离散均匀分布的估计| Serial Number Estimation

Serial Number Estimation

Content

Serial Number Estimation

Parameter description

Method 1: Probability of each sample
Method 2: Probability of maximum sample

Point Estimate

Estimators intuited from discrete uniform distribution

Estimator 1: 2*Mean-1
Estimator 2: Max + Avg GAP
Estimator3: Min+max estimator
Estimator 4: Mean + 3 Std estimator

MLE estimator
Estimator 5: An improvement from MLE: UMVUE

UMVUE under method 1
UMVUE under method 2

Estimator 6: Bayes Estimator
Point Estimation Conclusion

Interval Estimation
Reference

Parameter description

If $\theta$ is contained in a function, $n$ would be the total sample numbers, and $\hat{\theta}$ would be the estimator for actual maximum ID.
If $M$ is contained in a function, $k$ would be the total sample numbers, and $N$ would be the actual maximum ID. As for $M$ , that is a random variable of maximum ID in a random sample.

Method 1: Probability of each sample

The estimator used to predict the maximum value can also be determined by assuming that the probability of getting each sample is uniform where $\theta$ represents the actual maximum ID in each day.

$P(x) = \frac{1}{\theta}$

Method 2: Probability of maximum sample

According to the Assumption 1, we consider the observed maximum ID as a r.v M, and take the maximum ID we encountered in one specific day as m (i.e. $x_{n:n}$ ). Assume that the N is the actual maximum ID and k represents the number of ill sample, the probability mass function (PMF) of getting the maximum ID can be expressed as follows:

$P(M = m) = \frac{C_{k-1}^{m-1}}{C_k^N}$

Point Estimate

Estimators intuited from discrete uniform distribution

Estimator 1: 2*Mean-1

Consider continuous distribution for this problem, i.e. $UNIF(0,\theta)$

$For~UNIF(0,\theta),E(X)=\frac{\theta}{2},Var(X)=\frac{\theta^2}{12}$

We consider the following estimator:
$\widehat{\theta}_1=\frac{2}{n}\sum^n_{i=1}X_i-1~for~discrete ~distrubution\\ \widehat{\theta}_1=\frac{2}{n}\sum^n_{i=1}X_i~for~continuous ~distrubution\\ E(\widehat{\theta}_1)=E(\frac{2}{n}\sum^n_{i=1}X_i-1)=2E(\overline{X})=\theta\\ Var(\widehat{\theta}_1)=\frac{4}{n^2}\sum_{i=1}^nVar(X_i)=\frac{4}{n^2}\sum_{i=1}^n\frac{\theta^2}{12}=\frac{\theta^2}{3n}\\ \therefore \widehat{\theta}_1~is~an~unbiased~estimator ~with~Var=\frac{\theta^2}{3n}\\$

Estimator 2: Max + Avg GAP

Consider other form of improvement from MLE estimator, i.e. using average approach to estimate the GAP between maximum and the upper limit:

$\widehat{\theta}_2 = X_{n:n}+\frac{1}{n-1}\sum_{i>j}(X_i-X_j-1)\quad\dots for~discrete~case\\ \widehat{\theta}_2 = X_{n:n}+\frac{1}{n-1}\sum_{i>j}(X_i-X_j) \quad\dots for~continuous~case$

Calculate the expected value and variance to determine if this estimator is biased or not.
$E(\widehat{\theta}_2) = E(X_{n:n}) + \frac{1}{n-1}\sum_{i>j}E{(X_i-X_j)} = \frac{n\theta}{n+1} \\ Var(\hat{\theta}_2) = \frac{n\theta^2}{(n+1)(n-1)(n+2)}$
Therefore, $\theta_2$ is a biased estimator.

