Let us assume that $X_i = \theta t_i + e_i, \ \ i=1,2,...,n$ where $\theta \in \Theta$ which is unknown and $\Theta$ is a closed subset of $\mathbb{R}$. Suppose $e_i$'s are i.i.d. on the interval $[-\tau,\tau]$ with unknown $\tau >0$ and $E(e_i)=0$. Given $t_i$'s are fixed constants let
$$T_n= S_n(\tilde{\theta})=\min_{\gamma \in \Theta} S_n(\gamma)$$ where
$$S_n(\gamma) = 2 \max_{i \le n} \frac{|X_i - \gamma t_i|}{\sqrt{1+\gamma^2}}.$$
If we assume that $\sup_{i} |t_i| < \infty$ and $\sup_i t_i - \inf_i t_i > 2\tau$ show that $\{\tilde{\theta_n}, n =1,2,...\}$ is bounded almost surely.
I have really no clue how to tackle this one as there is too much detail. Judging by the expression of $S_n(\gamma)$, it looks like the perpendicular distance of a straight line from a point and it feels that out of $t_1,...,t_n$, I need to find that $t_i$ which is the most distance from the straight line $X_i = \gamma t_i$. Can anyone have a crack at this?
Another follow-up question :
Under the same assumptions mentioned, can we show that $T_n$ is a strongly consistent estimator of $\nu = \min_{\gamma \in \Theta} S(\gamma)$, where $S(\gamma)=\lim_{n \rightarrow \infty} S_n(\gamma)$ almost surely?
By strong consistency of an estimator, you can think of almost sure convergence of the estimator to the parameter of interest.
Again, I could show that for any sequence $\theta_n \rightarrow \theta$, $S_n(\theta_n)-S_n(\theta)=O(|\theta_n-\theta|)$ almost surely. But how can I use this fact to show the almost sure convergence of $T_n$ to $\nu$?
First observe that $$ S_n(\theta)=2\max_{i\le n} \frac{|e_i|}{\sqrt{1+\theta^2}}\le 2\tau. $$ By assumption there must exist $i$ such that $|t_i|>\tau$. Now for any $n\ge i$ and $\gamma$, for sure we have the lower bound $$ S_n(\gamma)\ge \frac{2|(\theta-\gamma)t_i+e_i|}{\sqrt{1+\gamma^2}}\underset{\gamma\to\pm\infty}{\longrightarrow} 2|t_i|. $$ Thus, for $\gamma$ large enough (uniformly in $n$), we have $S_n(\gamma)>S_n(\theta)$, which immediately implies that $(\tilde{\theta_n})$ must be bounded almost surely.