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whitejac
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Homework Statement
Let X∼Geometric(p). Using Markov's inequality find an upper bound for P(X≥a), for a positive integer a. Compare the upper bound with the real value of P(X≥a).
Then, using Chebyshev's inequality, find an upper bound for P(|X - EX| ≥ b).
Homework Equations
P(X≥a) ≤ Ex / a
P(|X - EX| ≥ b) ≤ Var(x) / b2
The Attempt at a Solution
Part 1:
So the first inequality is easy enough...
Markov's inequality:
E(geometric) = 1/p ⇒ P(X≥a) = 1/ap.
Real value:
P(X≥a) = 1 - P(X≤a) = P(X = 0) + P(X = 1) +... + P(X = a-1) + P(X = a) = p + pq +... +pqa-1
What I'm having trouble with is understanding the definition of markov's inequality and how I can explain it mathematically compared to the real value of P(X≥a).
A lot of people simply say that the real value is less than markov's inequality and therefore that is a comparison. This doesn't make much sense to me in the general form because all i'd be saying is:
1-P(X≤a) < 1/apPart 2:
By definition, the upperbound is Var(x) / b^2 = (1-p) / (b2p2)
I have a similar issue with finding Chebyshev's inquality... I can "do" it, but I don't really know what I'm doing. My book simply states that this is calculating the difference between X and EX is limited by the variance which is fine and intuitive, but I don't really think it should be this simple...
So, are my interpretations correct? And are they explained properly...
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