Alpha-Trimmed Mean and Median 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!
 A: 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!
