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1. Let X1,…,XnX_{1}, \ldots, X_{n} be a random sample from a continuous distribution...
May 25, 2024
Solution by Steps
step 1
To find the maximum likelihood estimator (MLE) of θ\theta, we start by writing the likelihood function. Given the CDF FX(x;θ)F_X(x; \theta), the probability density function (PDF) is obtained by differentiating the CDF with respect to xx
step 2
The PDF fX(x;θ)f_X(x; \theta) is given by: fX(x;θ)=ddxFX(x;θ)=ddx(1−(3x)θ)=θ(3x)θ3x2 f_X(x; \theta) = \frac{d}{dx} F_X(x; \theta) = \frac{d}{dx} \left(1 - \left(\frac{3}{x}\right)^\theta \right) = \theta \left(\frac{3}{x}\right)^\theta \frac{3}{x^2} for x≥3x \geq 3
step 3
The likelihood function for a sample X1,X2,…,XnX_1, X_2, \ldots, X_n is: L(θ;x1,x2,…,xn)=∏i=1nfX(xi;θ)=∏i=1nθ(3xi)θ3xi2 L(\theta; x_1, x_2, \ldots, x_n) = \prod_{i=1}^n f_X(x_i; \theta) = \prod_{i=1}^n \theta \left(\frac{3}{x_i}\right)^\theta \frac{3}{x_i^2} for xi≥3x_i \geq 3
step 4
Taking the natural logarithm of the likelihood function, we get the log-likelihood function: ℓ(θ)=ln⁡L(θ;x1,x2,…,xn)=nln⁡θ+θ∑i=1nln⁡(3xi)−2∑i=1nln⁡xi \ell(\theta) = \ln L(\theta; x_1, x_2, \ldots, x_n) = n \ln \theta + \theta \sum_{i=1}^n \ln \left(\frac{3}{x_i}\right) - 2 \sum_{i=1}^n \ln x_i
step 5
To find the MLE, we take the derivative of the log-likelihood function with respect to θ\theta and set it to zero: dℓ(θ)dθ=nθ+∑i=1nln⁡(3xi)=0 \frac{d\ell(\theta)}{d\theta} = \frac{n}{\theta} + \sum_{i=1}^n \ln \left(\frac{3}{x_i}\right) = 0
step 6
Solving for θ\theta, we get: θ^ML=−n∑i=1nln⁡(3xi) \hat{\theta}_{ML} = -\frac{n}{\sum_{i=1}^n \ln \left(\frac{3}{x_i}\right)}
step 7
To show that θ^ML\hat{\theta}_{ML} maximizes the likelihood function, we check the second derivative: d2ℓ(θ)dθ2=−nθ2 \frac{d^2\ell(\theta)}{d\theta^2} = -\frac{n}{\theta^2} Since \frac{d^2\ell(\theta)}{d\theta^2} < 0, θ^ML\hat{\theta}_{ML} is a maximum
Answer
The maximum likelihood estimator (MLE) of θ\theta is θ^ML=−n∑i=1nln⁡(3xi)\hat{\theta}_{ML} = -\frac{n}{\sum_{i=1}^n \ln \left(\frac{3}{x_i}\right)}.
Key Concept
Maximum Likelihood Estimation (MLE)
Explanation
The MLE is found by maximizing the likelihood function, which involves taking the derivative of the log-likelihood function, setting it to zero, and solving for the parameter.
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Solution by Steps
step 1
To find the Cramer-Rao lower bound (CRLB) for the variance of an unbiased estimator of θ\theta, we need the Fisher information I(θ)I(\theta)
step 2
The Fisher information is given by: I(θ)=−E[d2ℓ(θ)dθ2] I(\theta) = -E\left[\frac{d^2 \ell(\theta)}{d\theta^2}\right]
step 3
From the previous steps, we have: d2ℓ(θ)dθ2=−nθ2 \frac{d^2 \ell(\theta)}{d\theta^2} = -\frac{n}{\theta^2}
step 4
Therefore, the Fisher information is: I(θ)=−E[−nθ2]=nθ2 I(\theta) = -E\left[-\frac{n}{\theta^2}\right] = \frac{n}{\theta^2}
step 5
The Cramer-Rao lower bound for the variance of an unbiased estimator of θ\theta is: Var(θ^)≥1I(θ)=θ2n \text{Var}(\hat{\theta}) \geq \frac{1}{I(\theta)} = \frac{\theta^2}{n}
Answer
The Cramer-Rao lower bound for the variance of an unbiased estimator of θ\theta is θ2n\frac{\theta^2}{n}.
Key Concept
Cramer-Rao Lower Bound (CRLB)
Explanation
The CRLB provides a lower bound on the variance of unbiased estimators, indicating the best possible precision of an estimator.
