Let $\theta$ be an unknown constant. Let $W_1,…,W_n$ be independent exponential random variables each with parameter $1$. Let $X_i=θ+W_i$.
First, I need to find $\hat\theta _{ML}(x_1,\ldots ,x_ n)$. I know the likelihood functionis $L(\theta)=\prod_{i=1}^nf(\theta|x_i)$, but I am not sure what $f(\theta|x_i)$ look like.
The second question is to construct a 95% confidence interval of the particular form $[ \hat\Theta -c, \hat\Theta ]$ for $n=10$, where $\hat\Theta = \min _ i \{ X_ i \}$ and c is a constant that I need to choose.
Any help will be appreciated, thanks a lot.
Let $X=(X1,…,Xn)$ and $x=(x1,…,xn)$. To find $\hat{\theta }_{ML}$, we first find $f_X(x;θ)$. Since the $W_i$'s are independent, so are the $X_i$'s. Hence,$$\displaystyle f_ X(x;\theta )\displaystyle = \prod _{i=1}^{n}f_{X_ i}(x_ i;\theta )\displaystyle = \begin{cases} \displaystyle \prod _{i=1}^{n}e^{-(x_ i-\theta )}, & \mbox{if } x_ i \geq \theta ~ ~ \forall i, \\ 0, & \textrm{otherwise.} \end{cases}$$
Note that this quantity is nonzero only if $θ$ is no greater than each of the $x_i$'s. Moreover, $e^{-(x_ i-\theta )}$ is greater when $θ$ is closer to $x_i$. Therefore, this quantity is maximized when we push $θ$ as high as possible while keeping it no greater than each of the $x_i$'s. This means that $\hat\theta _{ML}(x) = \min _ i x_ i$ (Any larger choice of $θ$ would give $f_ X(x;\theta )=0$.)
For the second part of question, we wish to find $c$ such that ${\bf P}(\hat{\Theta }-c \leq \theta \leq \hat{\Theta }) \geq 0.95$.Since the $X_i$'s are independent, we have
${\bf P}(\hat{\Theta }-c\leq \theta )= {\bf P}(\min _ i\{ X_ i\} \leq \theta +c)$ $= 1-{\bf P}(\min _ i\{ X_ i\} \geq \theta +c)$ $= 1-\prod _{i=1}^{10}{\bf P}(X_ i\geq \theta +c)$
$= 1-\prod _{i=1}^{10}{\bf P}(W_ i\geq c)$ $= 1-\prod _{i=1}^{10}e^{-c}$ $= 1-e^{-10c}.$
To have a 95% confidence interval, we require $1-e^{-10c}\geq 0.95$.