Method of moments and Maximum likelihood...

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Method of moments Maximum likelihood Asymptotic normality Optimality Delta method Parametric bootstrap Quiz Method of moments and Maximum likelihood estimation Botond Szabo Leiden University Leiden, 11. 03. 2019

Transcript of Method of moments and Maximum likelihood...

Page 1: Method of moments and Maximum likelihood estimationpub.math.leidenuniv.nl/~szabobt/STAN/STAN9.pdf · 2019. 3. 8. · Method of moments Maximum likelihood Asymptotic normality Optimality

Method of moments Maximum likelihood Asymptotic normality Optimality Delta method Parametric bootstrap Quiz

Method of moments and Maximum likelihoodestimation

Botond Szabo

Leiden University

Leiden, 11. 03. 2019

Page 2: Method of moments and Maximum likelihood estimationpub.math.leidenuniv.nl/~szabobt/STAN/STAN9.pdf · 2019. 3. 8. · Method of moments Maximum likelihood Asymptotic normality Optimality

Method of moments Maximum likelihood Asymptotic normality Optimality Delta method Parametric bootstrap Quiz

Outline

1 Method of moments

2 Maximum likelihood

3 Asymptotic normality

4 Optimality

5 Delta method

6 Parametric bootstrap

7 Quiz

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Method of moments

Recall: the jth moment of the random variable X ∼ fθ is

αj(θ) := EθXj =

∫x j f (x ; θ)dx .

The jth sample moment is defined as

αj := n−1n∑

i=1

X ji .

For θ = (θ1, ...θk) the method of moments estimator θn isdefined as

α1(θ) = α1,

....

αk(θ) = αk ,

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Examples I

Example

Let X1, ...,Xn ∼ Bernoulli(p). Then α1(p) = EpX1 = p andα1 = n−1

∑ni=1 Xi . By equating these we get

pn = n−1n∑

i=1

Xi .

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Examples II

Example

Let X1, ...,Xn ∼ N(µ, σ2). Then α1 = EX1 = µ andα2 = EX 2

1 = µ2 + σ2. Furthermore α1 = n−1∑n

i=1 Xi andα2 = n−1

∑ni=1 X

2i . Solving the equations

µn = n−1n∑

i=1

Xi , and µ2 + σ2 = n−1n∑

i=1

X 2i ,

we get that

µn = n−1n∑

i=1

Xi , and σ2 = n−1n∑

i=1

(Xi − Xn)2.

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Properties

Theorem

Let θn denote the method of moments estimator. Underappropriate conditions on the model, the following statementshold:

The estimate θn exists with probability tending to one.

The estimate is consistent , i.e. θnP→ θ.

The estimate is asymptotically normal (the precise form isgiven in the book).

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Likelihood

Definition

Let X1, . . . ,Xn be IID with PDF (or PMF) f (x ; θ). The likelihoodfunction is defined by

Ln(θ) =n∏

i=1

f (Xi ; θ).

The log-likelihood function is defined by `n(θ) = logLn(θ).

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Maximum likelihhod estimator

Definition

The maximum likelihood estimator (MLE), denoted by θn, is thevalue of θ that maximises Ln(θ) (or, equivalently, logLn(θ)).

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Example

Example

Suppose that X1, . . . ,Xn ∼ Bernoulli(p). The PMF isf (x ; p) = px(1− p)1−x for x = 0, 1. The unknown parameter is p.Then

Ln(θ) =n∏

i=1

f (Xi ; θ) =n∏

i=1

pXi (1− p)1−Xi = pS(1− p)n−S ,

where S =∑n

i=1 Xi . Therefore,

`n(θ) = S log p + (n − S) log(1− p).

To find the maximum, take a derivative of `n(θ), set it to zero andfind that pn = S/n.

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Example

Likelihood for Bernoulli with n = 20 and∑n

i=1 Xi = 12.

The MLE is pn = 12/20 = 0.6.

