How Do You Calculate the Cramer-Rao Lower Bound for θ?

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SUMMARY

The discussion focuses on calculating the Cramer-Rao Lower Bound (CRLB) for the parameter θ in the given distribution function f(x) = θ * 2^θ * x^(-θ-1) for x > 2. The CRLB is defined as [-n * E[d^2/dθ^2 ln(f(y))]]^-1, where the second derivative of the log-likelihood function is computed as -θ^(-2). The expected value with respect to x is determined to be -θ^(-2), which simplifies the calculation of the CRLB. Participants seek clarification on the next steps for completing the assignment.

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Sander1337
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Hi there,

I've got a distribution function for an assignment for my school here and I don't get the hang of the question;

f(x) = θ* 2^θ * x^(-θ-1) for x>2
0 for else

The assignment is to calculate the Cramer-Rao lower boundary for consistent estimators of θ.

This is what we've got so far;

Cramer-Rao lower boundary:
[-n*E[d^2/dθ^2 ln(f(y))]]^-1

(d^2/dθ^2)ln(f(x)) = -θ^(-2)= CRLB function

Now since our professor didn't explain the Cramer Rao lower boundary we haven't got a clue of how to continue now. Is there someone here who knows how to continue now?

Greetings,

Tony, Siebe & Sander

(question might be in wrong (sub)forum, apoligies for that, don't bother rerouting this question to the right (sub)forum, thanks!)
 
Last edited:
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Well, the next step is to take the expected value (with respect to x). Since -θ^(-2) does not involve x, it is just a constant, so the expected value is -θ^(-2) itself.
 

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