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Tchebychev's inequality

The Tchebychev inequality can also be stated in the following way:
For any random variable x with mean equal to &#956; and variance equal to &#948;². The minimum probability of X belong to the interval X?[ &#956;-k, &#956;+k] is at least:
P( | X- &#956;|<k &#8805; 1-( &#948;/k²)

Suppose that the random variables x1, x2, x3... xn form a random sample of size n drawn from some unknown distribution, then the sample mean is expressed as:
<Xn>=(x1+x2+x3+....+xn)/n
The mathematical expectation of sample mean is equal to:
E[<Xn>]= &#956;
The variance of the sample mean is equal to:
Var[<Xn>=&#948;²/n

Now we are applying the Tchebyshev inequality to the sample mean <Xn> to estimate the probabilities:
a) show that P(|<Xn>-&#956;|>k &#8804; &#948;²/(nk²)
b) show that when the sample size increases, the probability of <Xn> outside k units from the mean &#956; decreases and asymptotically approaches to 0
c) suppose we know the variance &#948;²=4 and we don't know &#956; and we have observed the data x1, x2, x3... xn. How large the sample size n is required in order to make sure the probability of estimated &#956; will satisfy the following condition:
P(|<Xn>-&#956;|>1) &#8804; 0.01

Solution Summary

The solution contains an application of Tchebychev's inequality

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