You can do better. In the sense that among all the unbiased estimators, you can find the one with the smallest variance. Our goal is to estimate $\mathbb{P}(X<0)$. An unbiased estimator is $$\... View answer Accepted answer 3 votes (Nobody answers so I post this answer which I am not sure is correct) We show (Y_{(1)},Y_{(n)}) is a sufficient complete statistic, which implies (Y_{(1)},Y_{(n)}) is minimal sufficient. We first ... View answer Accepted answer 1 votes We have the likelihood of m by$$f(m)=\frac{m!}{(m-X)!X!}p^X(1-p)^{m-X}$$Here m\geq X. Now, let's consider when \frac{f(m+1)}{f(m)} is less than 1. When it starts to be less than 1, we know ... View answer Accepted answer 1 votes Since nobody answered, I'll try to answer my own question. Please let me know if anything is not correct. Denote S=\frac{1}{n-1}\sum_{i=1}^n[X_i-X_{(1)}]. We have$$S\sim \frac{b}{n-1}\text{Gamma}(n-...

The answer is simple, any statistic $\delta(X)$ satisfying $$\delta(\theta-1)+\delta(\theta)+\delta(\theta+1)=0, \forall\theta\in\mathbb{Z}$$ is an unbiased estimator of zero. Thus, the totality of ...

Let me directly find the UMVUE of $\theta$. However, I cannot answer "can the BLUE be the UMVUE", as I do not have enough knowledge of BLUE. It can be shown that $$\mathbb{E}(\bar{X})=\... View answer 1 votes Following @whuber's answer, we have the joint density of (Y_{(1)},Y_{(n)})$$f(y_1, y_n,a,b)=\frac{n(n-1)}{(b-a)^n}(y_n-y_1)^{n-1},\quad\forall a\leq y_1\leq y_n\leq b\text{ and }a,b\in\mathbb{R}$$... View answer 0 votes I finished the implementation myself. Here is the code. gaussian_MLE = function(S, omega_structure, initial_omega = initial_omega, maxit = 2e9, tol = 1e-15){ omegas = list() # S is sample ... View answer 0 votes I am answering my own question. Please correct me if I make any mistakes. Continuing with$$T=n[-\log(\hat{\beta})+\log(\beta) - 1 +\frac{\hat{\beta}}{\beta}],$$we can write$$T=n[\log\frac{\hat{\...

I got it. The correct formula for Moment generating function should be $$M_{T}(u)=\frac{\exp{A(\eta+u)}}{\exp{A(\eta)}},$$ not $$M_{X}(u)=\frac{\exp{A(\eta+u)}}{\exp{A(\eta)}}$$ Here, $T(X)=(\log X,X)$...

Method 1 $(X_{(1)},X_{(n)})$ is not complete because we can find $g\neq0$ but $\mathbb{E}\left[g(X_{(1)},X_{(n)})\right]=0,\forall\theta$. $g$ is \$(t_1,t_2)\rightarrow\frac{n+1}{n-1}t_2-\frac{n+1}{1-n}...