I'm currently working on a proof for my stochastics course, and I am a bit stuck. I tried to find some hints or help online, but so far I haven't found anything. So I figured I'd see if someone could help me here.
So the problem is the following: We have to use a proof by cases, one with an even number of observations and one with an odd number of observations in our sample, that the $\alpha$-truncated mean converges to the median, as $\alpha \rightarrow \frac{1}{2}$.
As the course is in German, and I'm not sure about the English term for this mean, I'll also leave the definition here. It also contains the problem I have with this proof:
$$x_{t,\alpha} := \frac{1}{n-2k}\sum^{n-k}_{j=k+1}x_{[j]} $$ with $0 < \alpha < 1/2$ and $k:= \lfloor n\alpha \rfloor$, and where our sample $(x_{[1]},\dots,x_{[n]})$ is already ordered.
Now, I am just looking for a starting point, not a complete solution, after all I would really love to figure it out by myself - at least partially. I'm just not sure how to deal with the floor function and the change in the size of the sum. Any good hints or references?
Thanks in advance!
OK, so I figured it out myself. Once you get there it's actually pretty easy. It just seems I've been approaching the whole thing in a wrong way. I'll post my answer, just in case other math students at the beginning of their studies might have the same problem. So, spoiler alert. ;)
So for the first case assume that $n$ is an uneven number. Then for any $0 < \alpha < 1/2$, we find that $0\leq \lfloor n\alpha \rfloor = k \leq \frac{n}{2} - \frac{1}{2} = \frac{n-1}{2}$, as $\lfloor \frac{n}{2} \rfloor = \frac{n}{2} - \frac{1}{2}$ for any uneven $n \in \mathbb{N}$. So now that we know that $k \in [0,\frac{n-1}{2}]$, we can just substitute the maximum for $k$ into the definition and find: $$\frac{1}{n-n+1}\sum_{j = \frac{n+1}{2}}^{\frac{n+1}{2}}x_{[j]} = x_{[\frac{n+1}{2}]} $$,
the definition of the median for an uneven sample size $n$.
Using a similar logic, we find for even $n$ and all $0<\alpha<1/2$ that $0\leq \lfloor n\alpha \rfloor = k \leq \frac{n}{2} - 1$, as $\lfloor \frac{n}{2} \rfloor = \frac{n}{2}$, so any smaller fraction would map to a smaller integer. Again, substituting the maximum of $k \in [0,\frac{n}{2} - 1]$ into the definition above results in the definition of the median for an even sample size $n$.
I'm also open for feedback, if you have comments!