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1. Let YY be a uniform random variable with support [0,1][0,1]. Let $I_{1}, I_{2}...
Oct 9, 2024
Solution by Steps
step 1
To determine whether Xn X_n converges in probability, we need to find P(Xn=1) P(X_n = 1) and analyze its limit as n→∞ n \to \infty . For odd n n , P(Xn=1)=P(Y∈In)=P(Y∈[0,1n+1])=1n+1 P(X_n = 1) = P(Y \in I_n) = P\left(Y \in \left[0, \frac{1}{n+1}\right]\right) = \frac{1}{n+1} . For even n n , P(Xn=1)=P(Y∈[1n,2n])=1n P(X_n = 1) = P\left(Y \in \left[\frac{1}{n}, \frac{2}{n}\right]\right) = \frac{1}{n} . Thus, we have: P(Xn=1)={1n+1amp;if n is odd1namp;if n is even P(X_n = 1) = \begin{cases} \frac{1}{n+1} & \text{if } n \text{ is odd} \\ \frac{1}{n} & \text{if } n \text{ is even} \end{cases}
step 2
As n→∞ n \to \infty , both P(Xn=1) P(X_n = 1) for odd and even n n converge to 0. Therefore, we conclude that Xn X_n converges in probability to 0
step 3
To determine whether Xn X_n converges almost surely, we need to check if P(Xn=1 infinitely often)=0 P(X_n = 1 \text{ infinitely often}) = 0 . The intervals In I_n shrink as n n increases, and the probability of Y Y falling into these intervals becomes negligible. Specifically, the series ∑n=1∞P(Xn=1) \sum_{n=1}^{\infty} P(X_n = 1) diverges, indicating that P(Xn=1 infinitely often)=0 P(X_n = 1 \text{ infinitely often}) = 0 . Thus, Xn X_n converges almost surely to 0
Answer
(a) Xn X_n converges in probability to 0. (b) Xn X_n converges almost surely to 0.
Key Concept
Convergence in probability and almost sure convergence of random variables.
Explanation
Both types of convergence indicate that the sequence of random variables Xn X_n approaches 0 as n n increases, but they do so under different conditions.
Solution by Steps
step 1
For part (a), we start with the probability density function of the normal distribution N(θ,θ) N(\theta, \theta) . The likelihood function for a random sample X1,X2,…,Xn X_1, X_2, \ldots, X_n is given by: $$ L(\theta) = \prod_{i=1}^{n} \frac{1}{\sqrt{2\pi \theta}} \exp\left(-\frac{(X_i - \theta)^2}{2\theta}\right)
step 2
Applying the Fisher-Neyman factorization theorem, we can factor the likelihood function into two parts: one that depends on the data only through a statistic T(X) T(X) and another that depends only on θ \theta . We find that T(X)=∑i=1nXi T(X) = \sum_{i=1}^{n} X_i is a sufficient statistic for θ \theta
step 3
For part (b), we consider the likelihood function for the normal distribution N(θ,θ3) N(\theta, \theta^3) : $$ L(\theta) = \prod_{i=1}^{n} \frac{1}{\sqrt{2\pi \theta^3}} \exp\left(-\frac{(X_i - \theta)^2}{2\theta^3}\right)
step 4
Again, using the Fisher-Neyman factorization theorem, we find that T(X)=∑i=1nXi T(X) = \sum_{i=1}^{n} X_i is a sufficient statistic for θ \theta in this case as well
step 5
For part (c), we need to check if T(X) T(X) is sufficient for estimating θ3+θ2 \theta^3 + \theta^2 . Since T(X) T(X) does not provide information about θ3 \theta^3 directly, it is not sufficient for estimating θ3+θ2 \theta^3 + \theta^2
step 6
For part (d), when considering a random sample from the Poisson distribution Poi(θ) \text{Poi}(\theta) , the likelihood function is: $$ L(\theta) = \prod_{i=1}^{n} \frac{e^{-\theta} \theta^{X_i}}{X_i!}
step 7
The sufficient statistic for θ \theta in this case is T(X)=∑i=1nXi T(X) = \sum_{i=1}^{n} X_i , which is also the sample sum. This statistic is sufficient for estimating the population variance of the Poisson distribution, which is equal to θ \theta
Answer
(a) T(X)=∑i=1nXi T(X) = \sum_{i=1}^{n} X_i is sufficient for θ \theta in N(θ,θ) N(\theta, \theta) . (b) T(X)=∑i=1nXi T(X) = \sum_{i=1}^{n} X_i is sufficient for θ \theta in N(θ,θ3) N(\theta, \theta^3) . (c) No, it is not sufficient for estimating θ3+θ2 \theta^3 + \theta^2 . (d) T(X)=∑i=1nXi T(X) = \sum_{i=1}^{n} X_i is sufficient for estimating the population variance in Poi(θ) \text{Poi}(\theta) .
