I'm stuck in this statistical inference problem. It states:
Let $X_1,X_2,...X_n$ independent random variables with density:
$f(x_i|\theta)=e^{(i\theta-x_i)}I_{[i\theta,\infty)}(x_i)$
a) Find the moment estimator (i.e., the estimator of $\theta$ using the moment method. Suugestion: Use the transformation $Y_i=(X_i-1)/i$ and verify the $Y_i$'s have the same expectancy.
b) Find de MLE
My attempts:
Inded, I proved that the $Y_i$'s have the same expectancy. In fact, every $Y_i$ has density:
$f(y_i|\theta)=ie^{i(\theta-y_i)}I_{[(i \theta-1)/i,\infty)}(y_i)$
Then, by the moment method I put:
$\overline{Y}=\theta \quad \Rightarrow \quad (1/n)\sum_{i=1}^n(X_i-1)/i=\theta$
And I suppose this doesn't have a nice form, just the second summand includes the Euler-Macheroni constant, to approximate the armonic partial series. So i dont know hot to continue.
For part b) I computed the likelihood function $L(\theta,\mathbf{x})=e^{(\theta n(n+1)/2-\sum_{i=1}^nX_i)}I_{[\theta,\infty)\times[2\theta,\infty)\times...} (\mathbf{x})I_\mathbb{R}(\theta)$.
But from here I don't know how to proceed, since the parameter is included in the indicator functions.
Hope you can help me in this one. Thanks in advance!
1) You find correct MME, I do not understand your doubts. No Euler-Macheroni constant is included in this finite sum.
2) Lef us rewrite the likelihood function $$ L(\theta,\mathbf{x})=\begin{cases}e^{\theta n(n+1)/2}e^{-\sum_{i=1}^nX_i}, & \theta\leq X_1,\, \theta\leq \frac{X_2}{2},\ldots,\theta\leq \frac{X_n}{n}\\ 0, & \text{else}\end{cases} =\begin{cases}e^{\theta n(n+1)/2}e^{-\sum_{i=1}^nX_i}, & \theta\leq \min_{1\leq i\leq n}\left\{\frac{X_i}{i}\right\}\\ 0, & \text{else}\end{cases} $$ The first term $e^{\theta n(n+1)/2}$ is increasing in $\theta$, and given all $X_i$ are fixed, the likelihood function attaines its maximum value at $\theta=\min_{1\leq i\leq n}\left\{\frac{X_i}{i}\right\}$. This is MLE.