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The problem I am working on is:

Let $ Y = \{{y_1,y_2,y_3,...y_n\}}$ and $c=median(Y)$. Prove that: $$ \text{min}\left[\sum_{i=1}^n \lvert y_{i}-c\rvert\right]=c $$

My question is:

Is the $\text{min}\left[\right]$ function here, asking for the which $y_i$ the distance between $y_i$ and $c$, $\lvert y_{i}-c\rvert$ is the smallest? Or is it asking, for the literal minimum of the arugments its given, like $\text{min}\left[99,1,4,7,121\right]=1$?

What I have tried so far:

1. First I tried an example to get a feel for things:

Let $ Y = \{{1,2,3,4,5\}}$ and $c=3$. Then we have:

$$\begin{align} \sum_{i=1}^5 \lvert y_{i}-3\rvert & = \lvert 1-3\rvert = 2 \\ & = \lvert 1-3\rvert = 2 \\ & = \lvert 2-3\rvert = 1 \\ & = \lvert 3-3\rvert = 0 \\ & = \lvert 4-3\rvert = 1 \\ & = \lvert 5-3\rvert = 2 \\ & = 2+1+0+1+2=6 \end{align}$$ So,

$$ \text{min}\left[\sum_{i=1}^5 \lvert y_{i}-3\rvert\right]=\text{min}\left[6\right]=6 $$

Since this would now be acounterexample to the statement I am supposed to prove, I must be having a misunderstanding?

2. If the $\text{min}\left[\right]$ function works the other way, I tried the following proof:

Proof:

Since the distance between a point and itself is $0$, $\left[c-c\right]=0$ and the fact that there can only be one median $c$ for a given set of $y$, clearly $\left[c-c\right] \lt \left[y_i-c\right]$ for all $y_i.$

Is this correct if thats the case?

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    $\begingroup$ linear algebra $\neq$ peanut butter $\endgroup$ – mathreadler Sep 29 '18 at 14:45
  • $\begingroup$ the summation is over all possible $y_i's$ $\endgroup$ – Yan Lai Sep 29 '18 at 14:48
  • $\begingroup$ Part of me wants to say that this statement may not be correct. If we have a sample where each $y_i$ is the same, then $c=y_i$ for all $i$, thus... $\endgroup$ – WaveX Sep 29 '18 at 14:50
  • $\begingroup$ What is the variable in your formula which minimizes it? It seems that all values are fixed. A clarification is needed. $\endgroup$ – callculus Sep 29 '18 at 15:04
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    $\begingroup$ I suspect that you are being asked to show that the $c$ which minimise(s) $\sum\limits_{i=1}^n \lvert y_{i}-c\rvert$ for given $\{y_i\}$ is (or includes) the median of the $\{y_i\}$ $\endgroup$ – Henry Sep 29 '18 at 15:06

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