE))

95% confidence interval 99% confidence interval

p(π[n+1] | φ, θ) ≈ N((φMLE, θMLE), I −1

MLE)

p(φ, θ | π[n+1]) ∝ p(π[n+1] | φ, θ)f (φ, θ) If the prior is smooth around the MLE then

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Log likelihood as a function of φ and θ

θ φ

150 200 250 0.40 0.45 0.50 0.55

(φMLE, θMLE)

• −

logL(φ,θ|π[n+1]) logN((φMLE, θMLE), Iobs

−1 (φMLE, θMLE))

95% confidence interval 99% confidence interval

p(π[n+1] | φ, θ) ≈ N((φMLE, θMLE), I −1

MLE)

p(φ, θ | π[n+1]) ∝ p(π[n+1] | φ, θ)f (φ, θ) If the prior is smooth around the MLE then p(φ, θ | π[n+1]) ≈ N((φMLE, θMLE), I −1

MLE).

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Log likelihood as a function of φ and θ

θ φ

150 200 250 0.40 0.45 0.50 0.55

(φMLE, θMLE)

• −

logL(φ,θ|π[n+1]) logN((φMLE, θMLE), Iobs

−1 (φMLE, θMLE))

95% confidence interval 99% confidence interval

p(π[n+1] | φ, θ) ≈ N((φMLE, θMLE), I −1

MLE)

p(φ, θ | π[n+1]) ∝ p(π[n+1] | φ, θ)f (φ, θ) If the prior is smooth around the MLE then p(φ, θ | π[n+1]) ≈ N((φMLE, θMLE), I −1

MLE).

It follows that E(Φ | Π[n+1] = π[n+1]) ≈ φMLE. That is

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p(φ, θ | π[n+1]) ≈ N((φMLE, θMLE), I −1

MLE).

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p(φ, θ | π[n+1]) ≈ N((φMLE, θMLE), I −1

MLE).

It follows that E(Φ | Π[n+1] = π[n+1]) ≈ φMLE.

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p(φ, θ | π[n+1]) ≈ N((φMLE, θMLE), I −1

MLE).

It follows that E(Φ | Π[n+1] = π[n+1]) ≈ φMLE. LR = n E(Φ | Π[n+1] = π[n+1]) ≈ n + 1 + θMLE 1 − αMLE

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Sorted relative frequencies: how good is our prior?

Comparison between the spectrum from a big database, and simulations from PD(α, θ) using MLE estimators of the parameters.

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Sorted relative frequencies: how good is our prior?

Comparison between the spectrum from a big database, and simulations from PD(α, θ) using MLE estimators of the parameters.

1 5 10 50 100 500 5000 5e−05 2e−04 5e−04 2e−03 5e−03 2e−02 5e−02 Rank Relative frequencies

• Database N=18925

αMLE = 0.51, θMLE = 216 asymptotic behavior

Thick black line: ranked relative frequencies in the database. Thin black lines: simulations from the PD(αMLE, θMLE). Dotted line: asymptotics.

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The LR when p is known

Imagine we know p.

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The LR when p is known

Imagine we know p. LR|p = p(πDb++

[n+2] |Hp, p)

p(πDb++

[n+2] |Hd, p)

=

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The LR when p is known

Imagine we know p. LR|p = p(πDb++

[n+2] |Hp, p)

p(πDb++

[n+2] |Hd, p)

= Applying Lemma =

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The LR when p is known

Imagine we know p. LR|p = p(πDb++

[n+2] |Hp, p)

p(πDb++

[n+2] |Hd, p)

= Applying Lemma = 1 E(pxn+1|πDb+

[n+1], p)

.

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The LR when p is known

Imagine we know p. LR|p = p(πDb++

[n+2] |Hp, p)

p(πDb++

[n+2] |Hd, p)

= Applying Lemma = 1 E(pxn+1|πDb+

[n+1], p)

. How is this compared to the one we get with our method when p is unknown?

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Test Dutch database (N=2085, 7 loci)

Database of 2085 Y-STR profiles form Dutch men.

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Test Dutch database (N=2085, 7 loci)

Database of 2085 Y-STR profiles form Dutch men. Test: Compare the distribution of log10(LR|p) and log10 LR obtained by 100 samples of size 100 from this population.

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Results

Compare the distribution of log10(LR|p) and log10 LR obtained by 100 samples of size 100 from this population

• 2.5

3.0 3.5 4.0 log10(LR) log10(LRp)

log10(LR)

(a) Comparison

−0.4 0.0 0.4 Error

(b) Error

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