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A random sample of size $\displaystyle n$ is drawn from the pdf,
$\displaystyle f_Y(y;\theta) = e^{-(y-\theta)}$, $\displaystyle \theta\le y$
I can show that $\displaystyle Y_{max}$ is not sufficient by an example, but I am having a hard time showing that $\displaystyle Y_{min}$ is sufficient.
So I got that the $\displaystyle f_{y_{min}}(y_{min};\theta) = n[e^{-(y-\theta)}]^n$, but I can't seem to factor this out of $\displaystyle \prod{e^{-(y-\theta)}}$
Hope this helps clarify my question.
It's sufficient by the Factorization Theorem.
$\displaystyle f_Y(y;\theta) = e^{-(y-\theta)}$, $\displaystyle \theta\le y$
should be rewritten with indicators.
BUT you need a sample here
$\displaystyle L(y_1,...,y_n) =\prod_{i=1}^n e^{-(y_i-\theta)}I(y_i>\theta)$
$\displaystyle =e^{n\theta}e^{-\sum_{i=1}^ny_i}I(y_{min}>\theta)$
I am not familiar with what you mean with $\displaystyle I$? Thanks for your help.
Indicator/characteristic function :
$\displaystyle I(y_i>\theta)=1$ if $\displaystyle y_i$ is indeed $\displaystyle >\theta$. 0 otherwise.
