Proof related to Chebychev's inequality

Question

I need to prove that in a set of $N$ data $x_1, x_2, \ldots, x_n$, for all $i$ between 1 and $N$, we have

$$\mu-\sigma \sqrt N \leq x_i \leq \mu+\sigma \sqrt N$$

where $\mu$ is the average and $\sigma$ the standard deviation.

I know I need to use Chebyshev's inequality

$$\Pr(\mu-k\sigma \leq X \leq \mu+k\sigma) > 1-1/k^2.$$

I did notice the $X$ changing to $x_i$, which from my understanding means that what I am trying to prove is about one data ($x_i$) whereas Chebichev's inequality is about a set of data.

IT feels like this problem is a bit undefined. Presumably, these are $\mu$ and $\sigma$ from this set of data points $x_1,\dots,x_N$? — Thomas Andrews, Mar 18 '13 at 02:31
I think $X$ is the random variable which is selecting an $i$ from $1$ to $N$ uniformly and choosing the value $x_i$? — Thomas Andrews, Mar 18 '13 at 02:36

score 2 · Accepted Answer · answered Mar 18 '13 at 02:42

2

If $X$ is the random variable gotten by selecting a random $i$ with equal probability from $1$ to $N$ and then selecting $x_i$, then for any $a,b$, the value of:

$$P(a\leq X\leq b)$$

Is of the form: $$\frac{M}{N}$$ for some $M\in\{0,1,2,\dots,N\}$.

Now, use this with Chebyshev's inequality when $k=\sqrt{N}$ to conclude that the probability that $X$ is in this range is equal to $1$, and therefore that all of the $x_i$ are in this range.

answered Mar 18 '13 at 02:42

Thomas Andrews

177,126

Thanks for the solution! When replacing in Chebyshev's inequality $k=\sqrt N$, we get 1−1/N whereas your conclusion seems to infer that we get 1. I suppose I then use $M/N > 1-1/N$. – Justin D. Mar 18 '13 at 03:25
1

No, you get that it is greater than $1-1/N=\frac{N-1}{N}$. So what do we know about $M$? – Thomas Andrews Mar 18 '13 at 03:32
We know that $M$ must be greater than $N-1$, and since it is an integer, it must be equal to $N$. Thus, $M/N=1/1$. Right? – Justin D. Mar 18 '13 at 03:35
Yes, that's it exactly. – Thomas Andrews Mar 18 '13 at 03:56
Thanks again, you're a great pedagogue ;) – Justin D. Mar 18 '13 at 04:02

Proof related to Chebychev's inequality

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