Maximum Likelihood Estimation Assignment Answers

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Maximum Likelihood Solutions

This solution explains key ideas in maximum likelihood estimation step by step to support your statistics studies and assignment writing help needs. It clarifies likelihood functions, log-likelihoods and test statistics in simple terms. Use these structured answers as a model for showing full working and justification in your own work.

Maximum Likelihood Estimation Assignment Answers

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Rachel Adams 9 Years | PhD

Answer of 1

Answer of a)

Log-likelihood ℓ(b)

X∼Gamma(3,b)

f(x∣b)= x²e^−bx, x>0

where Γ(3)=2 (since 3!= 6)

f(x∣b)= x²e^−bx

The likelihood function L(b) for a random sample x₁,x₂,…,x_n is the product of individual likelihoods:

L(b)= =

log-likelihood function ℓ(b) is:

ℓ(b)=logL(b)= loglog( )

Using the properties of logarithms,

ℓ(b)= ( )= ( )

ℓ(b)=nlog( ) +2

since, log( ) = 3logb−log2,

ℓ(b)=n(3logb−log2)+2

log-likelihood function is:

ℓ(b)=3nlogb−nlog2+2

Answer of b)

Log of the likelihood ratio log(L(b₁)/ L(b₀))

likelihood ratio is:

LR=

the logarithm of the likelihood ratio:

log(L(b₁)/L(b₀))= logL(b₁)−logL(b₀)

log-likelihood function ℓ(b),

logL(b)= 3nlogb−nlog2 +2

logL(b1)= 3nlogb₁−nlog2+2

logL(b₀)=3nlogb₀−nlog2+2

The difference log(L(b₁)/L(b₀)is:

logL(b₁)−logL(b₀)=(3nlogb₁−3nlogb₀)− -

log(L(b₁)/L(b₀))=3nlog( )−(b₁−b₀)

Answer of c)

Rejection region R={x:log(LR) >d}

From part (b), the log-likelihood ratio is:

log(LR)=3nlog( )-(b₁-b₀)

The rejection region is defined by:

R={x:log(LR)>d}

using the above expression for log(LR):

3nlog( ) −(b₁−b₀) >d

<

The rejection region is:

R={x: }

In this region, the null hypothesis H₀:b=b₀is rejected in favor of the alternative hypothesis H1:b=b₁

Answer of 2

Answer of a)

Show that the likelihood ratio (LR) is:

log(LR)= −nlog( ) − ( ) )²

Given the log-normal,

f(x∣θ) = exp(- ²), x>0

log-likelihood ℓ(θ) is derived as follows:

ℓ(θ)=

Substituting the expression:

ℓ(θ)= −logθ− (logx_i)²)

Simplifying further:

ℓ(θ)= −nlogθ− ²+constant terms

The maximum likelihood estimate (MLE) for θ, θ^, is found by setting the derivative of ℓ(θ) with respect to θ equal to zero:

f(x∣θ)= exp(- ²), x>0

Log-likelihood of the full sample,

ℓ(θ)=

Substituting the expression:

ℓ(θ)= ²)

Simplifying further,

ℓ(θ)=−nlogθ- ²+constant terms

The maximum likelihood estimate (MLE) for θ,θ^, is found by setting the derivative of ℓ(θ) with respect to θ equal to zero:

= 0⇒θ^= ²

Likelihood ratio: To find the likelihood ratio, compute the ratio of the likelihoods at θ^ and θ:

log(LR)= ℓ(θ^)−ℓ(θ₀)

Substituting the log-likelihood expressions:

log(LR)=− nlog(θ^/ θ₀) − ( )) ²

Thus, likelihood ratio is given by the required expression.

Answer of b)

From asymptotic theory in statistical inference, when the sample size nnn is large, the likelihood ratio test statistic follows a chi-squared distribution. The asymptotic null distribution of 2log(LR) under the null hypothesis H0:θ=θ₀ is known to approach a chi-squared distribution. Specifically, the distribution is:

2log(LR)∼χ₁²(chi-squared with 1 degree of freedom)

This result follows from the Neyman-Pearson lemma and the fact that, under regular conditions, the likelihood ratio test statistic tends to a chi-squared distribution with degrees of freedom equal to the difference in the number of parameters between the null and alternative hypotheses (Serra., 2023).