Estimator3: Min+max estimator

We know that maximum sample ID is what’s closed to the upper limit, and we could add more information to it. Intuitively, we first consider minimum sample ID + maximum sample ID:
$\widehat{\theta}_3=x_{1:n}+x_{n:n}\\ F_{X_{n:n}}(x) =[F_X(x)]^n=\frac{x^n}{\theta^n},f_{X_{n:n}}(x)=n\frac{x^{n-1}}{\theta^n}\\ E[X_{n:n}]=\int xn\frac{x^{n-1}}{\theta^n}dx=\frac{n}{n+1}\theta\\ E[X_{n:n}^2] =\int x^2n\frac{x^{n-1}}{\theta^n}dx=\frac{n}{n+2}\theta^2\\ F_{X_{1:n}}(x) =1-[1-F_X(x)]^n=1-(\frac{\theta-x}{\theta})^n,f_{X_{1:n}}(x)=\frac{n (\theta-x)^{n-1}}{\theta^n}\\ E[X_{1:n}]=\int x\frac{n (\theta-x)^{n-1}}{\theta^n}dx=\frac{1}{n+1}\theta\\ E[X_{1:n}^2] =\int x^2\frac{n (\theta-x)^{n-1}}{\theta^n}dx=\frac{2}{n(n+1)}\theta^2\\ E(\widehat{\theta}_3)=E(x_{1:n})+E(x_{n:n})=\theta\\ Var(\hat{\theta}_3)=Var(X_{1:n})+Var(X_{n:n})+2Cov(X_{1:n},X_{n:n})\\ =\frac{2}{n(n+1)}\theta^2-(\frac{1}{n+1}\theta)^2 +\frac{n}{n+2}\theta^2-(\frac{n}{n+1}\theta)^2+2Cov(X_{1:n},X_{n:n})\\ Since~the~ joint~ distribution~ of ~the~ order~ statistics~ of~ the~ uniform~ distribution~is\\ f_{u_i,v_j}(u,v)=n!\frac{u^{i-1}}{(i-1)!}\frac{(v-u)^{j-i-1}}{(j-i-1)!}\frac{(1-v)^{n-j}}{(n-j)!}\\ Cov(u_k,v_j)=\frac{j(n-k-1)}{(n-1)^2(n+2)}\\ Var(\hat{\theta}_3)=\frac{2\theta^2}{n(n+2)}+\frac{2n^2\theta^2}{(n+1)^2(n+2)}$ θ3=x1:n+xn:nFXn:n(x)=[FX(x)]n=θnxn,fXn:n(x)=nθnxn−1E[Xn:n]=∫xnθnxn−1dx=n+1nθE[Xn:n2]=∫x2nθnxn−1dx=n+2nθ2FX1:n(x)=1−[1−FX(x)]n=1−(θθ−x)n,fX1:n(x)=θnn(θ−x)n−1E[X1:n]=∫xθnn(θ−x)n−1dx=n+11θE[X1:n2]=∫x2θnn(θ−x)n−1dx=n(n+1)2θ2E(θ3)=E(x1:n)+E(xn:n)=θVar(θ^3)=Var(X1:n)+Var(Xn:n)+2Cov(X1:n,Xn:n)=n(n+1)2θ2−(n+11θ)2+n+2nθ2−(n+1nθ)2+2Cov(X1:n,Xn:n)Since the joint distribution of the order statistics of the uniform distribution isfui,vj(u,v)=n!(i−1)!ui−1(j−i−1)!(v−u)j−i−1(n−j)!(1−v)n−jCov(uk,vj)=(n−1)2(n+2)j(n−k−1)Var(θ^3)=n(n+2)2θ2+(n+1)2(n+2)2n2θ2
Therefore, $\theta_3$ is a biased estimator.