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Example

Likelihood for Bernoulli with n = 20 and∑n

i=1 Xi = 12.The MLE is pn = 12/20 = 0.6.

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Example

Example

Let X1, . . . ,Xn ∼ N(µ, σ2). The unknown parameter is θ = (µ, σ2)and the likelihood (up to some constants not depending on θ) is

Ln(θ) =n∏

i=1

1

σexp

(− 1

2σ2(Xi − µ)2

)

= σ−n exp

(− 1

2σ2

n∑i=1

(Xi − µ)2

)

= σ−n exp

(−nS2

2σ2

)exp

(−n(X n − µ)2

2σ2

),

where X n = n−1∑n

i=1 Xi and S2 = n−1∑n

i=1(Xi − X n)2.

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Example (continued)

Example

The log-likelihood is

`(µ, σ) = −n log σ − nS2

2σ2− n(X n − µ)2

2σ2.

Solving the equations

∂`(µ, σ)

∂µ= 0,

∂`(µ, σ)

∂σ= 0,

we conclude that µn = X n and σn = S (it can be verified thatthese are the global maxima of the likelihood).

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Properties of maximum likelihood estimators

Theorem

Under suitable conditions on f (x ; θ), the MLE θn is consistent:

θnP−→ θ.

Theorem

Let τ = g(θ) be a function of θ. Let θn be an MLE of θ. Thenτn = g(θn) is an MLE of τ.

Example

Let X1, . . . ,Xn ∼ N(θ, 1). The MLE for θ is θn = X n. Let τ = eθ.

Then the MLE for τ is τn = eX n .

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Properties of maximum likelihood estimators

Theorem

Under suitable conditions on f (x ; θ), the MLE θn is consistent:

θnP−→ θ.

Theorem

Let τ = g(θ) be a function of θ. Let θn be an MLE of θ. Thenτn = g(θn) is an MLE of τ.

Example

Let X1, . . . ,Xn ∼ N(θ, 1). The MLE for θ is θn = X n. Let τ = eθ.

Then the MLE for τ is τn = eX n .

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Properties of maximum likelihood estimators

Theorem

Under suitable conditions on f (x ; θ), the MLE θn is consistent:

θnP−→ θ.

Theorem

Let τ = g(θ) be a function of θ. Let θn be an MLE of θ. Thenτn = g(θn) is an MLE of τ.

Example

Let X1, . . . ,Xn ∼ N(θ, 1). The MLE for θ is θn = X n. Let τ = eθ.

Then the MLE for τ is τn = eX n .

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Score function and Fisher information

Definition

The score function is defined by

s(x ; θ) =∂ log f (x ; θ)

∂θ.

The Fisher information in n IID observations X1, . . . ,Xn ∼ f (x ; θ)is

In(θ) = Vθ

(n∑

i=1

s(Xi ; θ)

)

=n∑

i=1

Vθ(s(Xi ; θ))

= nVθ(s(Xi ; θ)).

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Properties

Theorem

For n = 1 write I (θ) instead of I1(θ). One has Eθ[s(X1; θ)] = 0 andhence Vθ(s(Xi ; θ)) = E[s2(X1; θ)].

Theorem

One has In(θ) = nI (θ). Also,

I (θ) = −Eθ[∂2 log f (X ; θ)

∂θ2

]= −

∫ (∂2 log f (x ; θ)

∂θ2

)f (x ; θ)dx .

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Properties

Theorem

For n = 1 write I (θ) instead of I1(θ). One has Eθ[s(X1; θ)] = 0 andhence Vθ(s(Xi ; θ)) = E[s2(X1; θ)].

Theorem

One has In(θ) = nI (θ). Also,

I (θ) = −Eθ[∂2 log f (X ; θ)

∂θ2

]= −

∫ (∂2 log f (x ; θ)

∂θ2

)f (x ; θ)dx .

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Asymptotic normality

Theorem

Let θn be an MLE and se =√Vθ[θn]. Under appropriate

conditions,

1 se ≈√

1/In(θ) and

θn − θse

N(0, 1).