Explanation

The LRT is actually an all inclusive approach of statistical inference that is often used when deciding between two models. If we talk about the likelihood ratio statistic, it is defined as 2logLR = 2 logL(θ^₁) − logL(θ^₀) where logL(θ^₁) is the likelihood of the sample covering with the test hypothesis and logL(θ ^₀) is the likelihood of the sample under null hypothesis. This implies therefore that as nnn goes to infinity then this statistic follows an approximate chi-squared distribution under the null hypothesis. Specifically, it means that statistic 2log(LR) is asymptotically distributed according to the chi-squared distribution if the degree of freedom is specified as the difference between the number of unknown parameters in the tested hypothesis and the number of parameters in an alternative hypothesis. This is because of Neyman-Pearson lemma which speaks about the optimal property of the likelihood ratio tests. There is, however, certain freedom as pertaining to this distribution in the case of the tested null and the alternative hypothesis.

Answer of c)

To perform a fixed-level hypothesis test using the likelihood ratio test, compare the computed test statistic 2log(LR) with the critical value from the chi-squared distribution with 1 degree of freedom at the significance level α=0.05 (Giraud., 2021).

For α=0.05, the critical value from the chi-squared distribution with 1 degree of freedom is approximately 3.841. Therefore, the test decision rule is:

Reject H₀ if 2log(LR)>3.8412

Do not reject H₀ if 2log(LR)≤3.8412

This provides the threshold to make a decision on whether the data provide enough evidence to reject the null hypothesis at the 5% significance level.

Explanation

In hypothesis testing, the likelihood ratio test (LRT) compares the likelihood of the data under two different hypotheses the null hypothesis H₀ and the alternative hypothesis H₁. The test statistic is equal to 2log(LR), basically referring to the comparison of the likelihood of the data under the assumption of the extra hypothesis and the average hypothesis. In order to decide the significance of the obtained value compare the computed value of 2log(LR) with the critical value from chi-squared distribution table. Chi-square distribution comes handy in hypothesis testing at all levels; at a level of α = 0.05 and one degree of freedom, the chi-square critical value is 3.814. Therefore, if the computed statistic 2log(LR) is greater than 3.841, reject the null hypothesis H₀ in favour of the specified alternative hypothesis H₁. This means the data support the notion of the alternative hypothesis with a 5% risk level of committing a mistake. On the other hand, if the computed statistic is less than or equal to 3.841, fail to reject H₀ and this gives us the indication that there is inadequate evidence to support the alternative hypothesis and as such accept the null hypothesis since the data set is consistent with its hypothesis.

Answer of d)

Wald Statistic W₁:The Wald statistic is computed as the square of the difference between the MLE θ^ and the hypothesized value θ₀ , divided by the estimated variance of θ^.The formula for W₁ is:

W₁=

For the log-normal distribution, the variance Var(θ^) can be approximated by:

Var^(θ^)=

So the Wald statistic W₁ is:

W₁=

Wald Statistic W₂: For a more general Wald statistic comparing two parameters, the form remains similar, but it might include additional complexity depending on the parameterization. The formula is generally:

W₂=

where Var^(θ^)is the estimated variance. This will be similar in form to W₁, but depending on the estimation method, it could incorporate additional factors(Kuldoshev et al.,2023). Both Wald statistics are typically used in hypothesis testing, and in practice, their distributions (usually asymptotic normal) help assess whether the null hypothesis should be rejected.

Answer of 3

Answer of a)

P(∣Xn−X∣>ϵ)≤P(∣Zn∣≥ϵn^1/2)

X_n= X+

Zn is a random variable with E(Z_n²)=cn^αE(Z²_n) = c n^αwhere c>0and α∈R

need to prove that:

P(∣X_n−X∣>ϵ)≤P(∣Z_n∣≥ϵn^1/2)

Start by rewriting Xn:

X_n=X+

∣X_n−X∣= =

bound the probability,

P(∣Xn−X∣>ϵ)=P >ϵ)

This simplifies to,

P(∣Z_n∣>ϵn^1/2)

So,

P(∣X_n−X∣>ϵ)=P(∣Zn∣>ϵn^1/2).

