Re: About Tchebycheff's inequality
- From: se16@xxxxxxxxxxxxxx
- Date: Tue, 6 Jan 2009 01:43:25 -0800 (PST)
On 29 Dec 2008, 12:26, toh...@xxxxxxxxxxx wrote:
On 29 Dic, 00:18, s...@xxxxxxxxxxxxxx wrote:
On 27 Dec, 13:27, toh...@xxxxxxxxxxx wrote:
In the third edition of Papoulis' Probability, Random Variables an
Stochastic Processes book
I read this (Chapter 8, Section 8-4) about measuring the length a of
an object which leads to an RV X=a+NI where NI is an RV (error term)
with zero mean and standard deviation sigma: "ApplyingTchebycheff's
inequality, we conclude that:
P{|X-a|<epsilon}>1-sigma^2 / epsilon^2
If, therefore, sigma << epsilon, then the probability that |X-a| is
less than that epsilon is close to 1. From this it follows that
"almost certainly" the observed X(zeta) is between a-epsilon and a
+epsilon, or equivalently, that the unknown a is between X(zeta)-
epsilon and X(zeta)+epsilon. In other words, the reading X(zeta) of a
single measurement is "almost certainly" a satisfactory estimate of
the length a as long as sigma << a. If sigma is not small compared to
a, then a single measurement does not provide ab adequate estimate of
a."
My question is: why should I consider sigma << a ?? Wasn't sigma <<
epsilon ?
It all depends on how small "<<" suggests.
For example, if sigma < a/1000 then by Chebyshev's inequality
P{|X-a|<a/100} > 0.99.
Similarly if sigma < a/100 then P{|X-a|<a/10} > 0.99.
But I would not use "almost certainly" this way.
Why did you replace epsilon with a (or better: with a/100, or a/10).
Why did you write P{|X-a|<a/100} starting from a/1000; or P{|X-a|<a/
10} starting from a/100?
I cannot see any relationship between a and epsilon... can you explain
me or give me some more insight? Thank you.- Hide quoted text -
- Show quoted text -
epsilon is a variable.
Chebyshev's inequality is true for all epsilon>0.
So you can replace epsilon with anything positive.
.
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