Let X1, X2, ..., Xn be i.i.d. random variables with common pdf

f(xθ)=(logθ)θxθ1, for

0 < x < 1 where θ > 1 is an unknown parameter. Then the statistic T=i=1nXi is

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CSIR-UGC (NET) Mathematical Science: Held on (26 Nov 2020)
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  1. sufficient, but not complete
  2. Sufficient, but not minimal sufficient
  3. Complete sufficient
  4. neither complete, nor sufficient

Answer (Detailed Solution Below)

Option 3 : Complete sufficient
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Detailed Solution

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Given:-

X1, X2, ..., Xn are i.i.d. random variables with common pdf

f(xθ)=(logθ)θxθ1

Concept Used:-

To check whether T = ∑ᵢ Xᵢ is a sufficient statistic for the parameter θ, we need to find the likelihood function of the sample.

Explanation:-

The joint pdf of the sample is given by

f(x,x,...,x|θ)=f(x|θ)f(x|θ)...f(x|θ)f(x,x,...,x|θ)=i=1nf(xiθ)f(x,x,...,x|θ)=i=1n(logθ)θxiθ1 f(x,x,...,x|θ)=(logθ)nθi=1nxi(θ1)n

The factorization theorem states that a statistic T is sufficient for a parameter θ if and only if the joint pdf of the sample can be factorized into two functions.

One which depends on the sample only through T, and the other that does not depend on θ.

We can rewrite the joint pdf of the sample as

f(x,x,...,x|θ)=h(x,x,...,x)g(T|θ),

Here,

h(x,x,...,x)=(logθ)n   andg(T|θ)=(θT)/(θ1)n  

Since the joint pdf can be factorized in this way, T is a sufficient statistic for θ.

To check whether T is complete, we need to show that T is able to capture all the information about θ contained in the sample.

In other words, any function of T that is independent of θ must have an expected value equal to 0.

Let h(T) be a function of T that is independent of θ. Then

E[h(T)]=h(T)g(T|θ)dTE[h(T)]=01h(T)θT(θ1)ndTE[h(T)]=(logθ)n(θ1)n01h(T)exp(Tlogθ)dT

Since θ > 1, \log θ > 0, and the integrand is positive on [0,1].

Therefore, if E[h(T)] = 0 for all θ > 1, then h(T) must be identically zero almost everywhere on [0,1].

This shows that T is a complete, sufficient statistic for θ.

Therefore, the statistic T=i=1nXi is completely sufficient.

Hence, the correct option is 3.

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