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1

Your condition does hold in the case where $x/n=1/2$, due to the symmetry in $f(p)$. That is, $$f(1/2 + h)=f(1/2 - h).$$ In the general case, I don't think there is a nice condition, since the growth/decay on one side of $x/n$ will be different than the decay/growth on the other side.


3

$$\forallϵ\gt0,\quad\exists\alpha\gt0,\quad[|T_n−θ|<\alpha]\subseteq[|f(T_n)−f(θ)|<ϵ]$$


1

The short answer is that the likelihood ratio test is just an inference scheme and is different than the maximum likelihood estimator. For example, in a normal population, the sample mean is the MLE of the population mean. Now lets say you want to determine if the population mean is zero vs not zero. One way is the usual "Z-test" using the sample mean, but ...


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One of the most classical examples might be the following: assume that $(X_i)_{1\leqslant i\leqslant n}$ is i.i.d. exponential with parameter $\lambda$, then the likelihood is $$\ell(x_1,\ldots,x_n)=\lambda^n\mathrm e^{-\lambda(x_1+\cdots+x_n)},$$ hence the MLE for $\lambda$ is $$\hat\lambda=\frac{n}{X_1+\cdots+X_n}.$$ For every $i$, $E(X_i)=1/\lambda$, ...



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