We have one observation from the random variable with \(f(x|\theta)\) where\(\theta \in \{1,2,3\}\) and
Find the mle of \(\theta\).
max{L(\(\theta\)|0} = 1/3, f(0|1)=1/3, so mle of \(\theta\) is 1
max{L(\(\theta\)|1} = 1/3, f(1|1)=1/3, so mle of \(\theta\) is 1
max{L(\(\theta\)|2} = 1/4, f(2|2)=f(2|3)=1/4, so mle of \(\theta\) is 2 or 3
max{L(\(\theta\)|3} = 1/2, f(3|3)=1/2, so mle of \(\theta\) is 3
max{L(\(\theta\)|4} = 1/4, f(4|3)=1/4, so mle of \(\theta\) is 3
we have a sample of size n from a rv with distribution given by
In parts b-e assume we know \(\alpha\)
\(L(\beta) = 0\) for \(\beta<\max\{x\}=x_{[n]}\). If \(\beta>\max\{x\}=x_{[n]}\)
\(L(\beta) = \alpha^n/\beta^{n\alpha}\prod x_i^{\alpha-1}\)
so L is strictly decreasing for \(\beta>\max\{x\}=x_{[n]}\), and so argmax{L(\(\beta\)|x)} = x[n]
and T=x[n] is a sufficient statistic for \(\beta\).
so T is not unbiased (but it is asymptotically unbiased)
because P(X[n]<\(\beta\)+\(\epsilon\))=P(X[n]<\(\beta\))=1
so T is consistent