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## Problem 3 Let X1,...,Xn be a sample from a distribution Fθ ...
Aug 24, 2024
## Problem 3 Let X1,...,Xn be a sample from a distribution Fθ belonging to a class F. In addition, T=ϕ(S) and S are two statistics.
Let X and Y be two random variable such that Y has the binomial distribution with size N and probability π and, given Y=y, X has the binomial distribution with size y and probability p.
### (a) If T is sufficient, show whether S is sufficient.
Solution by Steps
step 1
To determine if S is sufficient given that T is sufficient, we can use the factorization theorem. A statistic T is sufficient for parameter θ if the likelihood function can be factored as: L(θ∣x)=g(T(x)∣θ)h(x) for some functions g and h
step 2
Since T=ϕ(S), we can express the likelihood function in terms of S: L(θ∣x)=g(ϕ(S(x))∣θ)h(S(x)) This shows that the information in S is captured by T
step 3
If T is sufficient, then S must also be sufficient because S contains all the information needed to compute T. Thus, we conclude that if T is sufficient, then S is also sufficient
Answer
S is sufficient if T is sufficient, as S contains all the information needed to compute T.
Key Concept
Sufficient statistics and the factorization theorem.
Explanation
The sufficiency of S follows from the fact that it contains all the information needed to compute the sufficient statistic T.
### (b) If S is sufficient, show whether T is sufficient.
Solution by Steps
step 1
To determine if T is sufficient given that S is sufficient, we again use the factorization theorem. A statistic S is sufficient for parameter θ if the likelihood function can be factored as: L(θ∣x)=g(S(x)∣θ)h(x) for some functions g and h
step 2
Since T=ϕ(S), we can express the likelihood function in terms of S. If S is sufficient, then the likelihood can be factored as: L(θ∣x)=g(S(x)∣θ)h(x) which implies that T can also be expressed in terms of S
step 3
Therefore, since S contains all the information needed to compute T, we conclude that if S is sufficient, then T is also sufficient
Answer
If S is sufficient, then T is also sufficient.
Key Concept
The relationship between sufficient statistics and the factorization theorem.
Explanation
If one statistic is sufficient, any function of that statistic is also sufficient.
Could you do it again with this information of binomial distribution
Solution by Steps
step 1
To determine if S is sufficient given that T is sufficient, we apply the factorization theorem. A statistic T is sufficient for parameter θ if the likelihood function can be factored as: L(θ∣x)=g(T(x)∣θ)h(x) for some functions g and h
step 2
Since T=ϕ(S), we can express the likelihood function in terms of S. If S is sufficient, then the likelihood can also be factored as: L(θ∣x)=g(S(x)∣θ)h(x), which implies that S contains all the information needed to compute T
step 3
Therefore, if T is sufficient, then S must also be sufficient, as S provides the necessary information to compute T
Answer
If T is sufficient, then S is also sufficient.
Key Concept
Sufficiency of statistics and the factorization theorem.
Explanation
The sufficiency of one statistic implies the sufficiency of another if the first can be expressed as a function of the second.
step 1
To determine if T is sufficient given that S is sufficient, we again use the factorization theorem. A statistic S is sufficient for parameter θ if the likelihood function can be factored as: L(θ∣x)=g(S(x)∣θ)h(x) for some functions g and h
step 2
Since T=ϕ(S), we can express the likelihood function in terms of T. If S is sufficient, then the likelihood can be factored as: L(θ∣x)=g(S(x)∣θ)h(x), which implies that T can also be computed from S
step 3
Therefore, if S is sufficient, then T is also sufficient, as T is a function of S and retains the information about θ
Answer
If S is sufficient, then T is also sufficient.
Key Concept
Relationship between sufficiency of statistics.
Explanation
The sufficiency of one statistic implies the sufficiency of another if the second statistic is a function of the first.
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