Let X be a random variable with a finite expected value μ = E ( X ) and a finite variance Var (X) = σ 2 > 0 . Then the probability P ( μ − 3 σ < X < μ + 3 σ ) ≥ b a with absolute certainty. Find this maximum value b a with a , b coprime positive integers. Submit a + b .
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Consider the RV X . where P [ X = 0 ] = 9 8 and P [ X = 1 ] = P [ X = − 1 ] = 1 8 1 . Then E [ X ] = 0 and E [ X 2 ] = 9 1 . so that μ = 0 and σ = 3 1 , and P [ ∣ X − μ ∣ ≥ 3 σ ] = P [ X = ± 1 ] = 9 1 .
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Oh, this is wonderful!!! ,thank you very much...
Woah! How did you know what numbers to use?
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I was looking for zero mean (for simplicity), so a symmetric RV. The RV X I came up with was therefore about as simple as possible, with the requirement that P [ ∣ X ∣ > u ] = 9 1 for some u > 0 . The rest was a matter of checking details.
Of course, if we set P [ X = 1 ] = P [ X = − 1 ] = 2 a 2 1 and P [ X = 0 ] = 1 − a 2 1 , then we obtain an example for P [ ∣ X ∣ ≥ a σ ] = a − 2 for any a ≥ 1 .
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Let's call A = { x ∈ sample space ; X ( x ) ≥ a } , a > 0 and if B is a set, let's define I B ( x ) = { 1 if x ∈ B 0 if x doesn’t belong to B , then E ( X ) = μ = E ( X ⋅ I A ) + E ( X ⋅ I A C ) ≥ E ( X ⋅ I A ) ≥ E ( a I A ) = a E ( I A ) = a P ( A ) ⇒ a E ( X ) ≥ P ( X ≥ a ) . This is Markov's Inequality or general Chebychev inequality.
Hence, P ( ∣ X − E ( X ) ∣ ≥ k ) ≤ k n E ( ∣ X − E ( X ) ∣ n ) if n ≥ 1 and k > 0 . Take n = 2 . The previous inequality remains P ( ∣ X − E ( X ) ∣ ≥ k ) ≤ k 2 Var(X) ⇒ 1 − k 2 Var(X) ≤ 1 − P ( ∣ X − E ( X ) ∣ ≥ k ) = P ( ∣ X − E ( X ) ∣ < k ) Let's take now k = 3 σ , this implies that 1 − 9 σ 2 Var(x) = 1 − 9 1 = 9 8 ≤ P ( ∣ X − E ( X ) ∣ < 3 σ ) = P ( μ − 3 σ < X < μ + 3 σ )
Note .- P ( ∣ X − μ ∣ ≥ k σ ) ≤ k 2 1 is known as Chebichev's inequality . I still need to prove that there exists a random variable fulfilling 9 8 = P ( ∣ X − E ( X ) ∣ < 3 σ ) = P ( μ − 3 σ < X < μ + 3 σ )
Bonus.- Does there exist another similar inequality more accurate? or Can you give one example fulfilling the previous equality? or due to V a r ( X ) = σ 2 > 0 and the proof for Markov's inequality, this inequality can not be improved?
I'm going to give almost one example.
Consider an unfair coin such that your chance to get face is 1/11, and the random variable X having a Bernouilli distribution associated with this coin when it is launched once, then μ = E ( X ) = 1 1 1 and σ 2 = V a r ( X ) = 1 2 1 1 0 ⇒ σ = 1 1 1 0 , then P ( 1 1 1 − 1 1 3 1 0 < X < 1 1 1 + 1 1 3 1 0 ) = P ( X = 0 ) = 1 1 1 0 = 0 . 9 0 ^ ≈ 0 . 8 ^ = 9 8