Estimator 4: Mean + 3 Std estimator

$\widehat{\theta}_4=\frac{\sum_{i=1}^n X_i}{n}+3\sqrt{\frac{\sum_{i=1}^n(X_i-\bar{X})^2}{n-1}}= \frac{\sum_{i=1}^n X_i}{n}+3S\\ E(\widehat{\theta}_4)=E[\overline{X}]+3*E(S)=\frac{1}{2} \theta+3\frac{\theta}{2\sqrt{3}}\\ Var(\widehat{\theta}_4)>Var(\overline{X})=\frac{\theta^2}{12n}\\ \therefore \widehat{\theta}_4~is~a~biased~estimator\\$

MLE estimator

$f_{x_1,x_2,...,x_n}(x_1,x_2,...,x_n) = \prod_{i=1}^{n}f_{x_i}(x_i)\\ = \prod_{i=1}^{n}\frac{1}{\theta}\\ = \frac{1}{\theta^n}\\ ln(f_{x_1,x_2,...,x_n}(x_1,x_2,...,x_n))=-nln\theta\\ \therefore \widehat{\theta}_{MLE} = X_{n:n}\\ F_{X_{n:n}}(x) =[F_X(x)]^n=\frac{x^n}{\theta^n},f_{X_{n:n}}(x)=n\frac{x^{n-1}}{\theta^n}\\ E[X_{n:n}]=\int xn\frac{x^{n-1}}{\theta^n}dx=\frac{n}{n+1}\theta\\ E[X_{n:n}^2] =\int x^2n\frac{x^{n-1}}{\theta^n}dx=\frac{n}{n+2}\theta^2\\ E(\widehat{\theta}_{MLE})=E(X_{n:n})=\frac{n}{n+1}\theta\\ Var(\widehat{\theta}_{MLE})=\frac{n}{n+2}\theta^2-(\frac{n}{n+1}\theta)^2=\frac{n}{(n+1)^2(n+2)}\theta^2\\ \therefore the ~MLE~estimator~is~an ~biased~estimator.\\ f(x)= \left\{\begin{matrix} \frac{1}{\theta},~0<x<\theta\\ 0,~~~~~~~~~~0.w. \end{matrix}\right.\\ f(x)=\frac{1}{\theta}I_{0,\theta}(x)\\ f(x_1,...,x_n)=\frac{1}{\theta^n}\prod_{i=1}^{n}I_{0,\theta}(x_i)\\ = \frac{1}{\theta^n}I_{0,\theta}(x_{n:n})\\ = g(x_{n:n},\theta)*h(x_1,...,x_n)\\ \therefore~S=x_{n:n}~is~sufficient~for~\theta\\ \therefore the ~MLE~estimator~is~sufficient.$ fx1,x2,...,xn(x1,x2,...,xn)=i=1∏nfxi(xi)=i=1∏nθ1=θn1ln(fx1,x2,...,xn(x1,x2,...,xn))=−nlnθ∴θMLE=Xn:nFXn:n(x)=[FX(x)]n=θnxn,fXn:n(x)=nθnxn−1E[Xn:n]=∫xnθnxn−1dx=n+1nθE[Xn:n2]=∫x2nθnxn−1dx=n+2nθ2E(θMLE)=E(Xn:n)=n+1nθVar(θMLE)=n+2nθ2−(n+1nθ)2=(n+1)2(n+2)nθ2∴the MLE estimator is an biased estimator.f(x)={θ1, 0<x<θ0, 0.w.f(x)=θ1I0,θ(x)f(x1,...,xn)=θn1i=1∏nI0,θ(xi)=θn1I0,θ(xn:n)=g(xn:n,θ)∗h(x1,...,xn)∴ S=xn:n is sufficient for θ∴the MLE estimator is sufficient.

Estimator 5: An improvement from MLE: UMVUE

UMVUE under method 1

Consider an improved estimator from $\hat{\theta}$ , we have:

$\hat{\theta}_5 = \frac{n+1}{n}x_{n:n} - 1 \quad \dots for~discrete~case. \\ \hat{\theta}_5 = \frac{n+1}{n}x_{n:n} \quad \dots for~continuous~case.$
Get the variance to determine if the estimator has been improved to an unbiased estimator where $Var(x) = \mathbb{E}(x^2) - [\mathbb{E}(x)]^2$ .