2 Let se =√

1/In(θn). Then

θn − θse

N(0, 1).

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Confidence interval

Theorem

LetCn = (θn − zα/2se, θn + zα/2se).

Then Pθ(θ ∈ Cn)→ 1− α as n→∞.

In particular, for α = 0.05, zα/2 = 1.96 ≈ 2, and

θn ± 2se

is an approximate 95% confidence interval.

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Example

Example

Let X1, . . . ,Xn be IID Poisson(λ). Then it can be shown thatλn = X n and I (λ) = 1/λ, so that

se =1√

nI (λn)=

√λnn.

Therefore, an approximate 1− α confidence interval for λ is

λn ± zα/2

√λnn.

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MLE vs. median

Example

Let X1, . . . ,Xn be IID N(θ, 1). The MLE for θ is θn = X n. Anotherreasonable estimator for θ is the sample median θn.

The MLE satisfies

√n(θn − θ) N(0, 1).

The sample median can be shown to satisfy

√n(θn − θ) N

(0,π

2

).

The sample median converges to the right value, but has a largervariance than MLE.

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Efficiency of MLE

Theorem

Let θn be an MLE and θn (almost) any other estimator. Suppose

√n(θn − θ) N(0, σ2MLE),

√n(θn − θ) N(0, σ2tilde).

Define the asymptotic relative efficiency as

ARE(θn, θn) =σ2MLE

σ2tilde.

Then ARE(θn, θn) ≤ 1. Thus the MLE has the smallest(asymptotic) variance and we say that the MLE is optimal orasymptotically efficient.

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Delta method

Theorem

If τ = g(θ), where g is differentiable with g ′(θ) 6= 0, then forτn = g(θn),

τn − τse(τn)

N(0, 1).

Herese(τn) = |g ′(θn)|se(θn).

Hence, if

Cn = (τn − zα/2se(τn), τn + zα/2se(τn)),

then Pθ(τ ∈ Cn)→ 1− α as n→∞.

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Example

Example

Let X1, . . . ,Xn ∼ N(0, σ2). Suppose we want to estimateτ = log σ. The log-likelihood is

`(σ) = −n log σ − 1

2σ2

n∑i=1

X 2i .

Differentiate `(σ) and set the derivative to zero to conclude that

σn =

√∑ni=1 X

2i

n.

With some effort we can show that I (σ) = 2/σ2. Therefore,se(σn) = σn/

√2n.

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Example (continued)

Example

Now τn = log σn. Since (log σ)′ = 1/σ, we get that

se(τn) =1

σn

σn√2n

=1√2n,

and an approximate 95% confidence interval for τ is

τn ±√

2

n.

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Parametric bootstarp

The bootstrap is a method for estimating standard errors,computing confidence intervals and other quantities thatmight be difficult to compute otherwise.

In the parametric models we know that X1, . . . ,Xn ∼ f (x ; θ).

Suppose we want to approximate the standard error ofT (X1, . . . ,Xn), denoted by se =

√Vθ[T ].

The idea of the parametric bootstrap is to approximate se

with√

Vθn [T ], where θn is e.g. the MLE. However, in many

cases it is difficult to compute Vθn [T ]. Idea: approximate itvia simulation.

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Parametric bootstarp: sampling

Sample X ∗1 , ...,X∗n from f (x ; θn).

Compute the estimator T using the “new sample” X ∗1 , ...,X∗n :

T ∗ = T (X ∗1 , ...,X∗n ).

Repeat this procedure B times resulting T ∗1 , ...,T∗B .

Denote the average of the outomes by T ∗.

Approximate Vθn [T ] by

se2boot = B−1B∑

b=1

(T ∗b − T ∗

)2.

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Example

Example

Let X1, . . . ,Xn ∼ N(0, σ2). The MLE of σ is

σn =

√∑ni=1 X

2i

n.

