Estimation

$-\ln X_i$ ~ Exp$(\theta )$ with mean $1/\theta$, and $\sum \ln X_i$ is complete sufficient; the sample mean of $-\ln X_i$ is unbiased for $1/\theta$.

An unbiased estimator g(X) must satisfy $\sum g(k){\lambda }^k{e}^{-\lambda }/k$! $=h(\lambda )$; the left side is $e^{-\lambda }$ times a power series in $\lambda$, which can never equal $\lambda$⁻${}^1($unbounded at $0).\lambda ,{\lambda }^2$ and $e^{-\lambda }$ = P(X = 0) are all estimable.

(1), (3) are sample means of n i.i.d. terms. (2) converges to $2\mu .(4)W/n\to 0$ in probability since W is a fixed random variable, so consistency is preserved.

The joint density factorises through $\sum x_i($Fisher–Neyman), so X̄ is sufficient; the normal location family is a full exponential family, giving completeness; Lehmann–Scheffé then makes the unbiased X̄ the UMVUE. The median alone is not sufficient — the factorisation criterion fails, and conditioning on it does not free the likelihood of $\mu$.

(1) If $U_1$ and $U_2$ were two UMVUEs, then $(U_1+U_2)/2$ is unbiased with variance strictly smaller unless $U_1=U_2$ almost surely — the Cauchy–Schwarz step forces uniqueness. (2) This is the Rao–Blackwell theorem; the conditional expectation is a statistic precisely because the conditioning statistic is sufficient. (3) Fails: the bound is attained only in exponential families and only for particular parametric functions; many unbiased estimators, including some UMVUEs, sit strictly above it. (4) Fails — the two notions are independent; completeness is a property of the induced family of distributions and does not by itself say the statistic captures all information about the parameter.

2X̄ is unbiased but has larger variance than $(n+1)X_{(n)}/n$; it is not a function of the complete sufficient statistic.

Posterior is Beta(5 + 10, 5 + 15) = Beta(15, 20) with mean 15/35 = 3/7.

$E\bar{X}=\mu ($unbiased) by linearity, and $\bar{X}\to \mu$ in probability by the weak law of large numbers (consistent). The two properties are logically independent in general — this estimator happens to enjoy both.

The likelihood ${\theta }^{-n}$·1{$\theta \ge X_{(n)}$} increases as $\theta$ shrinks toward the largest observation, so the MLE is $X_{(n)}$ — a boundary maximum, not a stationary point, which is why differentiating fails here. $EX=\theta /2$ gives the moments estimator 2X̄. Since $E[X_{(n)}]=n\theta /(n+1)<\theta$, the MLE is biased low; the factor (n + 1)/n corrects it exactly.