$E(\hat{\theta}_5)=\theta\\ E(X_{n:n}^2) = \int x^2\cdot\frac{n\cdot x^{n-1}}{\theta^n}dx=\frac{n\theta^2}{n+2} \\ Var(\hat{\theta}_5) = \Big(\frac{n+1}{n}\Big)^2\cdot Var(X_{n:n}) \\ = \Big(\frac{n+1}{n}\Big)^2\Big(\frac{n\theta^2}{n+2}-\big(\frac{n\theta}{n+1}\big)^2\Big) = \frac{\theta^2}{n(n+2)}$
Therefore, $\hat{\theta}_5$ is an unbiased estimator.

To obtain the maximum value of $\theta$ , the best option is to get the maximum value of each random sample m (i.e. $\hat{\theta} = x_{n:n}$ ). According to the Ex 3.4, we know that

$F_{x_{n:n}}(x) = [F_x(x)]^n = \frac{x^n}{\theta^n}\quad\rightarrow\quad f_{x_{n:n}}(x) = \frac{n\cdot x^{n-1}}{\theta^n}$

where n is the total number of sample in each day. To obtain the expected value of m (i.e. $x_{n:n}$ ), we can determine if the estimator is an biased estimator such that:

$\mathbb{E}(x_{n:n}) = \int x\cdot \frac{n\cdot x^{n-1}}{\theta^n}dx = \frac{n\theta}{n+1} = \mathbb{E}(\hat{\theta})\neq \theta$

Therefore, the estimator is an biased estimator.
$f(x)=\left\{\begin{matrix} \frac{1}{\theta} ~0<x<\theta\\ 0 o.w. \end{matrix}\right.\\ f(x) = \frac{1}{\theta}I_{0,\theta}(x) \\ f(x_1,...,x_n) = \frac{1}{\theta^n}\prod_{i=1}^{n}I_{0,\theta}(x_i) = \frac{1}{\theta^n}I_{0,\theta}(x_{n:n}) \\ = g(x_{n:n},\theta)\cdot h(x_1,...,x_n)$
Therefore, $S = x_{n:n}$ is sufficient for $\theta$ . According to Lehmann-Scheffe Theorem, we have:

$T = \hat{\theta}_2 = \frac{n+1}{n}X_{n:n}$

which is unbiased for $\tau(\theta)=\theta$ . Thus, $T$ is UMVUE.

UMVUE under method 2

The expected value of $M = m$ can be calculated as follows where $C_k^N = \frac{N!}{k!(N-k)!}$ , $\sum_{m=k}^NC_k^m = C_{k+1}^{N+1}$ :

$\mathbb{E}(M=m) = \sum_{m=k}^Nm\cdot P(m) \\ = \sum_{m=k}^Nm\cdot\frac{\frac{(m-1)!}{(k-1)!(m-k)!}}{\frac{N!}{k!(N-k)!}} \\ = \sum_{m=k}^N\frac{m!k(N-k)!k!}{k!(m-k)!N!} \\ = \frac{k(N-k)!k!}{N!}\cdot\sum_{m=k}^NC_k^m \\ = \frac{k(N-k)!k!}{N!}\cdot \frac{(N+1)!}{(k+1)!(N-k)!} = \frac{k(N+1)}{k+1}$
Since we are looking for the maximum ID from our observation, the best guess of M should be the maximum ID of THAT particular day m. Therefore, we get:
$m = \frac{k(N+1)}{k+1}\quad\rightarrow\quad k\hat{N} = mk + m - k\quad\rightarrow\quad \hat{N} = m + \frac{m}{k} - 1$
By Finding the expected value of $\hat{N}$ , we have:

$\mathbb{E}(\hat{N}) = \mathbb{E}\Big[\mathbb{E}(M) + \frac{\mathbb{E}(M)}{k} - 1\Big] = \mathbb{E}(M) + \frac{\mathbb{E}(M)}{k} - 1 \\ = \frac{k(N+1)}{k+1} + \frac{N+1}{k+1} - \frac{k+1}{k+1} = \frac{N(k+1)}{k+1} = N$
Therefore, $\hat{N}$ is proved to be unbiased.