We use the parametric bootstrap to approximate se.

Simulate X ∗1 , . . . ,X∗n ∼ N(0, σ2n) and compute

σ∗ =√n−1

∑ni=1 X

∗2i . Next repeat this process B times to get

σ∗1, . . . , σ∗B . Finally, set

seboot =

√√√√√ 1

B

B∑b=1

σ∗b − 1

B

B∑j=1

σ∗j

2

.

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Question 1

In what sense is the MLE optimal?

Answers:

1 It has the smallest asymptotic variance.

2 It is unbiased.

3 It is a normally distributed random variable and therefore easyto work with.

4 It is the value minimizing the likelihood function.

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Question 1

In what sense is the MLE optimal?

Answers:

1 It has the smallest asymptotic variance.

2 It is unbiased.

3 It is a normally distributed random variable and therefore easyto work with.

4 It is the value minimizing the likelihood function.

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Question 2

What is the Delta method good for

Answers:

1 Compute the MLE for a function of the paramter τ = g(θ).

2 Construct a confidence set for a function of the paramterτ = g(θ).

3 Compute the Fisher information.

4 Compute the relative efficiency.

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Method of moments Maximum likelihood Asymptotic normality Optimality Delta method Parametric bootstrap Quiz

Question 2

What is the Delta method good for

Answers:

1 Compute the MLE for a function of the paramter τ = g(θ).

2 Construct a confidence set for a function of the paramterτ = g(θ).

3 Compute the Fisher information.

4 Compute the relative efficiency.

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Question 3

If τ = g(θ) and τn = g(θn), then what is the estimated standarderror of τn

Answers:

1 se(τn) = se(θn).

2 se(τn) = se(θn)/|g ′(θn)|.3 se(τn) = |g(θn)|se(θn).

4 se(τn) = |g ′(θn)|se(θn).

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Question 3

If τ = g(θ) and τn = g(θn), then what is the estimated standarderror of τn.

Answers:

1 se(τn) = se(θn).

2 se(τn) = se(θn)/|g ′(θn)|.3 se(τn) = |g(θn)|se(θn).

4 se(τn) = |g ′(θn)|se(θn).

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Question 4

What is the bootstrap method good for?

Answers:

1 To estimate standard error and compute confidence intervals.

2 To help put on boots.

3 To compute the MLE.

4 Compute the Fisher information.

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Question 4

What is the bootstrap method good for?

Answers:

1 To estimate standard error and compute confidence intervals.

2 To help put on boots.

3 To compute the MLE.

4 Compute the Fisher information.

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Question 5

What does the star denote in X ∗1 , ...,X∗n in the bootstrap method?

Answers:

1 They are a permutation of X1, ...,Xn.

2 They are a “new” sample (with replacement) from thenumbers X1, ...,Xn.

3 They are a “new” sample (without replacement) from thenumbers X1, ...,Xn.

4 They are new draws from the distribution F .

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Question 5

What does the star denote in X ∗1 , ...,X∗n in the bootstrap method?

Answers:

1 They are a permutation of X1, ...,Xn.

2 They are a “new” sample (with replacement) from thenumbers X1, ...,Xn.

3 They are a “new” sample (without replacement) from thenumbers X1, ...,Xn.

4 They are new draws from the distribution F .

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Question 6

How do we estimate the variance using the bootstrap sample?

Answers:

1 B−1∑B

b=1

(T ∗b − T ∗

)2.

2 B−1∑B

b=1

(Xb − T

)2.

3 B−1∑B

b=1

(T ∗b − T

)2.

4 B−1∑B

b=1

(Xb − T ∗

)2.

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Question 6

How do we estimate the variance using the bootstrap sampleseboot?

Answers:

1 B−1∑B

b=1

(T ∗b − T ∗

)2.

2 B−1∑B

b=1

(Xb − T

)2.

3 B−1∑B

b=1

(T ∗b − T

)2.

4 B−1∑B

b=1

(Xb − T ∗

)2.