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Solution by Steps
step 1
Given n=100n=100 and θ^=2.5\widehat{\theta}=2.5, we use the asymptotic normality of the MLE to construct a 95%95\% confidence interval for θ\theta
step 2
The asymptotic distribution of θ^ML\widehat{\theta}_{ML} is: θ^ML∼N(θ,θ2n) \widehat{\theta}_{ML} \sim N\left(\theta, \frac{\theta^2}{n}\right)
step 3
For a 95%95\% confidence interval, we use the critical value z0.025≈1.96z_{0.025} \approx 1.96
step 4
The confidence interval is given by: θ^±z0.025θ^2n \widehat{\theta} \pm z_{0.025} \sqrt{\frac{\widehat{\theta}^2}{n}}
step 5
Substituting the values, we get: 2.5±1.962.52100=2.5±1.96⋅0.25=2.5±0.49 2.5 \pm 1.96 \sqrt{\frac{2.5^2}{100}} = 2.5 \pm 1.96 \cdot 0.25 = 2.5 \pm 0.49
step 6
Therefore, the 95%95\% confidence interval for θ\theta is: (2.01,2.99) (2.01, 2.99)
Answer
The 95%95\% confidence interval for θ\theta is (2.01,2.99)(2.01, 2.99).
Key Concept
Confidence Interval
Explanation
A confidence interval provides a range of values within which the true parameter is expected to lie with a certain probability.
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Solution by Steps
step 1
Given the exponential prior distribution for θ\theta with PDF: f_{\Theta}(\theta) = e^{-\theta}, \quad \theta > 0 we find the posterior distribution using the observed data
step 2
The likelihood function is: L(θ;x1,x2,…,xn)=∏i=1nθ(3xi)θ3xi2 L(\theta; x_1, x_2, \ldots, x_n) = \prod_{i=1}^n \theta \left(\frac{3}{x_i}\right)^\theta \frac{3}{x_i^2}
step 3
The posterior distribution is proportional to the product of the prior and the likelihood: fΘ∣X(θ∣x)∝e−θ⋅θn(3x1)θ(3x2)θ⋯(3xn)θ f_{\Theta|\mathbf{X}}(\theta|\mathbf{x}) \propto e^{-\theta} \cdot \theta^n \left(\frac{3}{x_1}\right)^\theta \left(\frac{3}{x_2}\right)^\theta \cdots \left(\frac{3}{x_n}\right)^\theta
step 4
Simplifying, we get: fΘ∣X(θ∣x)∝θne−θ(3x1)θ(3x2)θ⋯(3xn)θ f_{\Theta|\mathbf{X}}(\theta|\mathbf{x}) \propto \theta^n e^{-\theta} \left(\frac{3}{x_1}\right)^\theta \left(\frac{3}{x_2}\right)^\theta \cdots \left(\frac{3}{x_n}\right)^\theta
step 5
The posterior mean of θ\theta is the expected value of the posterior distribution. Given the complexity, we use numerical methods or approximations to find the posterior mean
step 6
Comparing the posterior mean to the MLE θ^=2.5\widehat{\theta}=2.5, we check if they are close
Answer
The posterior distribution of θ\theta is proportional to θne−θ(3x1)θ(3x2)θ⋯(3xn)θ\theta^n e^{-\theta} \left(\frac{3}{x_1}\right)^\theta \left(\frac{3}{x_2}\right)^\theta \cdots \left(\frac{3}{x_n}\right)^\theta. The posterior mean can be approximated numerically and compared to the MLE θ^=2.5\widehat{\theta}=2.5.
Key Concept
Bayesian Inference
Explanation
Bayesian inference combines prior information with observed data to update the probability distribution of a parameter.
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Solution by Steps
step 1
To test the null hypothesis H0:θ=2H_0: \theta=2 against the alternative H_a: \theta>2, we use the asymptotic normality of θ^ML\widehat{\theta}_{ML}
step 2
The test statistic is: Z=θ^−θ0θ^2n Z = \frac{\widehat{\theta} - \theta_0}{\sqrt{\frac{\widehat{\theta}^2}{n}}} where θ0=2\theta_0 = 2
step 3
Substituting the values, we get: Z=2.5−22.52100=0.50.25=2 Z = \frac{2.5 - 2}{\sqrt{\frac{2.5^2}{100}}} = \frac{0.5}{0.25} = 2
step 4
The p-value is the probability that ZZ is greater than or equal to the observed value under the null hypothesis
step 5
Using the standard normal distribution, the p-value is: P(Z≥2)=1−Φ(2)≈1−0.9772=0.0228 P(Z \geq 2) = 1 - \Phi(2) \approx 1 - 0.9772 = 0.0228
step 6
Since the p-value 0.02280.0228 is greater than the 1%1\% significance level, we do not reject H0H_0
Answer
The p-value is 0.02280.0228. We do not reject H0H_0 at the 1%1\% significance level.
Key Concept
Hypothesis Testing
Explanation
Hypothesis testing involves comparing a test statistic to a critical value or p-value to decide whether to reject the null hypothesis.
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