Estimator 6: Bayes Estimator

The Bayesian approach is to consider the credibility $P(N=n|M=m, K=k)$ that the maximum random ID $N$ is equal to the number $n$ , and the maximum observed serial number $M$ is equal to the number $m$ . Consider Conditional probability rule instead of using a proper prior distribution.

$P(n|m, k)P(m|k)=P(m|n,k)P(n|k)=P(m,n|k)$

$P(m|n,k)$ answers the question: “What is the probability of a specific serial number $m$ being the highest number observed in a sample of $k$ patients, given there are $n$ in total?” The probability of this occurring is:
$P(m|n,k) =k\frac{(n-k)!}{n!}\frac{(m-1)!}{(m-k)!}=\frac{C_{k-1}^{m-1}}{C_k^n}I_{k\leq m}I_{m\leq n}$
$P(m|k)$ is the probability that the maximum serial number is equal to $m$ once $k$ tanks have been observed but before the serial numbers have actually been observed.
$P(m|k) = P(m|k)*\sum_{n=0}^\infty P(n|m,k)\\ =P(m|k)*\sum_{n=0}^\infty \frac{P(m|n,k)P(n|k)}{P(m|k)}\\ =\sum_{n=0}^\infty P(m|n,k)P(n|k)\\$
$P(n|k)$ is the credibility that the total number of tanks, $N$ , is equal to $n$ when the number $K$ patients observed is known to be $k$ , but before the serial numbers have been observed. Assume that it is some discrete uniform distribution:

$P(n|k) = \frac{1}{\Omega-k}I_{k \leq \Omega }I_{n \leq \Omega}\\ where~~\Omega ~is~the~upper ~limit~and ~it's~finite\\$