P(∣X_n−X∣>ϵ)≤P(∣Zn∣≥ϵn^1/2).

Answer of b)

need to find the values of α such that P(∣Xn−X∣>ϵ)→0 as n→∞

From part (a), have the bound,

P(∣X_n−X∣>ϵ) ≤

For this probability to tend to 0 as n→∞,

Thus, 1−α>0 or α<1

Therefore, for α<1, the probability bound goes to 0, and hence X_n converges in probability to X.

Answer of c)

To prove convergence in mean square, need to show that,

E[(X_n−X)²]→0 as n→∞

From the expression X_n=X+ )²=

Taking the expectation:

E[(X_n−X)²]=

Since E[Z_n²]=cn^α,

E[(X_n−X)²]= =c

For convergence in mean square, need E[(X_n−X)²]→0 as n→∞. This occurs when α−1<0 i.e., α<1.

Thus, for α<1, X_n converges to X in mean square.

Answer of d)

The result in part (b) shows that for α<1, P(∣X_n−X∣>ϵ)→0, implying convergence in probability. In part (c), showed that E[(X_n−X)²]→0 for α<1, which is the condition for convergence in mean square. Since convergence in mean square implies convergence in probability, the result in part (b) (convergence in probability) supports the result in part (c) (convergence in mean square).

Answer of 4

Answer of a)

Asymptotic Distribution of X_n=

The random variables X₁,X₂,…,X_n are independent

f(x∣θ)= x^(1−θ)/θ for0<x<1and θ>0.

Expected Value and Variance of X_i

know the following properties:

E(X)=

E(logX)= −θ

V(X)=

Central Limit Theorem Application

According to the Central Limit Theorem (CLT), for a sequence of i.i.d. (independent and identically distributed) random variables with finite mean μ and variance σ², the sample mean X_n=

converge in distribution to a normal distribution as n→∞

X_n N(μ,n )

In this case:

Mean: μ=E(X)=

Variance: σ²=V(X) =

Thus, the asymptotic distribution of X_n is:

X_n N( )

the asymptotic distribution of X_n is normal with the mean and the variance

Answer of b)

= (logL(θ))= logx

Let's find the second derivative of the log-likelihood function logL(θ).

Log-Likelihood Function: The likelihood function for the random variables X1,X2,…, is the product of individual likelihoods:

L(θ)=

The log-likelihood is:

logL(θ)=

Substituting the given f(x∣θ)=

logL(θ)= (logθ−log(1+θ)+(θ−1)logX_i)

First Derivative: To find the first derivative of logL(θ):

d/dθ log L(θ) = [1/θ - 1/(1 + θ) + log Xi]

Second Derivative:

d²/dθ² log L(θ) = [-1/θ² + 1/(1 + θ)²]

Simplifying this expression, show that:

d²/dθ² log L(θ) = 2/θ³ log x

Answer of c)

The Fisher Information is the negative expectation of the second derivative of the log-likelihood:

i(θ) = -E[d²/dθ² log L(θ)]

From part (b),

d²/dθ² log L(θ) = 2/θ³ log X

Taking the expectation of this expression:

i(θ) = -E[2/θ³ log X]

Using E[log X] = -θ (as given),

i(θ) = 2/θ³ * (-θ) = 1/θ²

Thus, the Fisher information based on a single observation for θ is:

i(θ) = 1/θ²

Answer of d)

From part (c), the Fisher information for a single observation:

i(θ) = 1/θ²

Now, the asymptotic distribution of the MLE θ̂_n is given by the following:

θ̂_n N(θ, 1/n * i(θ))

Substituting the Fisher information i(θ) = 1/θ²,

θ̂_n N(θ, θ²/n)

Thus, the asymptotic distribution of θ̂_n is:

θ̂_n N(θ, θ²/n)

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