Confidence interval for a density function parameter I am

Harley Ayers

Harley Ayers

Answered question

2022-03-28

Confidence interval for a density function parameter
I am trying to solve this problem: Given X1,,Xn a random sample of a population of random variables with p.d.f.
f(x,θ)=e(xθ)I(x)xθ, find a confidence interval of level 1α for θ.SNKFollowing a hint, I first looked into the distribution of Y=min(Xi)θ, which I found to be Exponential(n).
Taking into account that the indicator function in the pdf limits to values of x greater than θ, and using the exponential's cdf. and some bounds, I got that the confidence interval should be I1α(X1,,Xn)=[0,ln(α)n].
Would that be correct? I am not sure I am using the hint correctly. If it is wrong, what else could I try?

Answer & Explanation

Payten Reese

Payten Reese

Beginner2022-03-29Added 10 answers

I think you were on the right track initially, but may have confused the arithmetic of the exponential distribution
If YExp(n) then P(Yy)=1eny so
- there is a probability α that 0Y1nloge(1α).
- and also a probability α that 1nloge(1+α2)Y1nloge(1α2)
This gives confidence intervals for θ.
- [min(Xi)+1nloge(1α),min(Xi)]
- or [min(Xi)+1nloge(1α2),min(Xi)+1nloge(1+α2)]
The first of these alternatives might be seen as "one-sided" but is narrower than the second which might be seen as "two-sided". As an example, suppose α=95%,n=2,X1=5 and X2=6.2 then this would give confidence intervals for θ of about [3.502,5] or [3.156,4.987].

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