MLE estimator of $\theta$ for $f(x;\theta)=e^{i\theta -Xi}$

Question

$X_1...X_i$ are independent randomly distributed with pdf $$f(x_i|\theta)=e^{i\theta-x_i}$$$ $$x_i≥i\theta$

-Previous problems I had done where the range of x was dependent on $\theta$ just involved bounds ....$x_i\lt\theta$ than you just use $X_n$ and when $x_i\gt\theta$, I use $X_1$. I was just confused because this time the bounds are $i\theta$.

• My thoughts were to just use $X_1/i$ as the MLE of $\theta$ but I was not sure

Answer 1

The joint probabilty of the $X_s$'s is: $$f(x_1, \ldots, x_i | \theta) = e^{\sum_{s=1}^i (s\theta -x_s)}1_{\{\min_s\{x_s/s \} \ge \theta\}}$$

This is because: $$ \Pi_{s = 1}^i 1_{\{x_s \ge s \theta\}} = 1_{\{\min_s\{x_s/s \} \ge \theta\}}$$

Finding MLE means finding the argument $\theta$ that maximises this function. Hence we ask:

• $\theta \le \min_s\{x_s/s\}$

Since the exponential function is strictly increasing in $\theta,$ we find that the argument that maximises the distribution is exactly:

$$\theta: = \min_s\{x_s/s\}$$