My question is related to : Maximum likelihood estimation and the n-th order statistic estimation-and-the-n-th-order-statistic. I will try to answer the questions, and I would like to ask you guys to look at it, and criticize it, because I want all the answers to be correct obviously :)
Given $$ f_{\theta}(x)=e^{ (\theta -x)}$$for $x \ge \theta$, otherwise $f_\theta = 0$
I showed that the MLE $\hat \theta_n $of $\theta= min(x_1,x_2, ... x_n) = X_{(1)}$
I have to answer the following questions about this distribution:
a.) What is the cdf of $\hat \theta_n $
b.) What is the pdf of $\hat \theta_n $ and draw the distribution
c.) We take a sample of size $n$ from this distribution and by chance we took a very unusual value $\hat \theta_n = q_{1-\alpha}$ where $q_{1-\alpha}$ is the $1-\alpha$ quantile of the distribution of $\hat\theta_n$. Determine $q_{1-\alpha}$ as a function of of $\theta$ and $\alpha$
d.) Again we take sample of size $n$ and now we got very lucky by obtaining $\hat\theta_n=q_0$ where $q_0$ is the 0th quantile of the $\hat \theta_n$ distribution. Determine $q_o$
e.)We observe $X_1=3, X_2=4, X_3=10$. With the answers of c.) and d.) we should now be able to obtain a 95%-confidence interval for $\theta$
My try:
a.) the cdf is $F_{\hat\theta_n}1-(1-(F_{\theta}(x))^n= 1-e^{n\theta - nx}$
b.) $f_{\hat\theta_n}= ne^{n\theta - nx} = \frac{d}{dx}( 1-e^{n\theta - nx})$.
c.) $$1-\alpha =F(q_{1-\alpha})= 1-e^{n\theta -nq_{1-\alpha}} \implies \ln{(1-\alpha)}=n\theta-nq_{1-\alpha} \implies -nq_{1-\alpha}= \ln{(1-\alpha)}-n\theta \implies q_{1-\alpha}=-\frac{\ln{(1-\alpha)}}{n} +\theta$$
d.)$$0=F(q_0)= 1- e^{n\theta -nq_0} \implies 1=e^{n(\theta-q_0)} \implies q_0=\theta$$
e.) We need a $1-\alpha$-confidence interval for $\theta$. Let $c_l$ and $c_u$ denote the lower and upper boundaries of the confidence interval of $\theta$. I will solve $$P(c_l < \hat\theta_n < c_r)= 1-\alpha $$ by requiring $$P(\hat\theta_n \le c_l) =P(\hat\theta_n \ge c_r)= \frac{1}2{}\alpha $$ Now I find the following equations
$$1-e^{n\theta-nc_l} = \frac{\alpha}{2}\implies \ln(1-\frac{1}{2}\alpha)=n\theta-nc_l \implies c_l = -\frac{\ln(1-\frac{1}{2}\alpha)}{n}+\theta $$
Similarly
$$e^{n\theta-nc_u}= \frac{1}{2} \alpha \implies c_u=-\frac{\ln(\frac{1}{2}\alpha)}{n}+\theta $$
Now I have found that, whatever the value of $\theta$:
$$P(-\frac{\ln(1-\frac{1}{2}\alpha)}{n}+\theta <\hat\theta_n <-\frac{\ln(\frac{1}{2}\alpha)}{n}+\theta ) = 1-\alpha $$ By re-arranging the inequalities, we see this is equivalent to
$$ P(\frac{\ln(\frac{1}{2}\alpha)}{n}+\hat\theta_n <\theta < \frac{\ln(1-\frac{1}{2}\alpha)}{n}+\hat\theta_n ) = 1-\alpha$$
Therefore, a $100(1-\alpha) $% confidende interval for $\theta$ is given by: $$\left( \frac{\ln(\frac{1}{2}\alpha)}{n}+\hat\theta_n <\theta < \frac{\ln(1-\frac{1}{2}\alpha)}{n}+\hat\theta_n \right) $$
Looking at the observation, we have min(X1, X2, X3)= 3 and n = 3. So our 95% confidence interval should be: $$\left( \frac{\ln(\frac{0.05}{2})}{3}+3 <\theta < \frac{\ln(1-\frac{0.05}{2})}{3}+3 \right) $$This interval is [1.77, 2.99]
