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probability and variance?
Let X be defined as above for α=4.
Define the random variable Y63=∑63i=1Xi, Xi independent identically distributed according to X.
Using linearity of expectation, what is E[Y63]?
Using the fact that the Xi random variables are independent and identically distributed, what is var[Y63]?
What about the the standard deviation std[Y63]?
Using Chebyshev’s inequality, what is a bound on the P(|Y63|>27)?
1 Answer
- kbLv 78 years agoFavorite Answer
Assuming that (as in your previous post)
P(X = 0) = 0 and otherwise P(X = i) = 1/(Zα |i|^α), where
Zα = Σ(-∞ to ∞, nonzero) 1/|i|^α = 2 * Σ(i = 1 to ∞) 1/|i|^α.
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We found previously that
E[X] = 0 and E[X^2] = (2/Zα) Σ(i = 1 to ∞) 1/|i|^(α-2) = (1/Zα) * Z_(α-2).
Letting α = 4, we have E[X] = 0 and E[X^2] = Z₂/Z₄
So, μ = 0 and σ^2 = Z₂/Z₄ [or σ = √(Z₂/Z₄)].
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Hence, E[Y₆₃] = E[X₁] + ... + E[X₆₃] = 0 + ... + 0 = 0.
Var(Y₆₃) = 1^2 Var(X₁) + ... + 1^2 Var(X₆₃) = 63 Z₂/Z₄.
==> Std(Y₆₃) = √(63 Z₂/Z₄).
By Chebychev's Inequality,
P(|Y₆₃| > 27) ≤ (63 Z₂/Z₄) / 27^2 = (7/81) (Z₂/Z₄).
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Note: Just in case...
Z₂ = 2 * Σ(n = 1 to ∞) 1/n^2 = π^2/3
Z₄ = 2 * Σ(n = 1 to ∞) 1/n^4 = π^4/45
So, Z₂/Z₄ = 15/π^2; this can be used to simplify the probabilities above.
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I hope this helps!