$P(n|m, k)= \frac{P(m|n,k)P(n|k)}{P(m|k)}\\ =\frac{P(m|n,k)P(n|k)}{\sum_{n=0}^\infty P(m|n,k)P(n|k)},k\leq m,m\leq n,k \leq \Omega,n \leq \Omega\\ =\frac{P(m|n,k)}{\sum_{n=m}^{\Omega-1} P(m|n,k)}I_{m\leq n}I_{n\leq \Omega}\\ for~k \geq 2, ~P(n|m, k)= \frac{P(m|n,k)}{\sum_{n=m}^{\infty } P(m|n,k)}I_{m\leq n}\\ =\frac{\frac{C_{k-1}^{m-1}}{C_k^n}}{\sum_{n=m}^{\infty}\frac{C_{k-1}^{m-1}}{C_k^n}}I_{m\leq n}\\ = \frac{k-1}{k}\frac{C_{k-1}^{m-1}}{C_k^n}I_{m\leq n}\\ P(N>x|M=m, K=k)=\sum_{n=x+1}^{\infty }P(n|m, k)I_{m\leq x}\\ =I_{m< x}+I_{m\geq x}\sum_{n=x+1}^{\infty }\frac{k-1}{k}\frac{C_{k-1}^{m-1}}{C_k^n}\\ =I_{m< x}+I_{m\geq x}\frac{k-1}{k}\frac{C_{k-1}^{m-1}}{1}\sum_{n=x+1}^{\infty }\frac{1}{C_k^n}\\ =I_{m< x}+I_{m\geq x}\frac{k-1}{k}\frac{C_{k-1}^{m-1}}{1}\frac{k}{k-1}\frac{1}{C_{k-1}^x}\\ =I_{m< x}+I_{m\geq x}\frac{C_{k-1}^{m-1}}{C_{k-1}^x}\\ P(N\leq x|M=m, K=k)=1-P(N>x|M=m, K=k)\\ =I_{m\geq x}(1-\frac{C_{k-1}^{m-1}}{C_{k-1}^x})\\ \mu_{Bayes}=\sum_{n}nP(n|m, k)\\ =\sum_{n}\frac{k-1}{k}\frac{C_{k-1}^{m-1}}{C_k^n}I_{m\leq n}\\ =\sum_{n}\frac{m-1}{n}\frac{C_{k-2}^{m-2}}{C_{k-1}^{n-1}}I_{m\leq n}\\ =(m-1)C_{k-2}^{m-2}\sum_{n\geq n}\frac{1}{C_{k-1}^{n-1}}\\ =(m-1)C_{k-2}^{m-2}\frac{k-1}{k-2}\frac{1}{C_{k-2}^{m-2}}\\ =\frac{(m-1)(k-1)}{k-2}\\$ P(n∣m,k)=P(m∣k)P(m∣n,k)P(n∣k)=∑n=0∞P(m∣n,k)P(n∣k)P(m∣n,k)P(n∣k),k≤m,m≤n,k≤Ω,n≤Ω=∑n=mΩ−1P(m∣n,k)P(m∣n,k)Im≤nIn≤Ωfor k≥2, P(n∣m,k)=∑n=m∞P(m∣n,k)P(m∣n,k)Im≤n=∑n=m∞CknCk−1m−1CknCk−1m−1Im≤n=kk−1CknCk−1m−1Im≤nP(N>x∣M=m,K=k)=n=x+1∑∞P(n∣m,k)Im≤x=Im<x+Im≥xn=x+1∑∞kk−1CknCk−1m−1=Im<x+Im≥xkk−11Ck−1m−1n=x+1∑∞Ckn1=Im<x+Im≥xkk−11Ck−1m−1k−1kCk−1x1=Im<x+Im≥xCk−1xCk−1m−1P(N≤x∣M=m,K=k)=1−P(N>x∣M=m,K=k)=Im≥x(1−Ck−1xCk−1m−1)μBayes=n∑nP(n∣m,k)=n∑kk−1CknCk−1m−1Im≤n=n∑nm−1Ck−1n−1Ck−2m−2Im≤n=(m−1)Ck−2m−2n≥n∑Ck−1n−11=(m−1)Ck−2m−2k−2k−1Ck−2m−21=k−2(m−1)(k−1)

Hence, $\mu_{Bayes}=\frac{(X_{n:n}-1)(n-1)}{n-2}$ using the standard of writing in other chapter.
$E(\mu_{Bayes})=\frac{n}{n+2}\theta$ The Bayes estimator is a biased one.
To measure its uncertainty, we calculate its variance:
$\mu^2+\sigma^2-\mu=\sum_{n}n(n-1)P(n|m, k)\\ =\sum_{n}n(n-1)\frac{m-1}{n}\frac{m-2}{n-1}\frac{k-1}{k-2}\frac{C_{k-3}^{m-3}}{C_{k-2}^{n-2}}I_{m\leq n}\\ =(m-1)(m-2)\frac{k-1}{k-2}C_{k-3}^{m-3}\sum_{n\geq m}\frac{1}{C_{k-2}^{n-2}}\\ =(m-1)(m-2)\frac{k-1}{k-2}C_{k-3}^{m-3}\frac{k-2}{k-3}\frac{1}{C_{k-3}^{m-3}}\\ =\frac{(m-1)(m-2)(k-1)}{k-3}\\ \sigma^2_{Bayes}=\frac{(m-1)(m-2)(k-1)}{k-3}-(\frac{(m-1)(k-1)}{k-2})^2+\frac{(m-1)(k-1)}{k-2}\\ =\frac{(m-1)(k-1)(m+1-k)}{(k-3)(k-2)^2}\\ Var(\widehat{\theta}_{Bayes}) =\frac{(x_{n:n}-1)(n-1)(x_{n:n}+1-n)}{(n-3)(n-2)^2}$ μ2+σ2−μ=n∑n(n−1)P(n∣m,k)=n∑n(n−1)nm−1n−1m−2k−2k−1Ck−2n−2Ck−3m−3Im≤n=(m−1)(m−2)k−2k−1Ck−3m−3n≥m∑Ck−2n−21=(m−1)(m−2)k−2k−1Ck−3m−3k−3k−2Ck−3m−31=k−3(m−1)(m−2)(k−1)σBayes2=k−3(m−1)(m−2)(k−1)−(k−2(m−1)(k−1))2+k−2(m−1)(k−1)=(k−3)(k−2)2(m−1)(k−1)(m+1−k)Var(θBayes)=(n−3)(n−2)2(xn:n−1)(n−1)(xn:n+1−n)