Key Concept
The Fisher-Neyman factorization theorem helps identify sufficient statistics for parameter estimation.
Explanation
Sufficient statistics summarize the data without losing information about the parameter, making them crucial for efficient estimation.
Solution by Steps
step 1
The given distribution function is F(x;θ1,θ2)=1−(θ1x)θ2 F(x; \theta_1, \theta_2) = 1 - \left( \frac{\theta_1}{x} \right)^{\theta_2} for x≥θ1 x \geq \theta_1 . The likelihood function for a random sample X1,X2,…,Xn X_1, X_2, \ldots, X_n is given by: L(θ1,θ2)=∏i=1n(θ2θ1θ2Xi−θ2−1) for Xi≥θ1. L(\theta_1, \theta_2) = \prod_{i=1}^{n} \left( \frac{\theta_2}{\theta_1^{\theta_2}} X_i^{-\theta_2 - 1} \right) \text{ for } X_i \geq \theta_1.
step 2
To find the sufficient statistics, we can rewrite the likelihood function as: L(θ1,θ2)=θ2n⋅θ1−nθ2⋅∏i=1nXi−θ2−1. L(\theta_1, \theta_2) = \theta_2^n \cdot \theta_1^{-n\theta_2} \cdot \prod_{i=1}^{n} X_i^{-\theta_2 - 1}. This shows that the sufficient statistics are related to the minimum of the sample and the sum of the logarithm of the sample values
step 3
The sufficient statistics for θ1 \theta_1 and θ2 \theta_2 can be identified as: T1=min⁡(X1,X2,…,Xn)andT2=∑i=1nlog⁡(Xi). T_1 = \min(X_1, X_2, \ldots, X_n) \quad \text{and} \quad T_2 = \sum_{i=1}^{n} \log(X_i).
Answer
The sufficient statistics for estimating θ1 \theta_1 and θ2 \theta_2 are T1=min⁡(X1,X2,…,Xn) T_1 = \min(X_1, X_2, \ldots, X_n) and T2=∑i=1nlog⁡(Xi) T_2 = \sum_{i=1}^{n} \log(X_i) .
Key Concept
Sufficient statistics summarize the information in the sample relevant for estimating parameters.
Explanation
The statistics T1 T_1 and T2 T_2 capture all necessary information from the sample to estimate θ1 \theta_1 and θ2 \theta_2 effectively.
Solution by Steps
step 1
The given probability density function is f(x;θ)=Q(θ)M(x) f(x; \theta) = Q(\theta) M(x) for 0 < x < \theta . The likelihood function for a random sample X1,X2,…,Xn X_1, X_2, \ldots, X_n is given by: $$ L(\theta) = \prod_{i=1}^{n} f(X_i; \theta) = \prod_{i=1}^{n} Q(\theta) M(X_i)
step 2
Since Q(θ) Q(\theta) does not depend on x x , we can factor it out of the product: $$ L(\theta) = Q(\theta)^n \prod_{i=1}^{n} M(X_i)
step 3
To apply the Fisher-Neyman factorization theorem, we need to express the likelihood in the form g(T(X),θ)h(X) g(T(X), \theta) h(X) . Here, we can see that T(X)=max⁡(X1,X2,…,Xn) T(X) = \max(X_1, X_2, \ldots, X_n) is a sufficient statistic because it captures the information about θ \theta through the maximum value of the sample. Thus, we can write: $$ L(\theta) = Q(\theta)^n \cdot M(X_1) M(X_2) \cdots M(X_n)
Answer
The sufficient statistic for estimating θ \theta is T(X)=max⁡(X1,X2,…,Xn) T(X) = \max(X_1, X_2, \ldots, X_n)
Key Concept
The Fisher-Neyman factorization theorem helps identify sufficient statistics by factoring the likelihood function.
Explanation
The maximum of the sample captures all necessary information about the parameter θ \theta in this context.
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