Consider a mathematical sample $X_1, X_2,...X_n$ where $X_1 \sim \mathbb{P}_\lambda, \lambda \in \Theta$. For $ r>0$ and let $\psi_r: \mathbb{R} \rightarrow\mathbb{R}$ with $$ \psi(x)= \begin{cases} r, \,x<-r\\ x, \,|x|\leq r\\ r, \,x>r \end{cases} $$ Let $M_r$ be the set of zeros of the function $$ \Psi_r:\mathbb{R} \rightarrow\mathbb{R}, y \mapsto \Psi_r(y):= \frac{1}{n} \sum_{k=1}^n \psi_r(X_k - y), \, y \in \mathbb{R} $$ Define the random variable $\hat{m}_r = \inf M_r$
Now regarding the median - how do I show that $\hat{m}_r$ converges almost surely to a median of $X_1,...,X_n$ as $r \rightarrow 0$(as $r \rightarrow \infty$)?
I would very much appreciate your help!
This is not a solution but some thoughts I have had, this is a real interesting problem by the way. First that $\psi_r(x) \geq 0$ for all $x$ and $\psi_i(x) = 0$ iff $x=0$. For each sample $\omega \in \Omega$ then $$\Psi_r(y,\omega) =\frac{1}{n} \sum_{k=1}^n\psi_r(X_k(\omega)-y)$$ By positivity of the $\psi_r$, we can only have $\Psi_r(y,\omega) = 0$ iff $\psi_r(X_k(\omega)-y) = 0$ for all $k$ which occurs iff $X_k(\omega)=y$ for all $k$. This shows that $$M_r(\omega)=\{y: X_k=y\,\,\,\text{for all $k$}\} $$ The second thing I noticed is that $|\Psi_r(y,\omega)|\leq \frac{r}{n}$ and so $\Psi_r\to 0$ uniformly. The third thing is that unless there is a typo, I am not sure that it is true that $\mathbb{P}(M_r =\emptyset)\not=0$, as a consequence we could have a set of positive measure where $\hat{m}_r$ is not defined. Not sure how helpful these are, but good luck!