Point Estimation Conclusion

According to the distribution of the question, we find six possible estimators, in which four of them are intuitive from the distribution and the background of the question, one is maximum likelihood estimator (MLE) and one the Bayes estimator and the improved estimator from MLE. We proved that the improved estimator from MLE is exactly uniformly minimum-variance unbiased estimator (UMVUE) which is unbiased estimator with the smallest variance.

Also, what is most important in our findings is the $X_{n:n}$ plays an important role in estimating the upper limit of th discrete uniform distribution since the maximum sample give the closest information of the upper limit intuitively and we also proved that it’ s the sufficient statistics to estimate $N$ .

To compare the unbiasedness, effectiveness of the estimators we find, we summarize the results and give the following table:

No.	$Function$	$E(\widehat{\theta})$	$Var(\widehat{\theta})$
$\widehat{\theta}_1$	$\frac{2}{n}\sum^n_{i=1}X_i-1$	$\theta$	$\frac{\theta^2}{3n}$
$\widehat{\theta}_2$	$X_{n:n}+\frac{1}{n-1}\sum_{i>j}(X_i-X_j-1)$	$\frac{n\theta}{n+1}$	$\frac{n\theta^2}{(n+1)(n-1)(n+2)}$
$\widehat{\theta}_3$	$x_{1:n}+x_{n:n}$	$\theta$	$\frac{2\theta^2}{n(n+2)}+\frac{2n^2\theta^2}{(n+1)^2(n+2)}$
$\widehat{\theta}_4$	$\frac{\sum_{i=1}^n X_i}{n}+3\sqrt{ \frac{\sum E(X_i-\overline{X})^2}{n-1}}$	$\frac{1}{2}\theta+\frac{2\sqrt{3}}{3}\theta$	$Var(\widehat{\theta}_4)>\frac{\theta^2}{12n}$
$\hat{\theta}_5$	$\frac{n+1}{n}x_{n:n} - 1$	$\theta$	$\frac{\theta^2}{n(n+2)}$
$\widehat{\theta}_{MLE}$	$X_{n:n}$	$\frac{n}{n+1}\theta$	$\frac{n}{(n+1)^2(n+2)}\theta^2$
$\hat{\theta}_{Bayes}$	$\frac{(X_{n:n}-1)(n-1)}{n-2}$	$\frac{n}{n+2}\theta$	$\frac{(x_{n:n}-1)(n-1)(x_{n:n}+1-n)}{(n-3)(n-2)^2}$

Interval Estimation

In addition to point estimation, interval estimation can be carried out. Based on the observation that the probability that k observations in the sample will fall in an interval covering p of the range (0 ≤ p ≤ 1) is $p^k$ (assuming in this section that draws are with replacement, to simplify computations; if draws are without replacement, this overstates the likelihood, and intervals will be overly conservative).

Thus the sampling distribution of the quantile of the sample maximum is the graph $x^{1/k}$ from 0 to 1: the p-th to q-th quantile of the sample maximum m are the interval [ $p^{1/k}N$ , $q^{1/k}N$ ]. Inverting this yields the corresponding confidence interval for the population maximum of [ $m/q^{1/k}$ , $m/p^{1/k}$ ].

Reference

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离散均匀分布的估计| Serial Number Estimation